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Question: In the circuit shown current through inductor L as a function of time t is (if switch is closed at t...

In the circuit shown current through inductor L as a function of time t is (if switch is closed at t = 0)

A

V05R(1−e−5Rt2L)\frac{V_0}{5R}(1-e^{\frac{-5Rt}{2L}})

B

3V05R(1−e−5RtL)\frac{3V_0}{5R}(1-e^{\frac{-5Rt}{L}})

C

3V05R(1−e−5Rt2L)\frac{3V_0}{5R}(1-e^{\frac{-5Rt}{2L}})

D

V05R(1−e−5RtL)\frac{V_0}{5R}(1-e^{\frac{-5Rt}{L}})

Answer

(1) V05R(1−e−5Rt2L)\frac{V_0}{5R}(1-e^{\frac{-5Rt}{2L}})

Explanation

Solution

To determine the current through the inductor L as a function of time t, we need to find the steady-state current (I0I_0) through the inductor and the time constant (τ\tau) of the RL circuit. The general form for current growth in an RL circuit is I(t)=I0(1−e−t/τ)I(t) = I_0 (1 - e^{-t/\tau}).

1. Calculate the steady-state current (I0I_0) through the inductor:

At steady state (t → ∞), an inductor acts as a short circuit (its resistance becomes zero). The circuit simplifies as follows:

  • The resistor R (in series with V0V_0) is in series with the parallel combination of the other resistor R and the 2R resistor (which is in series with the shorted inductor).
  • The equivalent resistance of the parallel combination of R and 2R is: Rparallel=R×2RR+2R=2R23R=2R3R_{parallel} = \frac{R \times 2R}{R + 2R} = \frac{2R^2}{3R} = \frac{2R}{3}
  • The total equivalent resistance of the circuit at steady state is: Rtotal=R+Rparallel=R+2R3=5R3R_{total} = R + R_{parallel} = R + \frac{2R}{3} = \frac{5R}{3}
  • The total current drawn from the source is: Isource=V0Rtotal=V05R/3=3V05RI_{source} = \frac{V_0}{R_{total}} = \frac{V_0}{5R/3} = \frac{3V_0}{5R}
  • This source current IsourceI_{source} splits between the parallel R and 2R branches. The current through the inductor branch (containing 2R) is I0I_0. Using the current divider rule: I0=Isource×Rother_branchRother_branch+Rinductor_branchI_0 = I_{source} \times \frac{R_{other\_branch}}{R_{other\_branch} + R_{inductor\_branch}} I0=3V05R×RR+2R=3V05R×R3R=13×3V05R=V05RI_0 = \frac{3V_0}{5R} \times \frac{R}{R + 2R} = \frac{3V_0}{5R} \times \frac{R}{3R} = \frac{1}{3} \times \frac{3V_0}{5R} = \frac{V_0}{5R}

2. Calculate the Thevenin equivalent resistance (RthR_{th}) seen by the inductor:

To find RthR_{th}, we deactivate the independent voltage source (replace it with a short circuit) and look into the terminals where the inductor is connected.

  • When the voltage source V0V_0 is shorted, the resistor R (originally in series with V0V_0) becomes parallel to the other resistor R.
  • The equivalent resistance of this parallel combination is: Rparallel′=R×RR+R=R22R=R2R'_{parallel} = \frac{R \times R}{R + R} = \frac{R^2}{2R} = \frac{R}{2}
  • This Rparallel′R'_{parallel} is in series with the 2R resistor (which is directly connected to the inductor terminals).
  • Therefore, the Thevenin equivalent resistance seen by the inductor is: Rth=Rparallel′+2R=R2+2R=5R2R_{th} = R'_{parallel} + 2R = \frac{R}{2} + 2R = \frac{5R}{2}

3. Calculate the time constant (τ\tau):

The time constant for an RL circuit is given by τ=LRth\tau = \frac{L}{R_{th}}.

τ=L5R/2=2L5R\tau = \frac{L}{5R/2} = \frac{2L}{5R}

4. Formulate the current as a function of time:

Using the general formula I(t)=I0(1−e−t/τ)I(t) = I_0 (1 - e^{-t/\tau}):

I(t)=V05R(1−e−t/(2L5R))I(t) = \frac{V_0}{5R} \left(1 - e^{-t / \left(\frac{2L}{5R}\right)}\right)

I(t)=V05R(1−e−5Rt2L)I(t) = \frac{V_0}{5R} \left(1 - e^{-\frac{5Rt}{2L}}\right)

Comparing this result with the given options, it matches option (1).