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Standard XII

Physics

RLC Circuit

In the given ...

Question

In the given circuit, let $i_{1}$ be the current drawn from battery at time $t = 0$ i.e. when switch is just closed and $i_{2}$ be steady state current at $t = \infty$ , then the ratio $\frac{i_{1}}{i_{2}}$ is

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Solution

At $t = 0$ , inductor behave as open circuit so $i_{1} = \frac{10}{10} = 1 A$
At $t = \infty$ , inductor behaves as a short circuit, so
$i_{2} = \frac{10}{8} = \frac{5}{4} A$
Hence, $\frac{i_{1}}{i_{2}} = \frac{1}{5 / 4} = \frac{4}{5} = 0.8$

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In the given circuit, let i1 be the current drawn from battery at time t=0 i.e. when switch is just closed and i2 be steady state current at t=∞, then the ratio i1i2 is

At t=0, inductor behave as open circuit so i1=1010=1 A At t=∞, inductor behaves as a short circuit, so i2=108=54 A Hence, i1i2=15/4=45=0.8

In the given circuit, let $i_{1}$ be the current drawn from battery at time $t = 0$ i.e. when switch is just closed and $i_{2}$ be steady state current at $t = \infty$ , then the ratio $\frac{i_{1}}{i_{2}}$ is

At $t = 0$ , inductor behave as open circuit so $i_{1} = \frac{10}{10} = 1 A$
At $t = \infty$ , inductor behaves as a short circuit, so
$i_{2} = \frac{10}{8} = \frac{5}{4} A$
Hence, $\frac{i_{1}}{i_{2}} = \frac{1}{5 / 4} = \frac{4}{5} = 0.8$