- Ohm's Law (V = IR) relates the voltage (V) across a component or circuit, the current (I) flowing through it, and its resistance (R). - For a complete circuit with a voltage source and an equivalent resistance R_eq, the total voltage V supplied by the source is related to the total current I flowing out of the source by V = I R_eq. - The question provides possible pairs of total voltage (V) and total current (I) for an electric circuit. We can calculate the implied equivalent resistance (R_eq = V/I) for each pair: - A) V = 900 V, I = 18 A => R_eq = 900/18 = 50 Ω 54.55 Ω - C) V = 600 V, I = 1 A => R_eq = 600/1 = 600 Ω - D) V = 300 V, I = 2 A => R_eq = 300/2 = 150 Ω - The circuit diagram would define the value of R_eq based on the configuration and values of its components (resistors, etc.). Without the diagram, we must assume that one of these calculated R_eq values corresponds to the actual circuit's equivalent resistance. Assuming option D is correct, the circuit must have an equivalent resistance of 150 Ω when connected to a 300 V source or when a 2 A current flows through it resulting in a 300 V drop.
- Comparing the voltage across the bulb in each case: (a) V, (b) 2V, (c) V, (d) V/2. - Since the power dissipated (and thus brightness) is proportional to V², the bulb will glow brightest when the voltage across it is highest, which is in case (b) with two batteries in series (2V). 242. For an electric circuit given below, the correct combination of voltage (V) and current (I) is
Asked in UPSC NDA 2 2024 · Topic: Electric Current (Ohm's Law, resistance, resistivity, and conductivity)
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