A circuit is constructed with four resistors, one inductor, one battery, and a switch as shown.

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A circuit is constructed with four resistors, one inductor, one battery, and a switch as shown. The values for the resistors are: R1 = 770Ω, R2 = 660Ω, and R4 = 102Ω. The inductance is L = 635 mH and the battery voltage is V = 24V.
A circuit is constructed with four resistors, one inductor, one battery, and a switch as shown. The values for the resistors are: R1 = 770Ω, R2 = 660Ω, and R4 = 102Ω. The inductance is L = 635 mH and the battery voltage is V = 24V.

The Correct Answer and Explanation is:

Given Values:

  • Resistors:
    • R1=R2=77 ΩR_1 = R_2 = 77\,\OmegaR1​=R2​=77Ω
    • R3=66 ΩR_3 = 66\,\OmegaR3​=66Ω
    • R4=102 ΩR_4 = 102\,\OmegaR4​=102Ω
  • Inductance: L=635 mH=0.635 HL = 635\,\text{mH} = 0.635\,\text{H}L=635mH=0.635H
  • Battery voltage: V=24 VV = 24\,\text{V}V=24V

Circuit Configuration:

From the image:

  • R1R_1R1​ and R2R_2R2​ are in series in the top branch with the inductor LLL between them.
  • R3R_3R3​ and R4R_4R4​ are in the lower loop, also forming a branch when the switch is closed.
  • The battery supplies the entire circuit, and the switch controls the connection of the bottom branch.

Analysis Immediately After Switch is Closed (t = 0):

At t=0t = 0t=0, the inductor behaves like an open circuit (because inductors resist a change in current). Therefore, no current flows through the inductor branch (i.e., through R1R_1R1​, LLL, and R2R_2R2​).

So, the current only flows through the bottom branch: R4R_4R4​ and R3R_3R3​, in series. Req (t=0)=R4+R3=102 Ω+66 Ω=168 ΩR_{\text{eq (t=0)}} = R_4 + R_3 = 102\,\Omega + 66\,\Omega = 168\,\OmegaReq (t=0)​=R4​+R3​=102Ω+66Ω=168Ω Iinitial=VReq=24 V168 Ω≈0.143 AI_{\text{initial}} = \frac{V}{R_{\text{eq}}} = \frac{24\,\text{V}}{168\,\Omega} \approx 0.143\,\text{A}Iinitial​=Req​V​=168Ω24V​≈0.143A

After a Long Time (t → ∞):

After a long time, the inductor behaves like a short circuit (acts like a wire). So now current can flow through the inductor branch and the bottom branch.

  • Top branch resistance: R1+R2=77+77=154 ΩR_1 + R_2 = 77 + 77 = 154\,\OmegaR1​+R2​=77+77=154Ω
  • Bottom branch resistance: R3+R4=66+102=168 ΩR_3 + R_4 = 66 + 102 = 168\,\OmegaR3​+R4​=66+102=168Ω

These two branches are in parallel, so we compute the equivalent resistance: 1Req=1154+1168⇒Req≈80.0 Ω\frac{1}{R_{\text{eq}}} = \frac{1}{154} + \frac{1}{168} \Rightarrow R_{\text{eq}} \approx 80.0\,\OmegaReq​1​=1541​+1681​⇒Req​≈80.0Ω Ifinal=24 V80 Ω=0.30 AI_{\text{final}} = \frac{24\,\text{V}}{80\,\Omega} = 0.30\,\text{A}Ifinal​=80Ω24V​=0.30A

Conclusion:

  • Initial current (t = 0): ~0.143 A (only through R3R_3R3​ and R4R_4R4​)
  • Final current (t → ∞): ~0.30 A (distributed between two branches)

This demonstrates how inductors affect current flow over time, initially blocking and eventually allowing current.

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