Question

A coil of inductance $L=2\mu \text{H}$ and resistance $R_1=1\Omega$ is connected to a source of constant emf $E=3\text{V}$. A resistance of $R_2=2\Omega$ is connected in parallel with the coil. Find the amount of heat generated in the coil after the switch is disconnected. The internal resistance of the source is negligible,

• A. $3\mu \text{J}$
• B. $9\mu \text{J}$
• C. $4.5\mu \text{J}$
• D. $30\mu \text{J}$

Answer 1

The correct option is A $3\mu \text{J}$

The equivalent circuit is,


In steady state the coil will behave like a short circuit.

$\therefore$ The equivalent circuit in steady state is,


$\therefore$ The current flowing through the coil in steady state is,

$i_1=\frac{3}{1}=3\text{A}$

On disconnecting the switch, the current decays through the resistance of $2\Omega$ and $1\Omega$, which is in series combination.

As power is given by, $P=Vi$.

Since, in series combination current flowing through both the resistance is same. Therefore, the power and thus the heat generated in the coils will be divided in the same ration in which voltage is divided in series combination of two resistance.

Energy stored in inductor in steady state:

$E=\frac{1}{2}L{i}_1^2$

$=\frac{1}{2}\times 2\times {10}^{-6}\times 3^2=9\times {10}^{-6}\text{J}$

$\therefore$ Heat generated in the coil is,

$E_1=\frac{1}{2}L{i}_1^2\left(\frac{R_1}{R_1+R_2}\right)$

$=9\times {10}^{-6}\times \left(\frac{1}{1+2}\right)$

$=3\times {10}^{-6}\text{J}=3\mu \text{J}$

Hence, $\left(A\right)$ is the correct answer.