The current in LR circuit if reduced to half. What will be the energy stored in it?
The current in LR circuit if reduced to half. What will be the energy stored in it?
- 4 times
- 2 times
- half times
- $\left(\frac{1}{4}\right)^{\text {th }}$ times
Solution
Energy stored in LR circuit is
$\begin{aligned}
\mathrm{E} & =\frac{1}{2} \mathrm{LI}^2 ...(i)\\
\mathrm{I}^{\prime} & =\frac{\mathrm{I}}{2} \\
\mathrm{E}^{\prime} & =\frac{1}{2} \mathrm{~L} \times\left(\frac{\mathrm{I}}{2}\right)^2 \\
& =\frac{1}{4} \times \frac{1}{2} \mathrm{LI}^2 \\
\mathrm{E}^{\prime} & =\frac{1}{4} \times \mathrm{E}
\end{aligned}$
...(given)
...[From(i)]
Asked in: MHT CET 2024 (10 May Shift 1)
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