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Question: An inductor of inductance L = 400mH and resistors of resistances \({R_1} = 2\Omega \,\& \,2\Omega \)...

An inductor of inductance L = 400mH and resistors of resistances R1=2Ω & 2Ω{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.

Explanation

Solution

Hint
Use the concept of potential drop and the function of current with time i.e. I2=I∘(1−e−Tτ){I_2} = {I_ \circ }(1 - {e^{\dfrac{{ - T}}{\tau }}})
And the use Ohm’s law to find out the values of I∘{I_ \circ }, τ\tau and then use the formula of potential drop L=E−I2R2L = 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,
⇒I1=ER=124=3A\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,
⇒I2=I∘(1−e−Tτ)\Rightarrow {I_2} = {I_ \circ }(1 - {e^{\dfrac{{ - T}}{\tau }}})
Where I∘{I_ \circ } is the current steady state.
At steady state, the potential difference across an inductor is 0. Hence,
⇒I∘=ER2=122=6A\Rightarrow {I_ \circ } = \dfrac{E}{{{R_2}}} = \dfrac{{12}}{2} = 6A
Also, τ=LR2=400×10−32=0.2\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,
⇒I2=I∘(1−e−Tτ)\Rightarrow {I_2} = {I_ \circ }(1 - {e^{\dfrac{{ - T}}{\tau }}})
⇒I2=6(1−e−t0.2)\Rightarrow {I_2} = 6(1 - {e^{\dfrac{{ - t}}{{0.2}}}})
⇒I2=6(1−e−5t)\Rightarrow {I_2} = 6(1 - {e^{ - 5t}}) (putting the given values so obtained)
Now, potential drop across L is given by the formula,
⇒L=E−I2R2\Rightarrow L = E - {I_2}{R_2}
Putting the values from the question and the above steps we have,
⇒L=12−2×6(1−e−5t)\Rightarrow L = 12 - 2 \times 6(1 - {e^{ - 5t}})
⇒L=12e−5t\Rightarrow L = 12{e^{ - 5t}}
Hence, the equation of the inductor is given by L=12e−5tL = 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.