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(C) We apply Kirchhoff's Voltage Law while moving from point $A$ to point $B$ along the path of the current.The potential at $A$ is $V_A$. As we pass

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$15$

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(C) We apply Kirchhoff's Voltage Law while moving from point $A$ to point $B$ along the path of the current.The potential at $A$ is $V_A$. As we pass through the resistor $R = 1 \ \Omega$,the potential drops by $iR = 5 \ A 	imes 1 \ \Omega = 5 \ V$.Next,we pass through the battery of $15 \ V$ from the positive terminal to the negative terminal,which results in a potential drop of $15 \ V$.Finally,we pass through the inductor $L = 5 \ mH = 5 	imes 10^{-3} \ H$. The induced emf across the inductor is $\varepsilon = -L \frac{di}{dt}$. Since the current is decreasing,$\frac{di}{dt} = -10^3 \ A/s$. Thus,the potential change across the inductor is $-L \frac{di}{dt} = -(5 	imes 10^{-3} \ H) 	imes (-10^3 \ A/s) = +5 \ V$.The equation for the potential at $B$ is:$V_A - iR - 15 - L \frac{di}{dt} = V_B$$V_A - (5 	imes 1) - 15 - (5 	imes 10^{-3} 	imes -10^3) = V_B$$V_A - 5 - 15 + 5 = V_B$$V_A - 15 = V_B$$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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