Question

Consider the circuit shown in the figure. The sliding contact is being pulled towards the right so that the resistance in the circuit is increasing. Its value at the instant shown is $12\Omega$. The current flowing in the circuit is $i.$ Then :

• A. $i<0.5\text{A}$
• B. $i>0.5\text{A}$
• C. $i=0.5\text{A}$
• D. None of the above

Answer 1

The correct option is B $i>0.5\text{A}$

As the sliding contact is being pulled, the current in the circuit changes.

An induced emf, $E=-L\frac{di}{dt}$ is produced across the inductor.

The net emf in the circuits is,

$E_{net}=6\text{V}-L\frac{di}{dt}$

And hence the current is,

$i=\frac{E_{net}}{R}=\frac{6-L\frac{di}{dt}}{12}...\left(i\right)\left[\text{At the instant shown}\right]$


Now the resistance in the circuit is increasing, so the current is decreases and so $\frac{di}{dt}$ is negative.

Thus, the numerator of $\left(i\right)$ is more than $6$ and hence $i$ is greater than $\frac{6}{12}=0.50\text{A}$.

Thus, option $\left(B\right)$ is the correct answer.