A current \(I=10 \mathrm{~A}\) is passed through the part of a circuit shown in the figure. What will be the…
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 ?
\(-7.5 \mathrm{~V}\)
\(3.5 \mathrm{~V}\)
\(-3.5 \mathrm{~V}\)
\(4 \mathrm{~V}\)
Solution
$\begin{aligned}
& e = L \frac{d I}{d t} = 5 \times 10^{-3} \times 10^{2} \\
\Rightarrow & e = 0.5 \text{~V}
\end{aligned}$
Applying Kirchhoff's voltage law between point $A$ and $B$.
$\begin{aligned}
& V_{AB} + 2 \times 10 - 12 - 0.5 = 0 \\
\Rightarrow & V_{AB} + 20 - 12.5 = 0 \\
\Rightarrow & V_{AB} + 7.5 = 0 \\
\Rightarrow & V_{AB} = -7.5 \text{~V}
\end{aligned}$
