Electric Current: notes and MCQs for State PSC prelims

Electric Current is part of Physics. Below are notes on 9 sub-topics, followed by practice MCQs from State PSC and UPSC papers.

Definition of Electric Current and current density

Electric Current (I):

Definition: Rate of flow of electric charge through any cross-section of a conductor.
Formula: $I = \frac{dQ}{dt}$, where $dQ$ is charge flowing in time $dt$.
Unit: Ampere (A) or Coulomb/second (C/s).
Nature: Scalar quantity. (Direction of flow is indicated, but it doesn't follow vector addition rules).
Charge: Quantized, $Q = ne$, where $n$ is an integer and $e$ is elementary charge ($1.6 \times 10^{-19}$ C).
Flow of Charge: Electricity is the flow of electrons (or positive ions in electrolytes).
Conventional Direction: Opposite to the flow of electrons.
Calculating Charge (Q): If current (I) is constant, $Q = I \times t$ (time in seconds).
Continuity Equation: Expresses charge conservation, fundamental to current flow.

Current Density (J):

Definition: Electric current per unit cross-sectional area perpendicular to the direction of current flow.
Formula: $\vec{J} = \frac{I}{A} \hat{n}$ (or $I = \int \vec{J} \cdot d\vec{A}$ for non-uniform current), where $A$ is the area and $\hat{n}$ is the unit vector normal to the area.
Unit: Ampere/meter² (A/m²).
Nature: Vector quantity. Its direction is the direction of the flow of positive charge.

Types of Current:

Direct Current (DC): Flows in one constant direction. Used in devices like batteries, mobile chargers. Not suitable for charging storage batteries (requires specific DC charging).
Alternating Current (AC): Direction reverses periodically.

Frequency: Number of cycles per second (measured in Hertz, Hz).
Household Supply (India): Typically 220 V, 50 Hz. This means current changes direction 100 times per second (50 cycles/sec 2 direction changes/cycle).
Appliances using AC: Refrigerator, washing machine, air cooler, air conditioner, microwave oven, television.

Miscellaneous:

Cathode Rays: Streams of electrons originating from the cathode and moving towards the anode; invisible.
Lightning: Due to the flow of charges between clouds or between clouds and the ground.
Generator: Device used for producing electric current (converts mechanical energy to electrical energy).
Diode: Can convert alternating current (AC) to direct current (DC).
Solar Cooker: Does not use electric current.

Drift velocity and microscopic origin of current

Drift velocity (vd): Average velocity attained by free electrons in a conductor due to an applied electric field; typically very small (~mm/s).
Comparison: Drift velocity is significantly less than the average thermal speed of electrons (which is ~105 m/s).
Microscopic Origin of Current: Electric current in metals is due to the directed motion (drift) of free electrons under an electric field, superimposed on their random thermal motion.
Charged Particle Acceleration: When a charged particle (q) is accelerated through a potential difference (V), the kinetic energy gained is K.E. = qV.

For same q and V, K.E. is the same for all particles.
Since K.E. = 1/2 mv², the speed (v = √(2qV/m)) is inversely proportional to the square root of mass (m).
Therefore, lighter particles (e.g., electron) achieve higher speeds than heavier particles (e.g., proton) for the same potential difference (Ve > Vp).

Lorentz Force (Magnetic Field): A charge q moving with velocity v in a magnetic field B experiences a force F = q(v x B).

Magnitude: F = qvB sinθ, where θ is the angle between v and B.
If the velocity v is parallel or anti-parallel to the magnetic field B (θ = 0° or 180°), the force F = 0, and the particle continues its motion unchanged.
If v is perpendicular to B, the force F = qvB. Direction is given by the RightHand Rule for positive charges or Left-Hand Rule for negative charges.
In a perpendicular magnetic field, positive and negative charges moving in the same initial direction will experience forces in opposite directions, causing them to separate.

Ionic Current in Solution: The ionic current (I) due to an applied electric field depends on the square of the charge (q) and inversely on the ionic radius (R), i.e., I ∝ q2R-1.
Fleming's Right-Hand Rule (Induced Current): Used to determine the direction of induced current when a conductor moves in a magnetic field.

Thumb: Direction of motion of the conductor (Force).
Forefinger: Direction of the magnetic field.
Middle Finger: Direction of the induced current.

Right-Hand Thumb Rule (Magnetic Field by Current): Used to determine the direction of the magnetic field around a current-carrying straight conductor.

Thumb: Points in the direction of the current.
Curled fingers: Show the direction of the magnetic field lines.
An anti-clockwise magnetic field implies an upward current direction.

Current-Carrying Solenoid:

The magnetic field inside a long current-carrying solenoid is uniform and strong.
It behaves like a bar magnet, with clearly defined North and South poles.
The magnetic field strength inside the solenoid is directly proportional to the current (B ∝ I).
Inserting a soft iron bar (a ferromagnetic material) inside the solenoid significantly increases the magnetic field strength, creating an electromagnet. It does not remain the same.

Electrical Power and Energy, Joule heating effect

Electrical Power (P): Rate at which electrical energy is consumed or produced.
Formulas for Power:

P = VI (Voltage × Current)
P = I²R (Current² × Resistance)
P = V²/R (Voltage² / Resistance)

SI Unit of Power: Watt (W). 1 Watt = 1 Joule/ second (J/s).
Other Units of Power: Horsepower (1 HP ≈ 746 W).
Electrical Energy (E): The total amount of work done or energy consumed.
Formulas for Energy:

E = P × t (Power × Time)
E = VIt (Voltage × Current × Time)
E = I²Rt (Current² × Resistance × Time)
E = V²t/R (Voltage² × Time / Resistance)

SI Unit of Energy: Joule (J).
Commercial Unit of Electrical Energy: Kilowatt-hour (kWh), also known as "unit".

1 kWh = 1 unit = 3.6 × 10⁶ Joules.

Cost of Energy: Calculated by (Power in kW × Time in hours) × cost per kWh.
Joule Heating Effect (Heating Effect of Electric Current): When current flows through a resistor, electrical energy is converted into heat energy.

Heat produced (H) = I²Rt (Joule's Law of Heating).
If current (I) is doubled, heat produced quadruples (increases by 4 times).
Rise in temperature of water depends on supply voltage, current, and time.

Applications of Joule Heating Effect: Electric heaters, electric irons, incandescent light bulbs, fuses.
Incandescent Electric Bulbs: Convert electrical energy into light (approx. 20%) and heat (approx. 80%).

Filament resistance (R) determines power for a given voltage (P = V²/R). A dim bulb has higher resistance than a brighter one, assuming same voltage.
They contain an inert gas (e.g., Argon, Nitrogen) to prolong filament life.
Approximate lifespan: 1,000 hours.

CFL (Compact Fluorescent Lamp): Uses mercury vapour and phosphor to produce light.

More energy-efficient than incandescent bulbs.
Contains poisonous mercury, requiring safe disposal.
Average lifespan: generally shorter than LED lamps.

LED (Light Emitting Diode): Uses semiconductor material to produce light.

Highly energy-efficient.
Longer lifespan (approx. 50,000 hours or more).
Does not contain mercury.

Energy Conversion Devices:

Electric Motor: Converts electrical energy into mechanical energy.
Electric Generator / Dynamo: Converts mechanical energy into electrical energy.
Fuel Cell: Converts chemical energy into electrical energy. If pure hydrogen is used, emits heat and water as byproducts.
Solar Panel: Converts light energy into electrical energy.
Microphone: Converts sound energy into electrical energy.
Telephone Receiver: Converts electrical energy into sound energy.
Heater: Converts electrical energy into heat energy.
Television Set: Converts electrical energy into light, sound, and heat.
Candle: Converts chemical energy into light and heat.

Household Electrical Circuits:

Appliances are connected in parallel to ensure they receive the full supply voltage.
Overloading: Connecting too many appliances to a single socket draws excessive current, leading to overheating of domestic wiring and potential damage.
Short Circuit: Occurs when live and neutral wires touch directly, resulting in very low resistance and a sudden, large increase in current. This causes heavy heating and can lead to fires.
Indian Domestic Supply: 220-240 V, 50 Hz AC. Using a 110 V appliance on a 220 V supply will likely damage or burn it out.
AC vs. DC: 220 V AC is generally more dangerous than 220 V DC because the peak voltage of AC is higher than its RMS value (220V RMS AC has a peak of ~311V).

Safety Devices in Electrical Circuits:

Electric Fuse: A safety device connected in series with the live wire.

Protects circuits and appliances from high currents due to shortcircuiting or overloading.
Made of an alloy of lead and tin.
Must have a low melting point.
Must have high specific resistance (though total resistance of fuse wire is low for normal current).
When current exceeds its rated value, the fuse wire melts, breaking the circuit.

Earthing: Provides a low-resistance path for fault current to flow directly to the earth, preventing electric shock to users. Used for safety.

Transformers:

A device that converts high AC voltage to low AC voltage (step-down) or low AC voltage to high AC voltage (stepup).
Works only on Alternating Current (AC) based on the principle of electromagnetic induction.
Cannot work on Direct Current (DC).
Purpose: Used to minimize power loss (P=I²R) in transmission cables by transmitting electricity at high voltage and low current over long distances.
Core Lamination: The core is laminated to reduce energy loss due to eddy currents (which cause heating).
Frequency: The frequency of AC remains unchanged when passing through a transformer.
Mobile Phone Charger: Typically a step-down transformer and rectifier circuit.

Choke Coil (in fluorescent tubes): An inductor that reduces the current in the circuit and can momentarily increase the voltage during starting.
Work Done / Energy Gained by an Electron: The work done (W) in moving a charge (q) across a potential difference (V) is W = qV.

For an electron (charge e = 1.6 × 10⁻¹⁹ C) moving through V volts, energy gained is E = eV Joules.

Static Electricity: Walking on a woolen carpet can build up static charge on the body. Touching a metallic handle allows this charge to transfer, causing an electric shock.
Lightning: Contains a tremendous amount of electric energy, capable of causing significant damage.
Wattless Current: Occurs in AC circuits when the phase difference between voltage and current is 90° (π/2 radians), meaning no average power is consumed (purely reactive load).
Capacitor: An electronic component primarily used to store electric charge and electrical energy in an electric field between its plates.
DC Motor Maximum Mechanical Power: A DC motor develops maximum mechanical power when its back electromotive force (EMF) is equal to one-half of the applied EMF.
3-Phase Power Factor (Two-wattmeter method): If the readings of two wattmeters are equal and positive, the load power factor is unity.
Alternator Power Factor: If an alternator delivers power at unity power factor, the phase angle between the line voltage and line current is 0°.
Rectifier (Half-wave, controlled): The effective (RMS) value of the output voltage of a controlled rectifier depends on the input AC voltage and the firing angle (α).

Electromotive Force (EMF) and internal resistance of a cell

Electromotive Force (EMF) is the maximum 1. potential difference across the terminals of a cell when no current is drawn (open circuit). Unit: Volt (V). Internal Resistance (r) is the resistance 2. offered by the cell's electrolyte and electrodes. It causes a voltage drop (Ir) within the cell. Terminal Voltage (V) across a cell when 3. drawing current (I) is given by V = E - Ir, where E is EMF. Current (I) drawn from a cell with external 4. resistance (R) is I = E / (R + r). Internal resistance can be calculated as r = 5. (E - V) / I or r = (E/I) - R. Faraday's Laws of Electromagnetic Induc6. tion: An EMF is induced when magnetic flux linked with a coil changes. The magnitude of induced EMF is proportional to the rate of change of magnetic flux. Increasing the speed of a moving magnet 7. inside a coil increases the rate of change of magnetic flux, leading to increased induced EMF and electric current. Lenz's Law states that the direction of 8. induced EMF opposes the cause producing it. A DC generator (dynamo) works on the prin9. ciple of Faraday's laws of electromagnetic induction, converting mechanical energy into electrical energy. To convert an AC generator into a DC gener10. ator, a split-ring type commutator is used. Batteries/Cells convert chemical energy into 11. electrical energy. Photovoltaic cells convert solar energy into 12. electrical energy. An ordinary torch cell uses a Zinc casing 13. (negative terminal) and a Carbon rod (positive terminal). A motor car battery (lead-acid) typically 14. provides 12V, uses dilute sulfuric acid as electrolyte, lead and lead dioxide as electrodes, and its capacity is expressed in Ampere-hour (Ah). For 'n' cells of EMF 'E' in series, if 'm' cells 15. are wrongly connected (reversed polarity), the total EMF of the battery is (n - 2m)E. (e.g., 10 cells, 1V each, 2 wrongly connected: (10 - 22)1V = 6V). Nickel-Cadmium (Ni-Cd) batteries are re16. chargeable and commonly used in cordless electronic appliances. A Zener diode is used as a voltage regulator. 17. A Light Emitting Diode (LED) produces light 18. when it is forward biased. In a DC motor, the current (I) drawn is given 19. by I = (Supply Voltage - Back EMF) / Armature Resistance. In a DC compound generator, "short shunt" 20. and "long shunt" refer to specific configurations of connecting the shunt field winding. Load frequency control in power systems is 21. achieved by properly matching the individual machine's turbine inputs. Series capacitive compensation in Extra 22. High Voltage (EHV) transmission lines is used to improve the stability of the system.

Factors affecting resistance (length, area, material, temperature)

Resistance (R) of a conductor is determined by its material, length, and cross-sectional area.
Formula: R = ρ(L/A), where ρ (rho) is resistivity, L is length, and A is the area of crosssection.
Length (L): Resistance is directly proportional to length (R ∝ L).

If length is reduced, resistance decreases proportionally.
If a wire is stretched to 'n' times its original length (assuming constant volume), its new area of cross-section becomes A/n, and its new resistance R' = n²R.

Area of Cross-section (A): Resistance is inversely proportional to the area of crosssection (R ∝ 1/A).

For a cylindrical wire, A = πr² (where r is radius), so R ∝ 1/r².
If the radius of a wire is doubled (2r), its area of cross-section becomes four times (4A), and its resistance becomes one-fourth (R' = R/4).

Material (Resistivity, ρ): Resistance is directly proportional to the resistivity (ρ) of the material (R ∝ ρ).

Resistivity is an intrinsic property of the material, indicating its opposition to current flow.
Different materials have different resistivities (e.g., copper has lower resistivity than aluminum).
Unit of resistivity is Ohm-meter (Ω·m) or Ohm-centimeter (Ω·cm). Conversion: 1 Ω·cm = 10⁻² Ω·m.
Tungsten is used for electric bulb filaments due to its high melting point and appropriate resistivity.

Temperature (T):

For most pure metallic conductors, resistance increases with increasing temperature.
For semiconductors and insulators, resistance generally decreases with increasing temperature.

Factors NOT affecting resistance: The current flowing through the resistor or the voltage across it (R is an inherent property of the conductor, independent of the applied current or voltage, for ohmic materials).
Implications:

Ohm's Law: I = V/R. Reducing resistance (e.g., by shortening wire) increases current for a constant voltage.
Power Dissipation: P = I²R or P = V²/ R. For constant voltage, if R decreases, P increases. For constant current, if R decreases, P decreases.

Kirchhoff's Laws (Current Law and Voltage Law)

Kirchhoff's Laws (K.L.) are fundamental for analyzing electrical circuits.
Kirchhoff's Current Law (KCL) or Junction Rule (First Law):

States that the algebraic sum of currents entering a junction (node) is equal to the algebraic sum of currents leaving it.
Alternatively, the algebraic sum of all currents meeting at any junction in an electric circuit is zero (ΣI = 0).
Based on the principle of conservation of electric charge.
Applies to electrical nodes in circuits.

Kirchhoff's Voltage Law (KVL) or Loop Rule (Second Law):

States that the algebraic sum of all potential differences (voltages) including electromotive forces (EMFs) in any closed loop (mesh) of an electric circuit is zero (ΣV = 0).
Alternatively, the sum of emfs and potential differences around a closed loop equals zero.
Based on the principle of conservation of energy.
Applies to electrical loops in circuits.

Both laws are used for analyzing complex electrical loops and nodes to determine currents and voltages.

Ohm's Law, resistance, resistivity, and conductivity

Ohm's Law: States that the current (I) flow1. ing through a conductor is directly proportional to the potential difference (V) applied across its ends, provided all physical conditions (e.g., temperature) remain constant. Mathematical Form of Ohm's Law: $V = 2. IR$, where $R$ is the resistance of the conductor. Ohmic Conductors/Devices: Materials/ 3. devices that strictly obey Ohm's Law (e.g., metals at constant temperature, conducting copper coil, electric heating coil, rheostat). For these, the resistance (R) is independent of the magnitude and polarity of the applied potential difference or the magnitude and direction of the applied electric field. Non-Ohmic Conductors/Devices: Materials/ 4. devices that do not obey Ohm's Law (e.g., semiconductor diodes, transistors, vacuum tubes, electrolytes). Their resistance varies with voltage or current. I-V Graph (Current vs. Voltage): 5.

For ohmic conductors, it is a straight line passing through the origin.
The slope of an I-V graph is $1/R$ (conductance). A higher slope indicates lower resistance.
The slope of a V-I graph is $R$ (resistance). A higher slope indicates higher resistance.
If the graph is not a straight line, the device is non-ohmic. Resistance (R): The opposition offered by a 6. conductor to the flow of electric current.
Unit: Ohm ($\Omega$).
Formula: $R = V/I$.
Factors affecting resistance: Length (L) [R $\propto$ L], Area of crosssection (A) [R $\propto$ 1/A], material (resistivity), and temperature.
Resistance of metals generally increases with temperature. Resistivity ($\rho$): An intrinsic property of 7. a material that quantifies its opposition to current flow, independent of its dimensions.
Formula: $R = \rho (L/A)$, thus $\rho = R(A/L)$.
Unit: Ohm-meter ($\Omega \cdot m$).
Depends on the nature of the material and temperature. Good conductors have low resistivity, insulators have high resistivity. Conductivity ($\sigma$): The reciprocal of 8. resistivity, indicating how easily current flows through a material.
Formula: $\sigma = 1/\rho$.
Unit: Siemens per meter (S/m) or mho per meter ($\mho/m$). Applications of Ohm's Law: 9.
Current calculation: $I = V/R$.
Potential difference calculation: $V = IR$.
Resistance calculation: $R = V/I$.
Charge passing per second in a circuit is equivalent to current: $Q/t = I$.
To extend the range of a galvanometer to measure higher voltages, a high resistance is connected in series ($R{series} = V{range}/I{full\scale} R{meter}$).

Series and parallel combination of resistors

Resistors in Series Combination:

Equivalent Resistance (Req): Sum of individual resistances. `Req = R1 + R2 + ... + Rn`.
Current: Same current flows through each resistor. `Itotal = I1 = I2 = ... = In`.
Voltage: Total voltage divides across resistors. `Vtotal = V1 + V2 + ... + Vn`.
Effect: Total resistance is greater than the largest individual resistance.
Maximum Resistance: Achieved by connecting resistors in series.

Resistors in Parallel Combination:

Equivalent Resistance (Req): Reciprocal of Req is sum of reciprocals of individual resistances. `1/Req = 1/R1 + 1/R2 + ... + 1/Rn`.
For two resistors: `Req = (R1 R2) / (R1 + R2)`.
For 'n' identical resistors (R): `Req = R / n`.
Current: Total current divides among branches. `Itotal = I1 + I2 + ... + In`.
Voltage: Same voltage across each resistor. `Vtotal = V1 = V2 = ... = Vn`.
Effect: Total resistance is always less than the smallest individual resistance.
Minimum Resistance: Achieved by connecting resistors in parallel.

Power Dissipation (P):

`P = V^2 / R = I^2 R = V I`.
If Voltage (V) is constant (e.g., parallel connection), `P ∝ 1/R` (lower R means higher P).
If Current (I) is constant (e.g., series connection), `P ∝ R` (higher R means higher P).

Applications & Specific Cases:

Domestic Electrical Wiring: Primarily parallel connection. Ensures each appliance receives full supply voltage, allows independent operation, and if one fails, others function.
Brightness of Bulbs: For maximum brightness, bulbs (of same resistance) should be connected in parallel to a voltage source as each gets full voltage, maximizing power `(P=V^2/ R)`.
Wire Cut into Equal Parts: If a wire of resistance R is cut into 'n' equal parts, each part has resistance `R/n`. If these 'n' parts are connected in parallel, the equivalent resistance `Req = (R/n) / n = R / n^2`.
Circular Coil: If a circular coil of resistance R is cut and reconnected across its diameter, the two halves (each with resistance R/2) are in parallel. The equivalent resistance is `(R/2 R/2) / (R/2 + R/2) = (R^2/4) / R = R/ 4`.
Infinite Ladder Network: For a repeating unit of R in series with (R in parallel with the infinite ladder's equivalent resistance Req), the solution for Req is `Req = R(1 + √5) / 2`.

Superconductivity (basic concepts, critical temperature, applications)

Superconductivity: Phenomenon where certain materials exhibit zero electrical resistance when cooled below a characteristic critical temperature.
Critical Temperature (Tc): The specific temperature below which a material becomes a superconductor. Above Tc, it behaves as a normal conductor.
Meissner Effect: The complete expulsion of magnetic field lines from the interior of a superconductor as it transitions to the superconducting state. This makes superconductors perfect diamagnets.
Discovery: First observed in 1911 by Heike Kamerlingh Onnes in mercury at 4.2 K.
Types of Superconductors:

Type I (Soft): Exhibit abrupt transition to superconductivity, strong Meissner effect, and are typically pure metals (e.g., Mercury, Lead). Limited by low critical magnetic fields.
Type II (Hard): Exhibit a more gradual transition, tolerate higher magnetic fields and currents, and include alloys and ceramic materials (e.g., YBCO Yttrium Barium Copper Oxide, a hightemperature superconductor).

Applications of Superconductivity:

MRI (Magnetic Resonance Imaging): Utilizes strong, stable magnetic fields from superconducting magnets for medical diagnostics.
MAGLEV (Magnetic Levitation) Trains: Achieve levitation and propulsion through interactions with superconducting magnets, reducing friction.
SQUIDs (Superconducting Quantum Interference Devices): Extremely sensitive magnetometers used in biomagnetism, geophysics, and fundamental research.
High-Efficiency Power Transmission: Potential for lossless electricity transmission over long distances (currently facing technical challenges like refrigeration).
Supercomputers: High-speed, lowpower interconnects and processors.

11. Conductivity

Practice MCQs: Electric Current

Definition of Electric Current and current density

Current density is
Asked in: BPSC Prelims Exam 69th
1. a scalar quantity
2. a vector quantity
3. dimensionless
4. None of the above
Show answer
Answer: B. a vector quantity
- Current density (J) is current (I) divided by area (A). - Current (I) is a scalar, but current density (J) is a vector. - Its direction is the direction of charge flow.
Which one of the following statements is not correct?
Asked in: UPSC CDS 1 2017 · UPSC CDS 1 2022
1. The cathode rays originate from cathode and proceed towards the anode in a cathode- ray discharge tube.
2. The television picture tubes are nothing but cathode-ray tubes.
3. The cathode rays themselves are not visible.
4. The characteristics of cathode rays depend upon the nature of the gas present in the cathode-ray tube.
Show answer
Answer: D. The characteristics of cathode rays depend upon the nature of the gas present in the cathode-ray tube.
Statements A, B, and C are correct. Cathode rays originate from the negatively charged cathode and travel towards the positively charged anode (A). Older television and computer monitors used Cathode Ray Tubes (CRTs) to produce images (B). The beam of electrons itself is not visible to the naked eye; the light is produced when the electrons strike a fluorescent screen (C).
Which one of the following laws of electromagnetism does not give the direction of magnetic field ?
Asked in: UPSC CDS 1 2021
1. Right-hand thumb rule
2. Fleming's left-hand rule
3. Fleming's right-hand rule
4. Faraday's law of electromagnetic induction
Show answer
Answer: D. Faraday's law of electromagnetic induction
Faraday's Law quantifies the magnitude of induced EMF due to changing magnetic flux; the other rules/laws listed are primarily used to determine directions related to magnetic fields, current, force, or motion.
The magnetic field lines inside a current carrying long solenoid are in the form of
Asked in: UPSC CDS 2 2021
1. ellipse.
2. parabola.
3. hyperbola.
4. parallel straight lines.
Show answer
Answer: D. parallel straight lines.
The magnetic field inside an ideal long solenoid is uniform and axial.
A charged particle moves through a magnetic field B with a velocity v. Which one of the following statements is true for the force (F) experienced by the particle?
Asked in: UPSC CDS 2 2023
1. F is maximum when v and B are parallel to each other.
2. F is maximum when v and B are anti- parallel to each other.
3. F is maximum when v and B are perpendic‐ ular to each other.
4. The force F is independent of the angle between v and B.
Show answer
Answer: C. F is maximum when v and B are perpendic‐ ular to each other.
The sine function has a maximum value of 1, which occurs when the angle θ is 90 degrees (or π/2 radians). This means the magnetic force is maximum when the velocity of the charged particle is perpendicular to the magnetic field.
In a region, the electric potential varies as shown in the figure. Which one among the following is the correct representation of the magnitude of the associated electric field ?
Asked in: UPSC Geoscientist 2024
1. Graph (a)
2. Graph (b)
3. Graph (c)
4. Graph (d)
Show answer
Answer: B. Graph (b)
The electric field E is related to the electric potential V by the equation E = -dV/dx in one dimension. This means the electric field is the negative of the slope of the V-x graph.
Consider metallic spheres of different radii r, all at the same electrostatic potential. Then the magnitude E of the electric field just outside these spheres depends on r as :
Asked in: UPSC Geoscientist 2024
1. E ∝ r-2
2. E ∝ r-1
3. E ∝ r0
4. E ∝ r
Show answer
Answer: B. E ∝ r-1
Given that all spheres are at the same electrostatic potential V, we can express the charge Q in terms of V and r: Q = V (4πε₀r). Substitute this expression for Q into the formula for E.
Magnetic dipole moment of a square current loop of side 1 cm and carrying a current of 0.10 A is :
Asked in: UPSC Geoscientist 2024
1. 3.14 × 10-5 A m2
2. 1.00 × 10-5 A m2
3. 6.28 × 10-3 A m2
4. 4.00 × 10-3 A m2
Show answer
Answer: B. 1.00 × 10-5 A m2
The loop is a square with side length s = 1 cm. Convert the side length to meters: s = 1 cm = 0.01 m = 10⁻² m. The area of the square loop is A = s² = (10⁻² m)² = 10⁻⁴ m². The current is I = 0.10 A.
A small metallic sphere is held in a non- uniform electrostatic field as shown in the figure. Also shown are two points P and Q on the surface of the sphere. Which one of the following statements is true for the sphere?
Asked in: UPSC Geoscientist 2024
1. Sphere does not experience a force in the field.
2. Surface charge density at point P is positive while at point Q is negative.
3. Sphere experiences a force towards the left.
4. Sphere experiences a force towards the right.
Show answer
Answer: C. Sphere experiences a force towards the left.
A conductor in an electrostatic field is an equipotential surface, and the electric field lines are perpendicular to its surface. In a non-uniform field (indicated by varying density of field lines), the side facing the stronger field region (denser lines, like point P) will accumulate induced charge opposite to the field source, and the opposite side (sparser lines, like point Q) will accumulate the same type of charge as the source. The forces on these induced charges are F = qE. Since the field strength E is non-uniform, the force on the induced charge in the stronger field region will be greater.
Which of the following statement/s is/are correct regarding the electric current used in appliances used in your home? i. Microwave oven and Television use DC current. ii. Refrigerator and Washing machine use AC current. iii. Air cooler and Air conditioner use AC current. iv. Most appliances use 220 Ampere current. v. The amount of current flowing through different appliances is the same.
Asked in: HAS Prelims 2021
1. Only i, ii and iv
2. Only ii and iii
3. Only iii and iv
4. Only ii, iii, iv and v
Show answer
Answer: B. Only ii and iii
- Household power supply is AC. - Most appliances with motors (Refrigerator, Washing machine, Air cooler, Air conditioner) use AC. - Current drawn by appliances varies based on their power rating, not constant. - Household voltage is ~220-240V, not 220 Ampere.
Consider the following statements: Statement - I : Magnetic field lines emerge from the North pole of a bar magnet and merge into the South pole of the bar magnet. Statement - II : The direction of the magnetic field is considered to be the direction along which the North pole of a compass needle moves within that field. Statement - III : Inside the bar magnet, there is no magnetic field. Statement - IV : The relative strength of the magnetic field is shown by the degree of closeness of the field lines. Which of the above statements are correct?
Asked in: HAS Prelims 2021
1. Statement - I, II, III and IV
2. Statement - I, II and IV
3. Statement - I, II and IV
4. Statement - II, III and IV
Show answer
Answer: B. Statement - I, II and IV
- Magnetic field lines form closed loops. - Outside the magnet, they go from North to South. - Inside the magnet, they go from South to North. - Field direction is defined by the direction a North pole would move. - Density of field lines indicates field strength.

Drift velocity and microscopic origin of current

An electron and a proton starting from rest get accelerated through potential difference of 100 kV. The final speeds of the electron and the proton are V$_e$ and V$_p$ respectively. Which one of the following relations is correct?
Asked in: UPSC CDS 1 2019
1. V$_e$ > V$_p$
2. V$_e$ < V$_p$
3. V$_e$ = V$_p$
4. Cannot be determined
Show answer
Answer: A. V$_e$ > V$_p$
When a charged particle is accelerated from rest through a potential difference V, its kinetic energy is equal to the work done on it by the electric field, which is |qV|. So, (1/2)mv^2 = |qV|. The speed is given by v = sqrt(2|q|V/m). The electron and the proton have the same magnitude of charge (|q| = e) and are accelerated through the same potential difference V. Thus, their kinetic energies gained are equal: KE_e = KE_p = eV. (1/2)m_e V_e^2 = (1/2)m_p V_p^2 m_e V_e^2 = m_p V_p^2 V_e / V_p = sqrt(m_p / m_e) Since the mass of the electron (m_e) is significantly less than the mass of the proton (m_p), the ratio m_p / m_e is greater than 1. Therefore, V_e / V_p > 1, which means V_e > V_p. The lighter particle (electron) achieves a much higher speed than the heavier particle (proton) for the same kinetic energy.
An electron and a proton starting from rest are accelerated through a potential difference of 1000 V. Which one of the following statements in this regard is correct?
Asked in: UPSC CDS 2 2017
1. The kinetic energy of both the particles will be different.
2. The speed of the electron will be higher than that of the proton.
3. The speed of the proton will be higher than that of the electron.
4. The speed of the electron and the proton will be equal.
Show answer
Answer: B. The speed of the electron will be higher than that of the proton.
When an electron and a proton, both starting from rest, are accelerated through the same potential difference V, they gain the same kinetic energy (KE). The kinetic energy gained is equal to the work done by the electric field, which is qV, where q is the charge of the particle and V is the potential difference. The magnitude of charge is the same for both electron and proton (|q_e| = |q_p|). So, KE_electron = KE_proton = | q|V.
With reference to the electron drift speed in a current-carrying conductor, which one of the following statements is correct?
Asked in: UPSC Geoscientist 2022
1. It is much more than the average electron speed.
2. It is much lesser than the average electron speed.
3. It is very close to the average electron speed.
4. It is close to the speed of light.
Show answer
Answer: B. It is much lesser than the average electron speed.
In the absence of an electric field, free electrons in a conductor move randomly due to thermal energy, colliding with lattice ions. Their average velocity is zero, but their average speed is high. When an electric field is applied, the electrons experience a force that causes them to accelerate between collisions. Although collisions are frequent, resulting in a zig-zag path, there is a net average velocity in the direction of the force, which is the drift velocity. This drift velocity is responsible for the electric current.
Ions in a solution are subject to a uniform electric field. If each ion carries charge q and has radius R, then the ionic current I due to the applied field will depend on q and R as :
Asked in: UPSC Geoscientist 2024
1. I ∝ qR2
2. I ∝ q2R
3. I ∝ qR0
4. I ∝ q2R-1
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Answer: D. I ∝ q2R-1
The electric force on an ion is F_e = qE. Assuming Stokes' law for the viscous drag force in a solution, F_d = 6πηRv, where η is viscosity, R is the ion's radius, and v is its drift velocity. In steady state, F_e = F_d, so qE = 6πηRv. The drift velocity is v = qE / (6πηR). The ionic current I is proportional to the charge of each ion (q), the number density of ions (n), their drift velocity (v), and the cross-sectional area (A): I = nqvA.
If a charged particle (+q) is projected with certain velocity parallel to the magnetic field, then it will
Asked in: UPSC NDA 1 2015
1. trace helical path
2. trace circular path
3. continue its motion without any change
4. come to rest instantly
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Answer: C. continue its motion without any change
When the velocity vector (v) is parallel to the magnetic field vector (B), the cross product (v x B) is zero. Therefore, the magnetic force (F) on the charged particle is zero. With no force acting on it, the particle will continue to move in a straight line with constant velocity according to Newton's first law of motion.
According to Fleming's right-hand rule, if the forefinger indicates the direction of magnetic field and thumb shows the direction of motion of conductor, then the stretched middle finger will predict the direction of
Asked in: UPSC NDA 1 2022
1. force acting on the conductor
2. electric field
3. induced current
4. current
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Answer: C. induced current
Fleming's Right-Hand Rule is used to determine the direction of the induced current in a conductor moving in a magnetic field.
A positive charge is moving towards south in a space where magnetic field is pointing in the north direction. The moving charge will experience :
Asked in: UPSC NDA 1 2023
1. a deflecting force towards north direction.
2. a deflecting force towards east direction.
3. a deflecting force towards west direction.
4. no deflecting force.
Show answer
Answer: D. no deflecting force.
The direction of the velocity $\vec{v}$ is towards south, and the direction of the magnetic field $ \vec{B}$ is towards north. These two directions are opposite to each other. The angle between $ \vec{v}$ and $\vec{B}$ is 180 degrees.
Consider the following image: A proton enters a magnetic field at right angles to it, as shown above. The direction of force acting on the proton will be
Asked in: UPSC NDA 2 2020
1. to the right
2. to the left
3. out of the page
4. into the page
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Answer: D. into the page
The direction of the Lorentz force F on a positive charge is given by the Right Hand Rule applied to the cross product v x B. Standard convention for crosses (X) is 'into the plane of the diagram'.
Imagine a current-carrying straight conductor with magnetic field of lines in anti-clockwise direction. Then the direction of current is determined by
Asked in: UPSC NDA 2 2021
1. the Right-Hand Thumb rule and it would be in the downward direction.
2. the Left-Hand Thumb rule and it would be in the downward direction.
3. the Right-Hand Thumb rule and it would be in the upward direction.
4. the Left-Hand Thumb rule and it would be in the upward direction.
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Answer: C. the Right-Hand Thumb rule and it would be in the upward direction.
The Right-Hand Thumb Rule for a straight conductor states: Point your right thumb in the direction of the conventional current. Your fingers curl in the direction of the magnetic field. An anti-clockwise magnetic field (when looking down onto the wire) implies the current is flowing upwards.
Which one of the following statements regarding a current-carrying solenoid is not correct?
Asked in: UPSC NDA 2 2022
1. The magnetic field inside the solenoid is uniform.
2. The current-carrying solenoid behaves like a bar magnet.
3. The magnetic field inside the solenoid in‐ creases with increase in current.
4. If a soft iron bar is inserted inside the solenoid, the magnetic field remains the same.
Show answer
Answer: D. If a soft iron bar is inserted inside the solenoid, the magnetic field remains the same.
Ferromagnetic materials like soft iron dramatically increase the magnetic field strength when placed inside a solenoid or coil because they become strongly magnetized, effectively increasing the magnetic permeability of the core.
What will happen if a collection of positive and negative charges are passed at a high speed through a magnetic field which is perpendicular to the direction of motion of the charges ? (Assume that both kind of charges are NOT going to recombine)
Asked in: UPSC NDA 2 2023
1. Both kind of charges will stop moving
2. Positive charges and negative charges will separate out
3. Positive charges will stop but negative charges will continue moving uninterrupted
4. Both kind of charges will keep moving un‐ interrupted
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Answer: B. Positive charges and negative charges will separate out
The Lorentz force ($\vec{F} = q(\vec{v} \times \vec{B})$) acts on a moving charge in a magnetic field. The force is perpendicular to both the velocity ($\vec{v}$) and the magnetic field ($ \vec{B}$). For positive charges, the force direction is given by the right-hand rule applied to $ \vec{v} \times \vec{B}$. For negative charges (where $q$ is negative), the force is in the opposite direction. This difference in the direction of the force causes the positive and negative charges to be deflected in opposite directions, leading to separation.

Electrical Power and Energy, Joule heating effect

One should not connect a number of electrical appliances to the same power socket because (i) this can damage the appliances (ii) this can damage the domestic wiring due to overheating (iii) the appliances will not get full voltage Select the valid reason/reasons using the codes given below.
Asked in: APSC CCE Prelims Exam 2023
1. (ii) and (iii) only
2. (ii) only
3. (i) and (ii) only
4. (iii) only
Show answer
Answer: A. (ii) and (iii) only
- Overloading a circuit increases the total current drawn. - High current causes significant heat generation in the wiring (overheating). - High current also causes voltage drop across the wiring resistance (V=IR), leading to reduced voltage for appliances. - Damage to appliances (i) is a possible consequence, but (ii) and (iii) are the direct and most significant technical effects leading to the safety and performance issues.
In electrical motor
Asked in: BPSC Prelims Exam 64th
1. heat is converted into electrical energy
2. electrical energy is converted into heat
3. electrical energy is converted into mechanical energy
4. mechanical energy is converted into electrical energy
Show answer
Answer: C. electrical energy is converted into mechanical energy
Electric motors operate on the principle that a current-carrying conductor placed in a magnetic field experiences a force. This force produces torque, causing rotation and converting electrical energy input into mechanical energy output.
Working of safety fuses depends upon magnetic effect of the current chemical effect of the current magnitude of the current heating effect of the current Select the correct answer using the code given below.
Asked in: UPSC CDS 1 2018
1. 1, 2, 3 and 4
2. 1, 2 and 3 only
3. 3 and 4 only
4. 4 only
Show answer
Answer: C. 3 and 4 only
- The heating effect of electric current states that heat is produced when current flows through a resistance. - The amount of heat produced is proportional to the square of the current (I²), the resistance (R), and the time (t). - Fuse wire melts when the heat generated by excessive current raises its temperature to its melting point. - The specific melting current is determined by the material, length, and thickness of the fuse wire.
At the time of short-circuit, the current in the circuit :
Asked in: UPSC CDS 1 2022 · OPSC CS Prelims 2015
1. reduces substantially.
2. does not change.
3. increases heavily.
4. keeps on fluctuating.
Show answer
Answer: C. increases heavily.
According to Ohm's Law, the current (I) flowing through a circuit is directly proportional to the voltage (V) across it and inversely proportional to the resistance (R): I = V/R. During a short- circuit, the resistance of the path drops significantly, often close to zero. Assuming the voltage source remains relatively constant, a drastic decrease in resistance leads to a massive increase in the current flowing through the circuit.
A 100 W electric bulb is used for 10 hours a day. How many units of electrical energy are consumed by the bulb in 3 days? (1 unit = 1 kWh)
Asked in: UPSC CDS 1 2023
1. 3.00
2. 1.08
3. 2.16
4. 0.33
Show answer
Answer: A. 3.00
Electrical energy consumed is calculated as Power × Time. Energy unit (kWh) = Power (kW) × Time (h).
After using for some time, big transformers get heated up. This is due to the fact that 1. current produces heat in the transformers 2. hysteresis loss occurs in the transformers 3. liquid used for cooling gets heated Select the correct answer using the code given below.
Asked in: UPSC CDS 2 2017
1. 1 only
2. 2 and 3 only
3. 1 and 2 only
4. 1, 2 and 3
Show answer
Answer: C. 1 and 2 only
The main reasons for heating in a transformer are power losses occurring during its operation. These include copper losses (I²R losses) in the windings due to the resistance of the copper wires, and iron losses (core losses) in the core material. Iron losses consist of hysteresis loss (energy dissipated due to the repeated magnetization and demagnetization of the core by the alternating magnetic field) and eddy current loss (induced circulating currents in the core causing resistive heating). Statement 1 correctly identifies copper loss. Statement 2 correctly identifies hysteresis loss, which is one component of iron loss.
Which one of the following does not convert electrical energy into light energy?
Asked in: UPSC CDS 2 2018
1. A candle
2. A light-emitting diode
3. A laser
4. A television set
Show answer
Answer: A. A candle
- Identify the energy conversion process in each device. - Electrical energy to light energy conversion is the key process to look for. - Understand the working principle of each listed item.
A circuit has a fuse having a rating of 5 A. What is the maximum number of 100 W-220 V bulbs that can be safely connected in parallel in the circuit?
Asked in: UPSC CDS 2 2018
1. 20
2. 15
3. 11
4. 10
Show answer
Answer: D. 10
The circuit has a fuse rating of 5 A, which means the total current drawn by all connected devices must not exceed 5 A to prevent the fuse from blowing. Each bulb is rated at 100 W at 220 V. The current drawn by a single bulb is calculated using the formula P = V I, so I = P/V. Current per bulb = 100 W / 220 V = 10/22 A = 5/11 A. When bulbs are connected in parallel, the voltage across each bulb is the same (220 V), and the total current drawn from the source is the sum of the currents drawn by each bulb. Let N be the number of bulbs connected. The total current is N (5/11) A. For safe operation, the total current must be less than or equal to the fuse rating: N (5/11) A ≤ 5 A. N 5 ≤ 5 11 N ≤ 11. Since the number of bulbs must be a whole number, the maximum number of bulbs that can be safely connected in parallel is 11.
A 100 W electric bulb is used for 10 hours a day. How many units of energy are consumed in 30 days ?
Asked in: UPSC CDS 2 2019
1. 1
2. 10
3. 30
4. 300
Show answer
Answer: C. 30
Energy consumed is calculated as Power × Time. Power is measured in Watts (W) or Kilowatts (kW), and Time is measured in hours (h). Energy is typically measured in Watt-hours (Wh) or Kilowatt-hours (kWh). One unit of electrical energy is equal to 1 kWh.
An electric bulb is connected to a 110 V generator. The current is 0.2 A. What is the power of the bulb ?
Asked in: UPSC CDS 2 2021
1. 0.22 W
2. 2.2 W
3. 22 W
4. 220 W
Show answer
Answer: C. 22 W
Electrical power (P) is calculated using the formula P = VI, where V is the voltage in volts and I is the current in amperes. The unit of power is the Watt (W).
If potential difference and current flowing through a wire is increased twice each then electric power
Asked in: CGPSC Prelims Exam 2015
1. Will remain unaffected
2. Will increase four times
3. Will increase two times
4. Will decrease half
Show answer
Answer: B. Will increase four times
The formula for electric power (P) is given by the product of potential difference (V) and current (I): P = V × I. Let the initial potential difference be V₁ and the initial current be I₁. The initial power is P₁ = V₁ × I₁. If the potential difference is increased twice, the new potential difference V₂ = 2 × V₁. If the current is increased twice, the new current I₂ = 2 × I₁. The new electric power is P₂ = V₂ × I₂ = (2 × V₁) × (2 × I₁) = 4 × (V₁ × I₁). Since P₁ = V₁ × I₁, we have P₂ = 4 × P₁.

Electromotive Force (EMF) and internal resistance of a cell

If the speed of a moving magnet inside a coil increases, the electric current in the coil
Asked in: UPSC CDS 2 2019
1. increases
2. decreases
3. reverses
4. remains the same
Show answer
Answer: A. increases
- This phenomenon is described by Faraday's Law of Electromagnetic Induction. - Faraday's Law states that the magnitude of the induced electromotive force (EMF) in a circuit is proportional to the rate of change of magnetic flux through the circuit. - When a magnet moves near a coil, it causes a change in the magnetic flux passing through the coil. - Increasing the speed of the moving magnet increases the rate at which the magnetic flux changes. - A greater rate of change of magnetic flux induces a larger EMF in the coil. - According to Ohm's Law, a larger induced EMF results in a larger induced current, assuming the resistance of the coil remains constant.
A resistor of 10 $\Omega$ connected across a cell of emf 12 V, draws the current of 1.1 A. Find the internal resistance of the cell.
Asked in: MPSC CS Prelims 2018
1. 10 $\Omega$
2. 0.1 $\Omega$
3. 10.9 $\Omega$
4. 0.91 $\Omega$
Show answer
Answer: D. 0.91 $\Omega$
The terminal voltage of a cell is less than its emf when current is drawn from it due to the voltage drop across the internal resistance ($V = E - Ir$).
An electric circuit is given below. V₁ = 1 V and Resistance R = 1000 Ω. The current through the resistance R is very close to 1 mA and the voltage across point A and B, VAB = 1 V. Now the circuit is changed to : where value of V₂ = 5 V. The internal resistances of both the batteries are 0.1 Ω. The current through the resistance R is about :
Asked in: UPSC NDA 1 2024
1. 1.0 mA
2. 1.2 mA
3. 3.0 mA
4. 5.0 mA
Show answer
Answer: C. 3.0 mA
- The first part of the question establishes the context: R=1000Ω, V1=1V providing ~1mA and VAB=1V, suggesting R is connected across the 1V source. - The second part adds V2=5V and internal resistances 0.1Ω for both batteries. - Assuming a parallel connection of the two batteries and the resistor R between points A and B is the most likely configuration given the options and the structure implied by the first sentence. - When voltage sources are in parallel, they can be combined into an equivalent source V_eq and r_eq. - Formula for parallel sources: V_eq = (Σ Vi/ri) / (Σ 1/ri) and r_eq = 1 / (Σ 1/ri). - Calculate V_eq and r_eq for V1 (1V, 0.1Ω) and V2 (5V, 0.1Ω). - Calculate the current through R (1000Ω) when connected across the equivalent source. - Calculation gives current ≈ 3 mA.
A DC generator works on the principle of
Asked in: UPSC NDA 2 2022
1. Ohm's law
2. Joule's law of heating
3. Faraday's laws of electromagnetic induction
4. None of the above
Show answer
Answer: C. Faraday's laws of electromagnetic induction
Electromagnetic induction is the process where a conductor moving through a magnetic field (or a changing magnetic field near a conductor) induces an electromotive force (EMF), which can drive an electric current. Faraday's laws quantify this phenomenon.
Ms. Rani decides to convert her AC generator into a DC generator. Which one of the following she would need to use?
Asked in: UPSC NDA 2 2023
1. A split-ring type commutator
2. Slip rings and brushes
3. A stronger magnetic field
4. A rectangular wire loop
Show answer
Answer: A. A split-ring type commutator
- An AC generator produces alternating current due to the use of slip rings, which maintain continuous contact with the rotating coil. - To convert this to direct current (DC), the direction of the current in the external circuit needs to be maintained constant. - A split-ring commutator achieves this by reversing the connection of the coil to the external circuit every half rotation, ensuring that the current flows in the same direction in the external load.
A device which converts chemical energy into electrical energy is called
Asked in: RPSC RAS Prelims 2013
1. Motor
2. Generator
3. Moving coil meter
4. Battery
Show answer
Answer: D. Battery
A battery stores chemical energy and converts this stored chemical energy directly into electrical energy through electrochemical reactions when connected to a circuit.
Photovoltaic cell is related to
Asked in: RPSC RAS Prelims 2016
1. Solar Energy
2. Nuclear Energy
3. Wind Energy
4. Geo Thermal Energy
Show answer
Answer: A. Solar Energy
This process is fundamental to solar power generation. When photons from sunlight strike the cell, they excite electrons, creating an electric current.
Cycle dynamo converts
Asked in: TNPSC CS Prelims 2014
1. Mechanical energy into thermal energy
2. Mechanical energy into electrical energy
3. Electrical energy into light energy
4. Electrical energy into thermal energy
Show answer
Answer: B. Mechanical energy into electrical energy
- A dynamo is a type of electrical generator. - It operates based on Faraday's law of electromagnetic induction. - In a cycle dynamo, the rotation of the bicycle wheel provides the mechanical energy to spin a rotor (either magnets or coils) within the dynamo. - This rotation causes a change in magnetic flux through coils, inducing an electric current. - This generated electrical energy is then used to power the bicycle lights. - Therefore, the conversion is from the mechanical energy of rotation to electrical energy.
Nickel-Cadmium (Ni-Cd) battery is used in
Asked in: UKPSC CS Prelims 2016
1. Calculator
2. Cordless electronic appliances
3. Transistors
4. All of the above
Show answer
Answer: B. Cordless electronic appliances
- Ni-Cd batteries are rechargeable. - They provide high discharge rates and are suitable for applications requiring frequent recharging. - Common uses include portable power tools and cordless devices.
A battery consists of 10 cells, each of emf 1V. If 2 cells are wrongly connected, the emf of the battery becomes
Asked in: WBPSC CS Prelims 2021
1. 8V
2. 10V
3. 6V
4. 12V
Show answer
Answer: C. 6V
When cells are connected in series, their electromotive forces (emfs) add up. If a cell is connected incorrectly (reverse polarity), its emf opposes the total emf.
Which one of the following sets makes the terminals of the ordinary torch cell ?
Asked in: UPPSC Upper Mains 2010
1. Zinc - Carbon
2. Copper - Zinc
3. Zinc - Cadmium
4. Carbon - Copper
Show answer
Answer: A. Zinc - Carbon
The construction of a common dry cell involves a zinc casing (negative terminal) and a carbon rod (positive terminal).

Factors affecting resistance (length, area, material, temperature)

A copper wire of radius r and length 1 has a resistance of R. A second copper wire with radius 2r and length 1 is taken and the two wires are joined in a parallel combination. The resultant resistance of the parallel combination of the two wires will be
Asked in: UPSC CDS 1 2017
1. 5R
2. 5/4 R
3. 4/5 R
4. R/5
Show answer
Answer: D. R/5
- For the first wire: R₁ = R = ρ (l / πr²). - For the second wire: The radius is 2r, and the length is l. So, R₂ = ρ (l / π(2r)²) = ρ (l / 4πr²) = (1/4) [ρ (l / πr²)] = R/4. - When two resistors R₁ and R₂ are connected in parallel, the resultant resistance R_eq is given by 1/R_eq = 1/R₁ + 1/R₂ or R_eq = (R₁ R₂) / (R₁ + R₂). - Plugging in the values, R_eq = (R (R/4)) / (R + R/4) = (R²/4) / (5R/4). - R_eq = (R²/4) (4/5R) = R/5.
A wire of copper having length l and area of cross-section A is taken and a current I is flown through it. The power dissipated in the wire is P. If we take an aluminum wire having same dimensions and pass the same current through it, the power dissipated will be
Asked in: UPSC CDS 1 2018
1. P
2. < P
3. > P
4. 2P
Show answer
Answer: C. > P
- Power dissipated in a resistor is proportional to its resistance (P = I²R, or P = V²/R, or P = VI). - Resistance depends on the material's resistivity and the wire's dimensions (R = ρl/A). - Aluminum has higher resistivity than copper.
An electric circuit is consisting of a cell, an ammeter and a nichrome wire of length l. If the length of the wire is reduced to half (l/2), then the ammeter reading
Asked in: UPSC CDS 2 2021
1. decreases to one-half.
2. gets doubled.
3. decreases to one-third.
4. remains unchanged.
Show answer
Answer: B. gets doubled.
Resistance is directly proportional to length. Current is inversely proportional to resistance.
A metallic wire having length *l* and area of cross-section A has a resistance R. It is now connected in series with a wire of same metal but with length 2*l* and area of cross-section 2A. The total resistance of the combination will be:
Asked in: UPSC Geoscientist 2020
1. 2R
2. 4R
3. R
4. R/2
Show answer
Answer: A. 2R
The resistance of a wire is directly proportional to its length and inversely proportional to its area of cross-section. Resistances in series add up directly.
Consider a copper wire of length 1m and area of cross-section 1mm². Given the resistivity of the copper is 1.7 × 10⁻⁶ Ω cm. What is the resistance of this wire?
Asked in: UPSC Geoscientist 2021
1. 17 milli-ohm
2. 1.7 milli-ohm
3. 0.17 milli-ohm
4. 0.17 ohm
Show answer
Answer: A. 17 milli-ohm
Resistance depends on the material's resistivity and the dimensions (length and cross-sectional area) of the conductor. Ensure all units are consistent (e.g., meters and m², or centimeters and cm²) before calculation.
Which one of the following physical quantities does NOT affect the resistance of a cylindrical resistor ?
Asked in: UPSC NDA 1 2017
1. The current through it
2. Its length
3. The resistivity of the material used in the resistor
4. The area of cross-section of the cylinder
Show answer
Answer: A. The current through it
The resistance of a resistor is a property of the material and its geometry. For a cylindrical resistor made of a homogeneous material, its resistance (R) is determined by the formula R = ρ (L/A), where ρ (rho) is the resistivity of the material, L is the length of the resistor, and A is the area of its cross-section.
Let us consider a copper wire having radius r and length l. Let its resistance be R. If the radius of another copper wire is 2r and the length is l/2 then the resistance of this wire will be
Asked in: UPSC NDA 1 2019
1. R
2. 2R
3. R/4
4. R/8
Show answer
Answer: D. R/8
The resistance (R) of a wire is directly proportional to its length (l) and inversely proportional to its cross-sectional area (A). The formula is R = ρ(l/A), where ρ is the resistivity of the material. The cross-sectional area of a wire with radius r is A = πr².
Two resistors $R_1$ and $R_2$ arranged in parallel combination in an electrical closed circuit are made of the same material and of same thickness. If the length of $R_2$ is twice the length of $R_1$, then the total resistance $R$ satisfies
Asked in: UPSC NDA 1 2022
1. $3R = 2R_1$
2. $3R = 2R_2$
3. $2R = 3R_1$
4. $2R = 3R_2$
Show answer
Answer: A. $3R = 2R_1$
For resistors in parallel, the total resistance $R$ is given by $1/R = 1/R_1 + 1/R_2$. The resistance of a wire is $R = \rho L/A$, where $\rho$ is resistivity, $L$ is length, and $A$ is cross- sectional area.
In an electric circuit, a wire of resistance 10 Ω is used. If this wire is stretched to a length double of its original value, the current in the circuit would become :
Asked in: UPSC NDA 1 2023
1. half of its original value.
2. double of its original value.
3. one-fourth of its original value.
4. four times of its original value.
Show answer
Answer: C. one-fourth of its original value.
- Resistance R = $\rho \frac{L}{A}$, where $ \rho$ is resistivity, L is length, and A is cross- sectional area. - Volume of the wire V = A × L. When stretched, the volume remains constant: $A'L' = AL$. - Ohm's Law: Current I = Voltage V / Resistance R.
The resistance of a wire of length / and area of cross-section a is x ohm. If the wire is stretched to double its length, its resistance would become:
Asked in: UPSC NDA 2 2015
1. 2x ohm
2. 0.5 x ohm
3. 4x ohm
4. 6x ohm
Show answer
Answer: C. 4x ohm
When a wire is stretched, its length increases, and its cross-sectional area decreases while its volume remains constant. This significantly increases its resistance.
If the length of the filament of a heater is reduced by 10%, the power of the heater will
Asked in: WBPSC CS Prelims 2021
1. increase by about 9%
2. increase by about 11%
3. increase by about 19%
4. decrease by about 10%
Show answer
Answer: B. increase by about 11%
The resistance of a wire is directly proportional to its length (R ∝ L). Power dissipated by a resistor is given by P = V^2 / R, assuming constant voltage (V).

Kirchhoff's Laws (Current Law and Voltage Law)

An electrical circuit having combinations of resistances and capacitance is given below. The current, flowing through the circuit will be
Asked in: UPSC CDS 2 2022
1. 1 A
2. 2 A
3. 1·5 A
4. 0·5 A
Show answer
Answer: B. 2 A
- The diagram shows the 10 ohm resistor in series with a parallel combination. - The parallel combination consists of two branches: one with the 20 ohm resistor and the capacitor in series, and another branch which appears to be a simple wire (zero resistance). - In steady state DC, the capacitor acts as an open circuit, meaning the branch containing the 20 ohm resistor and capacitor has effectively infinite resistance to DC current flow. - The other branch in parallel has zero resistance (a wire). - Current always follows the path of least resistance. In parallel, a zero-resistance path effectively shorts out any parallel path with resistance. Thus, all the current passing through the 10 ohm resistor will flow through the wire branch, and no current will flow through the branch containing the 20 ohm resistor and capacitor in steady state. - The equivalent resistance of the parallel combination is 0 ohms. - The total resistance of the circuit is the sum of the series resistor and the equivalent parallel resistance: R_total = 10 ohms + 0 ohms = 10 ohms. - The total current flowing from the battery is given by Ohm's Law: I = V / R_total = 10V / 10 ohms = 1 A. This is the current flowing through the circuit from the battery.
Shown in the circuit is a 20 V battery connected to an arrangement of four resistors. If the potential difference between points P and Q in the circuit is zero then :
Asked in: UPSC Geoscientist 2024
1. R = 2 Ω
2. R = 8 Ω
3. R = 18 Ω
4. R = 4 Ω
Show answer
Answer: D. R = 4 Ω
Let the branches be from A to B. Branch 1 has resistors 10Ω and R in series, with P between them. Branch 2 has 5Ω and 10Ω in series, with Q between them. For the potential difference between P and Q to be zero (V_P = V_Q), the potential division must be the same in both branches relative to the terminals A and B. This occurs when the bridge is balanced, i.e., the ratio of resistances in the arms is equal: 10Ω / R = 5Ω / 10Ω.
Consider the following electric circuit : [Circuit diagram is present in the image] The current in the above electric circuit is :
Asked in: UPSC NDA 1 2024
1. 1 A
2. (10/15) A
3. 2 A
4. 1.5 A
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Answer: B. (10/15) A
The circuit consists of a 10V battery connected to two resistors, 5 ohms and 10 ohms, in series. For resistors connected in series, the total resistance (R_total) is the sum of individual resistances. R_total = 5 Ω + 10 Ω = 15 Ω. According to Ohm's Law, the current (I) flowing through the circuit is given by V = I R_total, where V is the voltage. Therefore, I = V / R_total = 10 V / 15 Ω = (10/15) A.
"The sum of emfs and potential differences around a closed loop equals zero" is a consequence of
Asked in: UPSC NDA 2 2019
1. Ohm's law.
2. Conservation of charge.
3. Conservation of momentum.
4. Conservation of energy.
Show answer
Answer: D. Conservation of energy.
- Kirchhoff's Voltage Law (KVL) is mathematically expressed as $\sum V = 0$ around any closed loop. - KVL is based on the conservative nature of the electric field and the conservation of energy. - Potential difference is defined as the change in potential energy per unit charge. Traversing a closed loop means returning to the initial potential, so the total change in potential (and potential energy) is zero.
Kirchoff's Laws are related to
Asked in: ASSISTANT ENGINEER(ELECTRICAL) 2016
1. Electromechanical systems
2. Hydraulic loops and nodes
3. Pneumatic loops and nodes
4. Electrical loops and nodes
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Answer: D. Electrical loops and nodes
Kirchhoff's Laws are foundational for understanding electrical circuits and are often tested in physics and engineering-related subjects.
Norton equivalent between A and B for the circuit shown in Fig. 1 is
Asked in: GS&EI ElectEngg 14 2024 1601 2025
1. 2 A and 2.5 $\Omega$
2. 0.5 A and 1 $\Omega$
3. 1 A and 2 $\Omega$
4. 0.4 A and 2 $\Omega$
Show answer
Answer: A. 2 A and 2.5 $\Omega$
The Norton equivalent circuit consists of a current source ($I_N$) in parallel with a resistance ($R_N$). The values of $I_N$ and $R_N$ are determined by analyzing the original circuit. $R_N$ is the equivalent resistance seen from the terminals with independent sources deactivated. $I_N$ is the current flowing through the terminals when they are short-circuited.
The voltage of the node $a$ with respect to $o$ for the circuit shown in Fig. 2 is
Asked in: GS&EI ElectEngg 14 2024 1601 2025
1. 20 V
2. 40 V
3. 50 V
4. 60 V
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Answer: C. 50 V
Nodal analysis is a powerful technique for solving circuit problems. It involves applying Kirchhoff's Current Law (KCL) at each node to establish a set of linear equations, which can then be solved for the unknown node voltages.
In a given circuit in Fig. 4, the value of R that will give critical damping is
Asked in: GS&EI ElectEngg 14 2024 1601 2025
1. 2 $\Omega$
2. 1 $\Omega$
3. 4 $\Omega$
4. 10 $\Omega$
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Answer: A. 2 $\Omega$
GS&EI ElectEngg 14 2024 1601-2025 The given circuit is a series RLC circuit. For a series RLC circuit, critical damping occurs when the resistance R is equal to the characteristic impedance of the circuit. The condition for critical damping in a series RLC circuit is given by $R = 2\sqrt{\frac{L}{C}}$.From the figure, assuming typical component values if not explicitly provided (e.g., by looking at typical circuit diagrams in textbooks or exams where such questions appear, or if the figure itself indicates values): Let's assume the inductance L and capacitance C are standard values that might be presented alongside a diagram. Without the diagram (Fig. 4), it is impossible to definitively determine L and C. However, based on the options provided (2 $\Omega$, 1 $\Omega$, 4 $ \Omega$, 10 $\Omega$), the question likely intends for the student to apply the formula $R = 2\sqrt{\frac{L}{C}}$.Let's work backward from the answer to see what L and C values would make R=2 $\Omega$ the critical damping resistance. If $R_{critical} = 2 \Omega$, then $2 = 2\sqrt{\frac{L}{C}}$, which implies $1 = \sqrt{\frac{L}{C}}$, so $\frac{L}{C} = 1$, or $L = C$. This is one possibility.Another common way critical damping is assessed is through the damping factor $\zeta$. For a series RLC circuit, the damping factor is $\zeta = \frac{R} {2}\sqrt{\frac{C}{L}}$. Critical damping occurs when $\zeta = 1$. So, $1 = \frac{R} {2}\sqrt{\frac{C}{L}}$, which means $R = 2\sqrt{\frac{L}{C}}$.Let's assume from context that the figure provides values for L and C. If we assume standard values that would lead to the answer 2 $\Omega$. For example, if L = 1 H and C = 1 F, then $R_{critical} = 2\sqrt{\frac{1}{1}} = 2 \Omega$. If L = 4 H and C = 4 F, then $R_{critical} = 2\sqrt{\frac{4}{4}} = 2 \Omega$. figure, we are relying on the premise that one of the options is correct, and option A (2 $\Omega$) is the intended answer. The condition for critical damping in a series RLC circuit is $R = 2\sqrt{\frac{L}{C}}$. If the values of L and C from the diagram satisfy this equation with R=2 $\Omega$, then this is the correct answer.

Ohm's Law, resistance, resistivity, and conductivity

An AC current can be produced by
Asked in: BPSC Prelims Exam 69th
1. choke coil
2. dynamo
3. transformer
4. None of the above
Show answer
Answer: B. dynamo
- A dynamo (generator) creates electrical energy from mechanical energy. - AC generators specifically produce alternating current. - The process involves electromagnetic induction.
The r.m.s. potential difference between the red live wire and black neutral wire in Indian domestic electric supply is
Asked in: UPSC CDS 1 2023
1. 160 V
2. 220 V
3. 300 V
4. 410 V
Show answer
Answer: B. 220 V
- The voltage of AC power is usually specified by its r.m.s. value. - The standard domestic supply voltage in India is 220-240 V AC at 50 Hz.
A wire with black insulation during electric supply is
Asked in: CGPSC Prelims Exam 2014
1. Live wire
2. Earth wire
3. Neutral wire
4. Fuse wire
Show answer
Answer: C. Neutral wire
In standard electrical wiring color codes, different colors are used to identify wires for safety and ease of installation. In older wiring standards (like the previous Indian and UK standards), the color code for mains electricity was typically: Red for the Live wire, Black for the Neutral wire, and Green for the Earth wire. While newer standards use Brown for Live and Blue for Neutral, black insulation is commonly found in existing installations as the Neutral wire. The Live wire carries the voltage from the source, the Neutral wire provides the return path for the current, and the Earth wire is a safety ground.
The basic difference between an AC generator and a DC generator is that
Asked in: CGPSC Prelims Exam 2018
1. AC generator has an electromagnet while a DC generator has permanent magnet.
2. AC generator has slip rings while the DC generator has a commutator.
3. DC generator will generate a higher voltage.
4. AC generator will generate a higher voltage.
Show answer
Answer: B. AC generator has slip rings while the DC generator has a commutator.
The primary difference between an AC generator and a DC generator lies in the mechanism used to collect the induced current. An AC generator uses slip rings, which allow the current to flow out as an alternating current. A DC generator uses a commutator (split rings), which reverses the connection to the external circuit every half rotation, thereby ensuring the current flows in only one direction in the external circuit, producing a direct current.
A straight conducting wire carries a DC current through it. Which one of the following statements is true regarding the magnetic field around the wire?
Asked in: UPSC Geoscientist 2020
1. The magnetic field does not depend upon the amount of current flowing through the wire
2. The magnetic field does not depend upon the distance from the wire
3. The magnetic field depends upon the material of the wire
4. The magnetic field decreases with an increase from the wire
Show answer
Answer: D. The magnetic field decreases with an increase from the wire
The magnitude of the magnetic field (B) at a distance 'r' from a long, straight wire carrying a current 'I' is given by the formula $B = \frac{\mu_0 I}{2\pi r}$ (in vacuum or air). This formula shows that B is directly proportional to the current I and inversely proportional to the distance r.
Which one of the following statements is correct regarding the magnetic field inside a current carrying solenoid?
Asked in: UPSC Geoscientist 2020
1. The magnetic field inside the solenoid is zero
2. The magnetic field inside the solenoid is uniform at all the points
3. The magnetic field inside the solenoid in‐ creases as we move towards the ends
4. The magnetic field inside the solenoid de‐ creases as we move towards the ends
Show answer
Answer: B. The magnetic field inside the solenoid is uniform at all the points
For an ideal, infinitely long solenoid, the magnetic field inside is uniform and given by the formula B = μ₀ n I, where μ₀ is the permeability of free space, n is the number of turns per unit length, and I is the current. This formula shows the field is constant everywhere inside, independent of the position.
A current carrying circular loop having n number of turns per unit length has a current I through it. If the current through it and the number of turns per unit length are doubled, then the magnetic field at the centre of the loop will:
Asked in: UPSC Geoscientist 2020
1. remain same.
2. increase by four times.
3. increase by two times.
4. decrease by two times.
Show answer
Answer: B. increase by four times.
The magnetic field produced by a current-carrying coil (like a loop or solenoid) is directly proportional to the number of turns in the coil and the current flowing through it.
The $I-V$ graph for two resistors, resistor 1 ($R_1$) and resistor 2 ($R_2$), are shown below : [Image of I-V graph] Which one of the following statements about these resistors is not correct?
Asked in: UPSC Geoscientist 2023
1. $R_1$ follows Ohm's law.
2. $R_2$ does not follow Ohm's law after voltage $V_1$.
3. Up to $V_1$, the resistance of $R_1$ is smaller than that of $R_2$.
4. Up to $V_1$, the resistance of $R_1$ is larger than that of $R_2$.
Show answer
Answer: D. Up to $V_1$, the resistance of $R_1$ is larger than that of $R_2$.
The resistance $R$ of a component can be determined from its I-V graph as $R = V/I$. This is equivalent to $R = 1 / (\text{slope of the I-V graph})$ if the graph plots I on the y-axis and V on the x-axis, as shown in the image. For resistor $R_1$, the graph is a straight line passing through the origin, indicating that $R_1$ is an ohmic resistor and follows Ohm's law. Its resistance is constant. The slope of the $R_1$ line (I/ V) is constant. For resistor $R_2$, the graph is a curve. Up to voltage $V_1$, the graph is approximately linear but less steep than $R_1$. The slope of the $R_2$ curve (I/V) is smaller than the slope of the $R_1$ line in this region. Since resistance $R = 1 / (\text{slope of I-V graph})$, a smaller slope corresponds to a larger resistance. Comparing the slopes up to $V_1$: (Slope of $R_1$) > (Slope of $R_2$). Therefore, (Resistance of $R_1$)
Core of a transformer is made of :
Asked in: HPPSC HPAS Prelims 2017
1. Steel
2. Soft iron
3. Copper
4. Aluminium
Show answer
Answer: B. Soft iron
A transformer relies on a changing magnetic field within its core to induce voltage in the secondary coil. The core material must readily conduct this magnetic field.
The core of a transformer is made of:
Asked in: HPPSC HPAS Prelims 2017
1. Of steel
2. Of soft iron
3. Of copper
4. Of aluminium
Show answer
Answer: B. Of soft iron
Soft iron is preferred due to its high magnetic permeability and low hysteresis loss, which minimizes energy losses when subjected to the alternating magnetic field produced by alternating current. This allows the magnetic flux to flow easily through the core, efficiently transferring energy from the primary to the secondary coil.
Consider the following statements: Statement - I : The magnetic field lines emerge from the north pole and merge at the south pole of a bar magnet. Statement - II : The direction of magnetic field is taken to be the direction in which the north pole of a compass needle moves. Statement - III : There is no field inside a bar magnet. Statement- IV : The relative strength of the magnetic field is shown by the degree of closeness of the field lines. Which of the above statement(s) is/are correct?
Asked in: HAS Prelims 2021
1. Statements - I, II, III and IV
2. Statements - I, II, and III
3. Statements - I, II and IV
4. Statements - II, III and IV
Show answer
Answer: C. Statements - I, II and IV
Statement I: Magnetic field lines emerge from the North pole and enter the South pole outside the magnet, forming closed loops. Inside the magnet, they go from South to North. The statement is conventionally understood to describe the field outside the magnet, making it correct. Statement II: The direction of the magnetic field at any point is defined as the direction in which the north pole of a compass needle would point if placed at that point. This statement is correct. Statement III: There is a magnetic field inside a bar magnet. The field lines continue from the South pole to the North pole within the magnet, completing the loops. This statement is incorrect. Statement IV: The density or closeness of magnetic field lines in a region indicates the strength of the magnetic field in that region. Where lines are closer, the field is stronger (e.g., near the poles); where they are farther apart, the field is weaker. This statement is correct.

Series and parallel combination of resistors

The total resistance of a circuit having two parallel resistors is 1·403 kilo-ohm. If one of the resistors is 2·0 kilo-ohm, then the other resistor will be
Asked in: BPSC Prelims Exam 66th
1. 1·403 kilo-ohm
2. 2·0 kilo-ohm
3. 3·403 kilo-ohm
4. 4·70 kilo-ohm
Show answer
Answer: D. 4·70 kilo-ohm
The reciprocal of the total resistance in a parallel circuit is the sum of the reciprocals of individual resistances.
Suppose voltage V is applied across a resistance R. The power dissipated in the resistance is P. Now the same voltage V is applied across a parallel combination of three equal resistors each of resistance R. Then the power dissipated in the second case will be
Asked in: UPSC CDS 1 2017
1. P
2. 3P
3. P/3
4. 2P/3
Show answer
Answer: B. 3P
In the first case, power dissipated is P = V²/R. In the second case, three equal resistors, each of resistance R, are connected in parallel across the same voltage V. The equivalent resistance of three resistors in parallel is R_eq = 1 / (1/R + 1/R + 1/R) = 1 / (3/R) = R/3. The power dissipated in the second case is P' = V²/R_eq. Substituting R_eq = R/3, we get P' = V²/(R/3) = 3 (V²/R). Since P = V²/R, the power dissipated in the second case is P' = 3P.
Three resistors of resistances 11 $\Omega$, 22 $\Omega$ and 33 $\Omega$ are connected in parallel. Their equivalent resistance is equal to
Asked in: UPSC CDS 1 2023
1. 66 $\Omega$
2. 22 $\Omega$
3. 12 $\Omega$
4. 6 $\Omega$
Show answer
Answer: D. 6 $\Omega$
- For resistors in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of individual resistances. - The equivalent resistance in a parallel circuit is always less than the smallest individual resistance. In this case, 6 $\Omega$ is less than 11 $\Omega$.
Consider the following combination of resistors : [Diagram of resistors] The equivalent resistance of the combination of resistors between P and Q is
Asked in: UPSC Geoscientist 2023
1. 2 Ω
2. 3 Ω
3. 3.5 Ω
4. 2.5 Ω
Show answer
Answer: D. 2.5 Ω
Resistors in series add directly (R_total = ΣR_i). Resistors in parallel combine as the reciprocal of the sum of reciprocals (1/R_total = Σ1/R_i).
Domestic electrical wiring is basically a :
Asked in: HPPSC HPAS Prelims 2014 · UPPCS Prelims 1996
1. Series connection
2. Parallel connection
3. Combinations of series and parallel connec‐ tion
4. Series connection within each room and parallel connection elsewhere
Show answer
Answer: B. Parallel connection
In parallel wiring, each appliance or light fixture is connected across the main voltage supply independently. This ensures that each device receives the full voltage and can be operated independently of others.
घरेलू ǒबजली के तार आधार रूप में कै सिे हैं ?
Asked in: HPPSC HPAS Prelims 2014
1. श्रेणीबद्ध सिंोजन
2. सिमानान्तर सिंोजन
3. श्रेणीबद्ध और सिमानान्तर सिंोजन का सिंोग
4. प्रत्ेक कक्ष में श्रेणीबद्ध सिंोजन और दिूसिरी जगह सिमानान्तर सिंोजन
Show answer
Answer: B. सिमानान्तर सिंोजन
सिमानांतर सिंोजन ह सिुǓनǔæचत करता है ͩक प्रत्ेक उपकरण को पूणर्य आपूǓतर्य वोल्टेज प्राप्त हो और एक उपकरण के खिराब होने ा बंदि होने पर भी अन् काम करते रहें।
Three resistors with magnitudes 2, 4 and 8 ohm are connected in parallel. The equivalent resistance of the system would be
Asked in: UPSC NDA 1 2016
1. less than 2 ohm
2. more than 2 ohm but less than 4 ohm
3. 4 ohm
4. 14 ohm
Show answer
Answer: A. less than 2 ohm
When resistors are connected in parallel, the reciprocal of the equivalent resistance (Req) is equal to the sum of the reciprocals of the individual resistances. Given resistances are R₁ = 2 ohm, R₂ = 4 ohm, and R₃ = 8 ohm. The formula for equivalent resistance in parallel is: 1/Req = 1/ R₁ + 1/R₂ + 1/R₃ Substituting the values: 1/Req = 1/2 + 1/4 + 1/8 To add these fractions, find a common denominator, which is 8: 1/Req = 4/8 + 2/8 + 1/8 1/Req = (4 + 2 + 1) / 8 1/Req = 7/8 Now, take the reciprocal to find Req: Req = 8/7 ohm To compare this value with the options, calculate the decimal value: Req ≈ 1.14 ohm. A key principle of parallel resistance is that the equivalent resistance is always less than the smallest individual resistance in the combination. In this case, the smallest resistance is 2 ohm, and 8/7 ohm (approx 1.14 ohm) is indeed less than 2 ohm.
Consider the following circuit : [Circuit diagram is present in the image] The equivalent resistance of the circuit will be
Asked in: UPSC NDA 1 2016
1. 12 Ω
2. 8 11/12 Ω
3. 9 1/11 Ω
4. 24/25 Ω
Show answer
Answer: C. 9 1/11 Ω
The circuit diagram typically associated with this question ID consists of a triangle formed by three nodes (say, A, B, C) with resistors between them. Specifically, R_AB = 4 Ω, R_BC = 4 Ω, and between A and C there are two resistors in parallel: 6 Ω and 3 Ω. The equivalent resistance between A and C from the parallel combination is (6 3) / (6 + 3) = 18 / 9 = 2 Ω. So, the triangle effectively has resistances R_AB = 4 Ω, R_BC = 4 Ω, and R_AC = 2 Ω. Calculating the equivalent resistance between terminals A and C involves the direct path (2 Ω) in parallel with the path through B (4 Ω + 4 Ω = 8 Ω). The equivalent resistance R_AC = (2 8) / (2 + 8) = 16 / 10 = 1.6 Ω = 16/10 = 8/5 Ω. However, 1.6 Ω (or 8/5 Ω) is not among the options. The fraction 9 1/11 Ω is equal to (9 11 + 1) / 11 = 100 / 11 Ω. There is a significant discrepancy between the standard calculation based on the typical diagram (1.6 Ω) and the provided option C (100/11 Ω ≈ 9.09 Ω). Assuming the diagram and options are correct as presented in the original source of the question, there might be an alternative interpretation of the circuit or a known mistake in the problem or options. However, given that option C is frequently cited as the correct answer for this specific question ID, it is listed here, despite the standard calculation yielding a different result. Without a clear alternative interpretation or correction, the discrepancy remains.
A circular coil of single turn has a resistance of 20 Ω. Which one of the following is the correct value for the resistance between the ends of any diameter of the coil?
Asked in: UPSC NDA 1 2017
1. 5 Ω
2. 10 Ω
3. 20 Ω
4. 40 Ω
Show answer
Answer: A. 5 Ω
Let the total resistance of the circular coil be R. The resistance of each semicircle is R/2. When these two semicircles are connected in parallel, the equivalent resistance (R_eq) is given by the formula for resistors in parallel: 1/R_eq = 1/R₁ + 1/R₂. Here, R₁ = R₂ = R/2. So, 1/R_eq = 1/(R/2) + 1/(R/2) = 2/R + 2/R = 4/R. Therefore, R_eq = R/4.
An electric wire of resistance 50 ohm is cut into five equal wires. These wires are then connected in parallel. What is the equivalent resistance of this combination?
Asked in: UPSC NDA 1 2022
1. 2 ohm
2. 10 ohm
3. 0.5 ohm
4. 5 ohm
Show answer
Answer: A. 2 ohm
The resistance of a uniform wire is directly proportional to its length. So, if the wire is cut into five equal parts, the resistance of each part is R' = R/5 = 50 ohm / 5 = 10 ohm. When n equal resistors (each of resistance r) are connected in parallel, the equivalent resistance (Req) is given by Req = r/n. In this case, r = 10 ohm and n = 5. Thus, Req = 10 ohm / 5 = 2 ohm.
What is the total resistance in the following circuit element ? [Image of a circuit diagram with three resistors labelled R. Two are in parallel, and this combination is in series with the third resistor.]
Asked in: UPSC NDA 1 2023
1. R/2
2. 3R
3. 3R/2
4. 2R/3
Show answer
Answer: C. 3R/2
- For resistors in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of individual resistances: $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + ...$ - For resistors in series, the equivalent resistance is the sum of individual resistances: $R_s = R_1 + R_2 + ...$

Superconductivity (basic concepts, critical temperature, applications)

Which of the following statements are incorrect? 1. LED lamps do not contain mercury 2. LED lights never fail abruptly, but there is a gradual decrease in the output 3. CFL lamps are more energy-efficient as compared to the LED lamps
Asked in: GPSC Prelims CSP 2 2019
1. 1 and 3 only
2. 2 and 3 only
3. 1 and 2 only
4. 1, 2 and 3
Show answer
Answer: B. 2 and 3 only
- Statement 1: LED lamps do not contain mercury. This is correct. Unlike CFLs which contain mercury, LEDs are considered more environmentally friendly in this regard. - Statement 2: LED lights never fail abruptly, but there is a gradual decrease in the output. This is incorrect. While LEDs typically degrade gradually (lumen depreciation), they can and sometimes do fail abruptly due to component failure (like the driver circuit). - Statement 3: CFL lamps are more energy-efficient as compared to the LED lamps. This is incorrect. LEDs are generally significantly more energy-efficient than CFLs.
Which cells use the sunlight to make electricity?
Asked in: GPSC Prelims CSP 2 2023
1. Photovoltaic cells
2. Photo reactive cells
3. Photo synthetic cells
4. Photolytic cells
Show answer
Answer: A. Photovoltaic cells
The photovoltaic effect is a process where a voltage or electric current is produced when a material is exposed to light. This effect is harnessed in solar panels to generate renewable electricity.
Which one of the following is the best shape of a solid metal rod to form the top end of a lightning conductor ?
Asked in: UPSC NDA 1 2024
1. Figure (a)
2. Figure (b)
3. Figure (c)
4. Figure (d)
Show answer
Answer: A. Figure (a)
A sharp point concentrates the electric field lines from a charged cloud. This high electric field at the tip promotes corona discharge, which is a continuous leakage of charge from the conductor into the air. This discharge helps to neutralize the surrounding air and reduce the potential difference between the cloud and the ground, thereby reducing the likelihood of a direct lightning strike to the protected structure.
Lightning conductors are used to protect building from lightning strikes. Which of the following statements is / are true about lightning conductors? 1. Lightning conductors create an electric field at its top so that lightning strikes it preferentially 2. Lightning conductors reduce the effect of the strike by uniformly distributing the charge (current) over the surface of the building 3. Lightning conductors take all charge (current) to deep down in the earth 4. Lightning conductors must be installed at a place taller than the building Select the correct answer using the code given below:
Asked in: UPSC NDA 2 2015
1. 1 and 2
2. 3 and 4 only
3. 1, 3 and 4
4. 4 only
Show answer
Answer: C. 1, 3 and 4
- Statement 1 is correct. The pointed tip of a lightning conductor concentrates electric charge, creating a strong electric field. This strong field can ionize the air, making the conductor the preferred point of strike if lightning occurs nearby. - Statement 2 is incorrect. Lightning conductors provide a direct, low-resistance path to the earth, diverting the lightning current away from the building's structure, not distributing it over the surface. - Statement 3 is correct. The conductor system is connected to grounding rods or plates buried deep in the earth, allowing the immense electrical charge from a lightning strike to dissipate safely into the ground. - Statement 4 is correct. For a lightning conductor to effectively intercept a strike, it must be the highest point of the structure it is protecting.
Two coils are placed close to each other. The mutual inductance of the pair of coils depends upon
Asked in: TNPSC CS Prelims 2017
1. the materials of wires of the coils
2. the currents in the two coils
3. the rates at which currents are changing in the two coils
4. relative position and orientation of the two coils
Show answer
Answer: D. relative position and orientation of the two coils
Mutual inductance is a measure of how much magnetic flux from one coil links with another coil. This linkage is determined by the geometric arrangement of the coils, including their shape, size, separation distance, and relative angle (orientation). It also depends on the magnetic properties (permeability) of the medium between the coils.
The principle of transformer is
Asked in: TNPSC CS Prelims 2024
1. Self induction
2. Mutual induction
3. Lenz law
4. Fleming's Right hand Rule
Show answer
Answer: B. Mutual induction
- A transformer transfers energy between circuits using magnetic coupling. - The key phenomenon is the induction of a voltage in one coil due to a changing magnetic flux produced by another coil. - Mutual induction is the process where a change in current in one coil induces an EMF in a neighboring coil.
Which of these connected in parallel in a circuit will produce undamped oscillations?
Asked in: TNPSC CS Prelims 2014
1. R, L (Resistance, Inductance)
2. R, C (Resistance, Capacitance)
3. C, L (Capacitance, Inductance)
4. R, L and C (Resistance, Inductance and Ca‐ pacitance)
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Answer: C. C, L (Capacitance, Inductance)
- Undamped oscillations in a circuit occur when there is no energy dissipation. - Resistance (R) in a circuit is a component that dissipates energy as heat (Joule heating). - A circuit containing only Capacitance (C) and Inductance (L) is an ideal resonant circuit where energy is continuously exchanged between the electric field in the capacitor and the magnetic field in the inductor. - In a purely LC circuit (assuming ideal components with zero resistance), there is no energy loss, leading to oscillations that would theoretically continue indefinitely (undamped). - Including resistance (R) in the circuit (RL, RC, or RLC circuits) introduces damping, causing the oscillations to decay over time.

Definition of Electric Current and current density

Drift velocity and microscopic origin of current

Electrical Power and Energy, Joule heating effect

Electromotive Force (EMF) and internal resistance of a cell

Factors affecting resistance (length, area, material, temperature)

Kirchhoff's Laws (Current Law and Voltage Law)

Ohm's Law, resistance, resistivity, and conductivity

Series and parallel combination of resistors

Superconductivity (basic concepts, critical temperature, applications)

11. Conductivity

Practice MCQs: Electric Current

Definition of Electric Current and current density

Drift velocity and microscopic origin of current

Electrical Power and Energy, Joule heating effect

Electromotive Force (EMF) and internal resistance of a cell

Factors affecting resistance (length, area, material, temperature)

Kirchhoff's Laws (Current Law and Voltage Law)

Ohm's Law, resistance, resistivity, and conductivity

Series and parallel combination of resistors

Superconductivity (basic concepts, critical temperature, applications)

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