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The network shown in the figure is part of complete circuit. If at a certain instant the current $i$ is $5 \mathrm{~A}$ and is decreasing at the rate of $10^3 \mathrm{~A} / \mathrm{s}$, then $\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}$ is
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The correct answer is:
15V
Consider the diagram below:
Given, $=5 \mathrm{~A}, \mathrm{R}=1 \Omega, \mathrm{L}=5 \mathrm{mH}, \frac{\mathrm{di}}{\mathrm{dt}}=10^3 \mathrm{~A} / \mathrm{s}$
$\begin{aligned}
& V_A-i R+E-L\left(\frac{d i}{d t}\right)=V_B \\
& \Rightarrow V_A-V_B=\left(i R+L\left(\frac{d i}{d t}\right)-E\right)
\end{aligned}$
$\begin{aligned}
& \mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=5 \mathrm{~A} \times 1 \Omega+5 \times 10^{-3} \mathrm{H} \times 10^{-3} \frac{\mathrm{A}}{\mathrm{s}}-15 \mathrm{~V} \\
& \mathrm{~V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}=-(5+5-15) \mathrm{V}=5 \mathrm{~V}
\end{aligned}$
Given, $=5 \mathrm{~A}, \mathrm{R}=1 \Omega, \mathrm{L}=5 \mathrm{mH}, \frac{\mathrm{di}}{\mathrm{dt}}=10^3 \mathrm{~A} / \mathrm{s}$
$\begin{aligned}
& V_A-i R+E-L\left(\frac{d i}{d t}\right)=V_B \\
& \Rightarrow V_A-V_B=\left(i R+L\left(\frac{d i}{d t}\right)-E\right)
\end{aligned}$
$\begin{aligned}
& \mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=5 \mathrm{~A} \times 1 \Omega+5 \times 10^{-3} \mathrm{H} \times 10^{-3} \frac{\mathrm{A}}{\mathrm{s}}-15 \mathrm{~V} \\
& \mathrm{~V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}=-(5+5-15) \mathrm{V}=5 \mathrm{~V}
\end{aligned}$
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