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In the given circuit, if $\frac{\mathrm{dI}}{\mathrm{dt}}=-1 \mathrm{~A} / \mathrm{s}$ then the value of $\left(V_A-V_B\right)$ at this instance will be
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The correct answer is:
$30 \mathrm{~V}$
Applying KVL in the given circuit from point $A$ to $\mathrm{B}$,
$$
\begin{array}{ll}
& \mathrm{V}_{\mathrm{AB}}-\mathrm{IR}-\mathrm{L} \frac{\mathrm{dI}}{\mathrm{dt}}-12=0 \\
& \mathrm{~V}_{\mathrm{AB}}-(2)(12)-6(-1)-12=0 \\
& \mathrm{~V}_{\mathrm{AB}}=24+12-6 \\
\therefore \quad & \mathrm{V}_{\mathrm{AB}}=30 \mathrm{~V}
\end{array}
$$
$$
\begin{array}{ll}
& \mathrm{V}_{\mathrm{AB}}-\mathrm{IR}-\mathrm{L} \frac{\mathrm{dI}}{\mathrm{dt}}-12=0 \\
& \mathrm{~V}_{\mathrm{AB}}-(2)(12)-6(-1)-12=0 \\
& \mathrm{~V}_{\mathrm{AB}}=24+12-6 \\
\therefore \quad & \mathrm{V}_{\mathrm{AB}}=30 \mathrm{~V}
\end{array}
$$
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