Question

A $12\text{V}$ battery connected to a $6\Omega$, $10\text{H}$ coil through a switch drives a constant current in the circuit. The switch is suddenly opened. Assuming that it took $1\text{ms}$ to open the switch, The average emf induced across the coil is-

• A. ${10}^4\text{V}$
• B. $1.5\times {10}^4\text{V}$
• C. $2\times {10}^4\text{V}$
• D. $5\times {10}^4\text{V}$

Answer 1

The correct option is C $2\times {10}^4\text{V}$

The steady state current flowing in the circuit until opening the switch is,

$i=\frac{E}{R}=\frac{12}{6}=2\text{A}$

The final current becomes zero, when the switch is opened.

Thus, change in current is,

$\frac{di}{dt}=-\frac{2}{1\times {10}^{-3}}=-2\times {10}^3{\text{As}}^{-1}$

$-ve$ sign indicates that $\frac{di}{dt}$ is decreasing.

The induced emf is,

$E=-L\frac{di}{dt}$

$E=-\left(10\right)\times (-2\times {10}^3)=2\times {10}^4\text{V}$

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

Note: Such a high emf may cause sparks across the open switch.