Question 238: Electromagnetic Induction

Question

The current in LR circuit if reduced to half What will be the energy stored in it?

Accepted Answer

In an LR (inductor-resistor) circuit, the energy stored in the inductor is given by the expression:
$$ E = \frac{1}{2} L I^2 $$
where:
$ E $ is the energy stored in the inductor,
$ L $ is the inductance, and
$ I $ is the current through the inductor.
If the current ($ I $) is reduced to half of its initial value, the new current becomes $ \frac{I}{2} $.
Substituting the new current into the expression for energy:
$$ E_{	ext{new}} = \frac{1}{2} L \left(\frac{I}{2}ight)^2 = \frac{1}{2} L \cdot \frac{I^2}{4} = \frac{1}{8} L I^2 $$
Comparing this with the original energy $ E = \frac{1}{2} L I^2 $, the new energy $ E_{	ext{new}} $ is:
$$ E_{	ext{new}} = \frac{1}{4} \cdot \left(\frac{1}{2} L I^2ight) = \frac{1}{4} E $$
Therefore, the energy stored in the inductor is reduced to $\frac{1}{4}$ times of its original value.
Hence, the correct option is:
Option D: $\left(\frac{1}{4}ight)^{	ext{th}}$ times

Answer

4 times

Answer

2 times

Answer

half times

Answer

$\left(\frac{1}{4}ight)^{	ext {th }}$ times

Electromagnetic Induction

The current in LR circuit if reduced to half What will be the energy stored in it?

Choose the correct answer: