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(B) Given: Current $I = 2 \ A$,rate of change of current $\frac{dI}{dt} = -10^2 \ A \ s^{-1}$ (since it is decreasing),resistance $R = 2 \ \Omega$,$EM

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$8.5$

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(B) Given: Current $I = 2 \ A$,rate of change of current $\frac{dI}{dt} = -10^2 \ A \ s^{-1}$ (since it is decreasing),resistance $R = 2 \ \Omega$,$EMF$ $\varepsilon = 12 \ V$,and inductance $L = 5 \ mH = 5 	imes 10^{-3} \ H$.Applying Kirchhoff's voltage law from point $A$ to $B$:$V_A - IR + \varepsilon - L \frac{dI}{dt} = V_B$$V_B - V_A = -IR + \varepsilon - L \frac{dI}{dt}$Substituting the values:$V_B - V_A = -(2 	imes 2) + 12 - (5 	imes 10^{-3} 	imes (-10^2))$$V_B - V_A = -4 + 12 + 0.5$$V_B - V_A = 8.5 \ V$

The network shown in the figure is a part of a circuit. (The battery has negligible internal resistance). At a certain instant,the current is $2 \ A$ and it is decreasing at the rate of $10^2 \ A \ s^{-1}$. What is the potential difference between the points $B$ and $A$ (in $V$)?

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