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(C) The current $i = 5 \, A$ is flowing from $A$ to $B$ and is decreasing at a rate of $\frac{di}{dt} = -10^3 \, A/s$.The potential drop across the re

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$15$

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(C) The current $i = 5 \, A$ is flowing from $A$ to $B$ and is decreasing at a rate of $\frac{di}{dt} = -10^3 \, A/s$.The potential drop across the resistor is $V_R = iR = 5 	imes 1 = 5 \, V$.The potential difference across the battery is $15 \, V$ (from positive to negative terminal,it is a drop).The induced $emf$ across the inductor is $\epsilon = -L \frac{di}{dt} = -(5 	imes 10^{-3} \, H) 	imes (-10^3 \, A/s) = 5 \, V$.Applying Kirchhoff's voltage law from $A$ to $B$:$V_A - iR - 15 - L \frac{di}{dt} = V_B$$V_A - (5 	imes 1) - 15 + 5 = V_B$$V_A - 5 - 15 + 5 = V_B$$V_A - V_B = 15 \, V$.

The network shown in the figure is a part of a complete circuit. If at a certain instant the current $i$ is $5 \, A$ and is decreasing at the rate of $10^3 \, A/s$,then $V_A - V_B$ is....... $V$.

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