This question is identical to ID 80853. The problem asks for the line charging current of a transmission line when the load is disconnected. This current is the flow of charge due to the line's capacitance. The generalized ABCD parameters of the line are given.The relation between sending-end ($S$) and receiving-end ($R$) quantities for a transmission line is: $V_S = A V_R + B I_R$ $I_S = C V_R + D I_R$Given that the load is disconnected, the receiving-end current $I_R = 0$. The equations become: $V_S = A V_R$ $I_S = C V_R$We need to find $I_S$, which represents the per-phase sending-end current (line charging current) at no load. We are given: $A = 90^{\circ} \text{ S}/\text{phase}$ Sending-end voltage $V_{SE} = 230 \text{ kV}$ (line-to- line).First, calculate the per-phase sending-end voltage: $V_{phS} = \frac{230 \text{ kV}} {\sqrt{3}}$. Let's assume a phase angle of $0^{\circ}$ for $V_{phS}$. $V_{phS} = \frac{230}{\sqrt{3}} \angle 0^{\circ} \text{ kV} $.Now, find the receiving-end phase voltage $V_R$ from $V_S = A V_R$: $V_R = \frac{V_S} {A} = \frac{(230/\sqrt{3}) \angle 0^{\circ} \frac{230}{\sqrt{3} \times 0.938} \angle (0 - \angle -1.2^{\circ} \text{ kV}$.Next, calculate the sending-end current $I_S$ using $I_S = C V_R$: $I_S = (0.001 \angle 90^{\circ} \text{ S}) \times ing Siemens (S) by kilovolts (kV) gives kilo‐ (90 - 1.2)^{\circ} \text{ kA}$ $I_S = 0.14157 \angle 88.8^{\circ} \text{ kA}$.Converting kA to A: $0.14157 \text{ kA} = 141.57 \text{ A}$. The magnitude of the line charging current per phase is approximately 141.5 A.