Decay Of Current In An Lr Circuit
A coil of inductance 2 H and resistance 4 (\Omega) carries a steady current that is suddenly disconnected from its battery and short-circuited at one instant. After how long does the current decay to roughly 37 percent of its initial value?
Select the correct option:
Solution
0.5 s
When an LR circuit is short-circuited, the inductor sustains the current, which decays exponentially as (I = I_0 e^{-t/\tau}), where the time constant (\tau = L/R) sets the pace of decay. The current falls to (1/e \approx 37%) of its initial value precisely when the elapsed time equals one time constant. Computing (\tau = L/R = 2/4 = 0.5) s, so the current reaches about 37 percent after 0.5 s. The option 2 s mistakenly uses the inductance value alone as the time constant. The option 8 s multiplies L by R instead of dividing. The option 0.25 s inverts the ratio to (R/L). This is the NCERT exponential-decay description of inductive transients, mirroring how charge decays on a discharging capacitor through an (RC) time constant. The decay law itself follows from solving (L,dI/dt + RI = 0), whose solution is a falling exponential with rate set by (R/L); during this decay the energy once stored in the magnetic field is gradually dissipated as heat in the resistance. A plausibility check confirms the units H/(\Omega) reduce to seconds, and a larger inductance or a smaller resistance would prolong the decay, consistent with the inductor's tendency to oppose any change in current; after roughly five time constants the current is essentially zero.
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About This Question
- Subject
- physics
- Chapter
- electromagnetic induction and alternating currents
- Topic
- decay of current in an lr circuit
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
0.5 s
When an LR circuit is short-circuited, the inductor sustains the current, which decays exponentially as (I = I_0 e^{-t/\tau}), where the time constant (\tau = L/R) sets the pace of decay. The current falls to (1/e \approx 37%) of its initial value precisely when the elapsed time equals one time constant. Computing (\tau = L/R = 2/4 = 0.5) s, so the current reaches about 37 percent after 0.5 s. The option 2 s mistakenly uses the inductance value alone as the time constant. The option 8 s multiplies L by R instead of dividing. The option 0.25 s inverts the ratio to (R/L). This is the NCERT exponential-decay description of inductive transients, mirroring how charge decays on a discharging capacitor through an (RC) time constant. The decay law itself follows from solving (L,dI/dt + RI = 0), whose solution is a falling exponential with rate set by (R/L); during this decay the energy once stored in the magnetic field is gradually dissipated as heat in the resistance. A plausibility check confirms the units H/(\Omega) reduce to seconds, and a larger inductance or a smaller resistance would prolong the decay, consistent with the inductor's tendency to oppose any change in current; after roughly five time constants the current is essentially zero.
This hard difficulty physics question is from the chapter electromagnetic induction and alternating currents, covering the topic of decay of current in an lr circuit. It appeared in the 2025 exam.
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