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QPages 306-313: Q83: HC Verma: Concepts of Physics - Volume 2 Electromagnetic Induction Pages 306-313: Q83 Solution: A solenoid having inductance 4.0 H and resistance is connected…
⚠️ Common Exam Traps & Conceptual Pitfalls
Students often assume the entire battery power $P=VI$ is dissipated as Joule heat at $t=\tau$, using $P=V^2/R$; however the inductor still has a voltage drop, so only $i^2R$ is converted to heat while the remainder $L i\,di/dt$ stores magnetic energy. Remember to compute resistor power with $i^2R$ and recognize that $P_{battery}=VI$ is split between heating and energy storage.
A solenoid having inductance 4.0 H and resistance is connected to a 4.0 V battery at t = 0. Find a) the time constant, b) the time elapsed before the current reaches 0.63 of its steady-state value, c) the power delivered by the battery at this instant and d) the power dissipated in Joule heating at this instant.
Step-by-Step Verified Solution
**Answer:** (a) 0.4 s (b) 0.4 s (c) 1.0 W (d) 0.63 W.
**Step-by-Step Explanation:**
_ = L/R = 4.0/10 = 0.4 s.
(b) i = 0.63i_: 1 - e^(-t/_) = 0.63, t � _ = 0.4 s.
(c) i = 0.63 _ 0.4 = 0.252 A. P = Ei = 4 _ 0.252 = 1.0 W.
(d) P_R = i_R = (0.252)_ _ 10 = 0.63 W.
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