Question

An inductor of inductance L = 400mH and resistors of resistances ${R_1} = 2\Omega \,\& \,2\Omega $ are connected to a battery of emf 12V as shown in the figure. The internal resistance of the battery is negligible. The switch S is closed at t = 0. What is the potential drop across L as a function of time.

Answer 1

Hint
Use the concept of potential drop and the function of current with time i.e. ${I_2} = {I_ \circ }(1 - {e^{\dfrac{{ - T}}{\tau }}})$
And the use Ohm’s law to find out the values of ${I_ \circ }$, $\tau $and then use the formula of potential drop $L = E - {I_2}{R_2}$ ( across inductor ) to solve the problem.

Complete step by step answer
In an inductor, when the switch is turned on, the potential drop across it = 0. Since the current does not change with time hence it acts as an open circuit.
Therefore,
$\Rightarrow {I_1} = \dfrac{E}{R} = \dfrac{{12}}{4} = 3A$
Now let’s consider after some time t. Now the inductor stores energy as time passes by.
Its function is given by the following formula,
$\Rightarrow {I_2} = {I_ \circ }(1 - {e^{\dfrac{{ - T}}{\tau }}})$
Where ${I_ \circ }$ is the current steady state.
At steady state, the potential difference across an inductor is 0. Hence,
$\Rightarrow {I_ \circ } = \dfrac{E}{{{R_2}}} = \dfrac{{12}}{2} = 6A$
Also, $\tau = \dfrac{L}{{{R_2}}} = \dfrac{{400 \times {{10}^{ - 3}}}}{2} = 0.2$
Hence, the equation of current in the inductor as a function of time becomes,
$\Rightarrow {I_2} = {I_ \circ }(1 - {e^{\dfrac{{ - T}}{\tau }}}) $
$\Rightarrow {I_2} = 6(1 - {e^{\dfrac{{ - t}}{{0.2}}}}) $
$\Rightarrow {I_2} = 6(1 - {e^{ - 5t}}) $ (putting the given values so obtained)
Now, potential drop across L is given by the formula,
$\Rightarrow L = E - {I_2}{R_2}$
Putting the values from the question and the above steps we have,
$\Rightarrow L = 12 - 2 \times 6(1 - {e^{ - 5t}}) $
$\Rightarrow L = 12{e^{ - 5t}} $
Hence, the equation of the inductor is given by $L = 12{e^{ - 5t}}$.

Note
Properties of an inductor are useful in solving this kind of problem. The properties are-
i) At t = 0, the current does not flow through the inductor. Hence it acts as an open circuit.
ii) At t = t when the inductor has reached steady state, the rate of change of current with voltage is 0. Hence, the inductor acts as a short circuit with zero potential difference across it.