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A current \(I=10 \mathrm{~A}\) is passed through the part of a circuit shown in the figure. What will be the potential difference between \(A\) and \(B\) when \(I\) is decreased at constant rate of \(10^2 \mathrm{As}^{-1}\) at the beginning ?
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Verified Answer
The correct answer is:
\(-7.5 \mathrm{~V}\)
The given circuit diagram is shown below,
Given, \(I=10 \mathrm{~A}\)
\(\therefore \quad \frac{d I}{d t}=10^2 \mathrm{As}^{-1}\)
Induced emf across inductor coil,
\(\begin{array}{lll}
& e =L \frac{d I}{d t}=5 \times 10^{-3} \times 10^2 \\
\Rightarrow & e =0.5 \mathrm{~V}
\end{array}\)
Applying Kirchhoff's voltage law between point \(A\) and \(B\).
\(\begin{aligned}
& V_{A B}+2 \times 10-12-0.5 =0 \\
\Rightarrow & V_{A B}+20-12.5 =0 \\
\Rightarrow & V_{A B}+7.5 =0 \\
\Rightarrow & V_{A B} =-7.5 \mathrm{~V}
\end{aligned}\)
Given, \(I=10 \mathrm{~A}\)
\(\therefore \quad \frac{d I}{d t}=10^2 \mathrm{As}^{-1}\)
Induced emf across inductor coil,
\(\begin{array}{lll}
& e =L \frac{d I}{d t}=5 \times 10^{-3} \times 10^2 \\
\Rightarrow & e =0.5 \mathrm{~V}
\end{array}\)
Applying Kirchhoff's voltage law between point \(A\) and \(B\).
\(\begin{aligned}
& V_{A B}+2 \times 10-12-0.5 =0 \\
\Rightarrow & V_{A B}+20-12.5 =0 \\
\Rightarrow & V_{A B}+7.5 =0 \\
\Rightarrow & V_{A B} =-7.5 \mathrm{~V}
\end{aligned}\)
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