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Byju's Answer

Standard XII

Physics

Wheatstone Bridge

In the circui...

Question

In the circuit shown switch $S$ is closed at $t = 0$ . Let $i_{1}$ and $i_{2}$ be the currents at any finite time $t$ , the ratio $i_{1} / i_{2}$ .

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Solution

We know that the current in $R C$ circuit at any time $t$ is $i = \frac{V_{0}}{R} e^{- t / R C}$

Here, $i_{1} = \frac{V}{2 R} e^{- t / (6 R C)} and i_{2} = \frac{V}{R} e^{- t / R C}$

$∴ \frac{i_{1}}{i_{2}} = \frac{1}{2} e^{- t / R C (1 - 1 / 6)} = \frac{1}{2} e^{5 t / 6 R C}$

Thus, we can see that this ratio is increasing with time.

Hence, option (b) is the correct answer.

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In the circuit shown switch S is closed at t=0. Let i1 and i2 be the currents at any finite time t, the ratio i1/i2.

We know that the current in RC circuit at any time t is i=V0Re−t/RC Here, i1=V2Re−t/(6RC) and i2=VRe−t/RC ∴i1i2=12e−t/RC(1−1/6)=12e5t/6RC Thus, we can see that this ratio is increasing with time. Hence, option (b) is the correct answer.

In the circuit shown switch $S$ is closed at $t = 0$ . Let $i_{1}$ and $i_{2}$ be the currents at any finite time $t$ , the ratio $i_{1} / i_{2}$ .