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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.

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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. $i=\frac{\epsilon }{{R}_{net}}\phantom{\rule{0ex}{0ex}}i=\frac{\epsilon }{\frac{{R}_{1}×{R}_{2}}{{R}_{1}+{R}_{2}}}=\frac{\epsilon \left({R}_{1}+{R}_{2}\right)}{{R}_{1}{R}_{2}}$ (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, i0 = $\frac{\epsilon }{{R}_{1}+{R}_{2}}$ ∴ i = $\frac{\epsilon }{{R}_{1}+{R}_{2}}×\frac{1}{e}$

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