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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 $\left(\mathrm{t}_{0}\right)$, the switch $\mathrm{S}_{1}$ is opened and $\mathrm{S}_{2}$ is closed. the behaviour of the current I as a function of time ' $\mathrm{t}$ ' is given by:
Options:
the switch $S_{1}$ is closed at time $t=0$ and the switch $S_{2}$ is kept open. At some later time $\left(\mathrm{t}_{0}\right)$, the switch $\mathrm{S}_{1}$ is opened and $\mathrm{S}_{2}$ is closed. the behaviour of the current I as a function of time ' $\mathrm{t}$ ' is given by:
- A
- B
- C
- D
Solution:
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Verified Answer
The correct answer is:
The current will grow for the time $\mathrm{t}=0$ to $\mathrm{t}=\mathrm{t}_{0}$ and after that decay of current takes place.
Growth and decay of current is of exponential nature
$\mathrm{i}=\mathrm{i}_0\left(1-\mathrm{e}^{-\mathrm{t} / \mathrm{\tau}}\right) \rightarrow \text { during growth }$
$\mathrm{i}=\mathrm{i}_{\max } \mathrm{e}^{-\mathrm{t} / \mathrm{\tau}} \rightarrow \text { during decay }$
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