Verified Solution • 2025–26 Session moderate

QPages 306-313: Q90: HC Verma: Concepts of Physics - Volume 2 Electromagnetic Induction Pages 306-313: Q90 Solution: Consider the circuit shown in figure. a) Find the current…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often substitute the steady‑state current $i_{\text{ss}}=E/R$ into the decay law $i(t)=i_0 e^{-t/\tau}$, writing $i(\tau)=i_{\text{ss}}/e$. In reality $i_0$ is the initial current at the moment the switch opens (the same $E/R$ for a fully charged RL circuit), so after one time constant the current is $i(\tau)=i_0 e^{-1}=\frac{E}{R}\,e^{-1}$, i.e. 37% of the initial value, not $1/e$ of some other reference. Remember to use $\tau=L/R$ and apply the exponential to the *initial* current, not to the final steady‑state value.

Consider the circuit shown in figure. a) Find the current through the battery a long time after the switch S is closed. b) Suppose the switch is again opened at t = 0. What is the time constant of the discharging circuit? c) Find the current through the inductor after one time constant.

Step-by-Step Verified Solution

**Answer:** (a) i = E/R (b) _ = L/R (c) i = i_e^(-1). **Step-by-Step Explanation:** (a) Long time after switch closed, inductor acts as short. i = E/R. (b) After opening, the circuit has resistance R and inductance L. _ = L/R. (c) After one time constant, i = i_e^(-1) [citation:4].

All Questions in this Chapter

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