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
