MHT CET · Physics · Alternating Current
The current in LR circuit if reduced to half. What will be the energy stored in it?
- A 4 times
- B 2 times
- C half times
- D \(\left(\frac{1}{4}\right)^{\text {th }}\) times
Answer & Solution
Correct Answer
(D) \(\left(\frac{1}{4}\right)^{\text {th }}\) times
Step-by-step Solution
Detailed explanation
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)]
\(\begin{aligned}
\mathrm{E} & =\frac{1}{2} \mathrm{LI}^2 ...(i)\\
\mathrm{I}^{\prime} & =\frac{\mathrm{I}}{2} \\
\mathrm{E}^{\prime} & =\frac{1}{2} \mathrm{~L} \times\left(\frac{\mathrm{I}}{2}\right)^2 \\
& =\frac{1}{4} \times \frac{1}{2} \mathrm{LI}^2 \\
\mathrm{E}^{\prime} & =\frac{1}{4} \times \mathrm{E}
\end{aligned}\)
...(given)
...[From(i)]
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