Consider the circuit shown in figure: 1. Find the current through the battery a long time after the switch S is closed. 2. Suppose the switch is again opened at t = 0. What is the time constant of the discharging circuit? 3. Find the current through the inductor after one time constant.

1. 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=i_0 (e^ -t ) i_0 = Steady state current = Time constant = L R After a long time, t Now, Current in the inductor, i = i_0 (1 - e^0) = 0 Thus, the effect of inductance vanishes. i= E R_ net = R_1 _2 R_1+R_2 = (R_1+R_2 ) R_1R_2 2. When the switch is opened the resistors are in series. The time constant is given by = L R_net = L R_1+R_2 . 3. The inductor will discharge through resistors R_1 and R_2. The current through the inductor after one time constant is given by t= Current, i=i_0 (e^ -t ) Here, i_0= R_1+ R_2 i= R_1+ R_2 1 e

Consider the circuit shown in figure: 1. Find the current through…

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Question

Consider the circuit shown in figure:

Find the current through the battery a long time after the switch S is closed.
Suppose the switch is again opened at $t = 0$. What is the time constant of the discharging circuit?
Find the current through the inductor after one time constant.

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Answer

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

$\text{i}=\text{i}_0\Big(\text{e}^{\frac{-\text{t}}{\tau}}\Big)$
$i_0$ = Steady state current
$\tau=$ Time constant $=\frac{\text{L}}{\text{R}}$
After a long time, $\text{t}\rightarrow\infty$
Now,
Current in the inductor, $i = i_0 (1 - e^0) = 0$
Thus, the effect of inductance vanishes.
$\text{i}=\frac{\text{E}}{\text{R}_{\text{net}}}$
$=\frac{\in}{\frac{\text{R}_1\times\text{R}_2}{\text{R}_1+\text{R}_2}}=\frac{\in\big(\text{R}_1+\text{R}_2\big)}{\text{R}_1\text{R}_2}$

When the switch is opened the resistors are in series.

The time constant is given by
$\tau=\frac{\text{L}}{\text{R}_\text{net}}=\frac{\text{L}}{\text{R}_1+\text{R}_2}.$

The inductor will discharge through resistors $R_1$ and $R_2$.

The current through the inductor after one time constant is given by
$\text{t}=\tau$
$\therefore$ Current, $\text{i}=\text{i}_0\Big(\text{e}^{\frac{-\text{t}}{\tau}}\Big)$
Here,
$\text{i}_0=\frac{\in}{\text{R}_1+{\text{R}_2}}$
$\therefore \ \text{i}=\frac{\in}{\text{R}_1+{\text{R}_2}}\times\frac{1}{\text{e}}$