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Electromagnetic Induction — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsElectromagnetic Induction5 Marks
Question
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.
✓

Answer

  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
$\text{i}=\text{i}_0\Big(\text{e}^{\frac{-\text{t}}{\tau}}\Big)$
$i_0=$ 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.
$\text{i}=\frac{\text{E}}{\text{R}_{\text{net}}}$
$=\frac{\in}{\frac{\text{R}_1\times\text{R}_2}{\text{R}_1+\text{R}_2}}=\frac{\in\big(\text{R}_1+\text{R}_2\big)}{\text{R}_1\text{R}_2}$
  1. When the switch is opened the resistors are in series.
The time constant is given by
$\tau=\frac{\text{L}}{\text{R}_\text{net}}=\frac{\text{L}}{\text{R}_1+\text{R}_2}.$
  1. 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,
$\text{i}_0=\frac{\in}{\text{R}_1+{\text{R}_2}}$
$\therefore \ \text{i}=\frac{\in}{\text{R}_1+{\text{R}_2}}\times\frac{1}{\text{e}}$

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