In the circuit shown, switch S is closed at t = 0. Let i1 and i2 be the currents at any finite time t, then the ratio i1i2.

From the given situation, i1 = V2R e−t6RC and i2 = VR e−tRCi1i2 = 12 e5t6RC, which increases with time

In the circuit shown, switch S is closed at t = 0. Let i1 and i2 be the currents at any finite time t, then the ratio i1i2.

From the given situation, i1 = V2R e−t6RC and i2 = VR e−tRCi1i2 = 12 e5t6RC, which increases with time

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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, then the ratio $\frac{i_{1}}{i_{2}} .$

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is constant

Increases with time

decreases with time

first increases and then decreases

answer is B.

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Detailed Solution

From the given situation, $i_{1} = \frac{V}{2 R} (e^{\frac{- t}{6 RC}}) and i_{2} = \frac{V}{R} e^{\frac{- t}{RC}}$

$\frac{i_{1}}{i_{2}} = \frac{1}{2} {(e)}^{\frac{5 t}{6 RC}},$ which increases with time

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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, then the ratio i1i2.

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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, then the ratio $\frac{i_{1}}{i_{2}} .$