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(A) Given: Current $I = 2 \ A$, Resistance $R = 7 \ \Omega$, $EMF$ $E = 4 \ V$, Inductance $L = 9 \ mH = 9 	imes 10^{-3} \ H$, Rate of change of curr

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(A) Given: Current $I = 2 \ A$, Resistance $R = 7 \ \Omega$, $EMF$ $E = 4 \ V$, Inductance $L = 9 \ mH = 9 	imes 10^{-3} \ H$, Rate of change of current $\frac{dI}{dt} = -10^3 \ A \ s^{-1}$ (since current is decreasing).Applying Kirchhoff's voltage law from $A$ to $B$:$V_A - I R - E - L \frac{dI}{dt} = V_B$$V_A - V_B = I R + E + L \frac{dI}{dt}$Substituting the values:$V_{AB} = (2 \ A)(7 \ \Omega) + 4 \ V + (9 	imes 10^{-3} \ H)(-10^3 \ A \ s^{-1})$$V_{AB} = 14 \ V + 4 \ V - 9 \ V$$V_{AB} = 18 \ V - 9 \ V = 9 \ V$Wait, let's re-evaluate the polarity of the inductor. The induced $EMF$ in the inductor is $\varepsilon = -L \frac{dI}{dt}$. Since current is decreasing, $\frac{dI}{dt} = -10^3 \ A \ s^{-1}$, so $\varepsilon = - (9 	imes 10^{-3})(-10^3) = +9 \ V$. The potential drop across the inductor is $V_L = L \frac{dI}{dt} = -9 \ V$. Path $A 	o B$: $V_A - I R + E - V_L = V_B$ (considering the battery polarity in the image: $A 	o$ resistor $	o$ negative terminal $	o$ positive terminal $	o$ inductor $	o B$).$V_A - (2)(7) + 4 - (9 	imes 10^{-3})(-10^3) = V_B$$V_A - 14 + 4 + 9 = V_B$$V_A - 1 = V_B \Rightarrow V_A - V_B = 1 \ V$.

$A$ branch of a circuit is shown in the figure. If the current is decreasing at the rate of $10^3 \ A \ s^{-1}$, then the potential difference between $A$ and $B$ is (in $V$)

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