An inductor of inductance L and resistors of resistances R and 2R are connected to a battery of emf E. The internal resistance of battery is negligible. The switch S is closed at t = 0. The current through resistor R as a function of time is

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$\frac{E}{2 R} e^{- \frac{2 R t}{3 L}}$

$\frac{E}{3 R} e^{- \frac{2 R t}{3 L}}$

$\frac{E}{2 R} [1 - e^{- \frac{2 R t}{3 L}}]$

$\frac{E}{3 R} [1 - e^{- \frac{2 R t}{3 L}}]$

answer is B.

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Detailed Solution

Current Through Resistor R as a Function of Time in RL Circuit

Consider an inductor of inductance L and resistors of resistances R and 2R connected to a battery of emf E. The battery's internal resistance is negligible. The switch S is closed at time t = 0.

The current through the resistor R as a function of time t is given by:

i_R(t) = (E / R) × (1 - e^{-(2R t) / L})

Explanation:

At the moment the switch is closed (t = 0), the inductor opposes any sudden change in current, so almost no current flows through R.
As time progresses, the current increases exponentially, eventually reaching a steady value as the inductor behaves like a short circuit.
The exponential term e^{-(2R t) / L} describes how quickly the current builds up through the inductor-resistor branch.

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