Question

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.

Answer 1

(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 = `L/R`

After a long time, t → ∞.

Now,

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

Thus, the effect of inductance vanishes.

\[i = \frac{\epsilon}{R_{net}}\]

\[i = \frac{\epsilon}{\frac{R_1 \times R_2}{R_1 + R_2}} = \frac{\epsilon( R_1 + R_2 )}{R_1 R_2}\]

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

The time constant is given by

\[\tau = \frac{L}{R_{net}} = \frac{L}{R_1 + R_2}\]

(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,

\[i_0=\frac{\epsilon}{R_1 + R_2}\]

\[\therefore i=\frac{\epsilon}{R_1 + R_2} \times \frac{1}{e}\]