Question

An average emf of $20\text{V}$ is induced in an inductor when the current in it is changed from $2.5\text{A}$ in one direction to $1.5\text{A}$ in the opposite direction in $0.1\text{s}.$ Find the self-inductance of the inductor.

• A. $1\text{H}$
• B. $2\text{H}$
• C. $\frac{1}{2}\text{H}$
• D. $\frac{1}{4}\text{H}$

Answer 1

The correct option is C $\frac{1}{2}\text{H}$

Given, $E=20\text{V};\Delta t=0.1\text{s}$

The initial current through the solenoid is, $i_1=2.5\text{A}$

Final current through the solenoid is, $i_2=-1.5\text{A}$

Here, the negative sign represents that the direction of current is reversed.

The induced emf in the coil is,

$E=-L\frac{\Delta i}{\Delta t}$

$\Rightarrow 20=\frac{-L\times (-1.5-2.5)}{0.1}$

$\Rightarrow L=\frac{20}{40}=\frac{1}{2}\text{H}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.