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A branch of a circuit is shown in the figure. If current is decreasing at the rate of $10^3 \mathrm{As}^{-1}$, then the potential difference between $A$ and $B$ is
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$1\ V$
$\begin{aligned} & \text { Voltage of source, } \mathrm{V}=4 \mathrm{~V} \\ & \mathrm{~V}_{\mathrm{L}}=\mathrm{L} \cdot \frac{d l}{d t}=9 \times 10^{-3} \times 10^3 \mathrm{~V}=9 \mathrm{~V} \\ & \mathrm{~V}_{\mathrm{AB}}-\mathrm{V}_{\mathrm{R}}+\mathrm{V}+\mathrm{V}_{\mathrm{L}}=0 \\ & \text { or, } \mathrm{V}_{\mathrm{AB}}-14+4+9=0 \\ & \Rightarrow \quad \mathrm{V}_{\mathrm{AB}}-1=0 \\ & \therefore \quad \mathrm{V}_{\mathrm{AB}}=1 \mathrm{~V}\end{aligned}$
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