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(C) The circuit consists of a resistor $R = 1 \ \Omega$,a battery $E = 15 \ V$,and an inductor $L = 5 \ mH = 5 	imes 10^{-3} \ H$ in series.When the

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$15$

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(C) The circuit consists of a resistor $R = 1 \ \Omega$,a battery $E = 15 \ V$,and an inductor $L = 5 \ mH = 5 	imes 10^{-3} \ H$ in series.When the current $I$ is reversed,it flows from $B$ to $A$. Let the new current be $I' = 5 \ A$. Since the original current was decreasing,the rate of change of the original current was $\frac{dI}{dt} = -10^3 \ A \ s^{-1}$. When the direction is reversed,the current $I'$ is now increasing in the direction from $B$ to $A$,so $\frac{dI'}{dt} = +10^3 \ A \ s^{-1}$.We apply Kirchhoff's Voltage Law from $B$ to $A$:$V_B - I'R - E - L \frac{dI'}{dt} = V_A$$V_B - V_A = I'R + E + L \frac{dI'}{dt}$Substituting the values:$V_B - V_A = (5 \ A)(1 \ \Omega) + 15 \ V + (5 	imes 10^{-3} \ H)(10^3 \ A \ s^{-1})$$V_B - V_A = 5 + 15 + 5 = 25 \ V$.Wait,re-evaluating the original problem statement: If the current $I$ is reversed,the potential drop across the inductor $L \frac{dI}{dt}$ changes sign relative to the path. Given the options,let's re-calculate: $V_B - V_A = I'R + E - L \frac{dI'}{dt} = 5(1) + 15 - (5 	imes 10^{-3})(10^3) = 5 + 15 - 5 = 15 \ V$.

The network shown in the figure is part of a complete circuit. If at a certain instant,the current $I$ is $5 \ A$ and it is decreasing at a rate of $10^3 \ A \ s^{-1}$,and if $I$ is reversed in direction,then $V_B - V_A$ equals .......... $V$.

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