Electromagnetic Induction and Alternating Current 4 Question 4
####4. In the circuit shown,
The switch $S_{1}$ is closed at time $t=0$ and the switch $S_{2}$ is kept open. At some later time $\left(t_{0}\right)$, the switch $S_{1}$ is opened and $S_{2}$ is closed. The behaviour of the current $I$ as a function of time ’ $t$ ’ is given by
(Main 2019, 11 Jan I)
(a)
(b)
(c)
(d)
Show Answer
Answer:
Correct Answer: 4. (b)
Solution:
- Initially in the given $R L$ circuit with a source, when $S_{1}$ is closed and $S_{2}$ is open at $t \leq t_{0}$.
$$ I_{1}=\frac{V}{R} \quad 1-\exp \frac{-R}{L} t $$
In this case, inductor $L$ is charging.
When switch $S_{2}$ is closed and $S_{1}$ is open (after $t>t_{0}$ ), the inductor will be discharged through resistor.
In this case $\left(t>t_{0}\right)$,
$$ I_{2}=\frac{V}{R} \exp -\frac{R}{L}\left(t-t_{0}\right) $$
Thus, the variation of $I$ with $t$ approximately is shown below
