Consider the circuit shown in figure. (a) Find the current through the battery a long time after the switch S is closed. (b) Suppose the switch is again opened at t = 0. What is the time constant of the discharging circuit? (c) Find the current through the inductor after one time constant.

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Solution

(a) Because the switch is closed, the battery gets connected across the L‒R circuit.
The current in the L‒R circuit after t seconds after connecting the battery is given by
i = i₀ (1 − e^−t^/τ)
Here,
i₀= Steady state current
τ = Time constant = $\frac{L}{R}$ LR
After a long time, t $\to \infty$ .

Now,
Current in the inductor, i = i₀ (1 − e⁰) = 0

Thus, the effect of inductance vanishes.

$i = \frac{ε}{R_{n e t}} i = \frac{ε}{\frac{R_{1} \times R_{2}}{R_{1} + R_{2}}} = \frac{ε (R_{1} + R_{2})}{R_{1} R_{2}}$

(b) When the switch is opened, the resistance are in series.

The time constant is given by
$τ = \frac{L}{R_{n e t}} = \frac{L}{R_{1} + R_{2}}$

(c) The inductor will discharge through resistors R₁ and R₂.
The current through the inductor after one time constant is given by
t = τ
∴ Current, i = i₀ e^−τ^/τ
Here,
i₀ = $\frac{ε}{R_{1} + R_{2}}$
∴ i = $\frac{ε}{R_{1} + R_{2}} \times \frac{1}{e}$

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