Verified Solution • 2025–26 Session moderate

QPages 198-203: Q58: HC Verma: Concepts of Physics - Volume 2 Electric Current in Conductors Pages 198-203: Q58 Solution: a) Find the current in the resistor shown in the figure. b) If…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often mistakenly calculate the voltage across the capacitor by treating it as a short circuit or ignoring the potential drop across series resistors. In steady state, the capacitor acts as an open circuit, so the current through its branch is zero, and the voltage across it equals the potential difference between the terminals, which must be found by analyzing the rest of the closed loop.

a) Find the current in the

20 \ \Omega

resistor shown in the figure. b) If a capacitor of capacitance 4 _F is joined between the points A and B, what would be the electrostatic energy stored in it in steady state?

Step-by-Step Verified Solution

**Answer:** (a)

i = 0.5

A (b)

U = 2 \\times 10^{-6}

J. **Step-by-Step Explanation:** (a) Current through

20\\ \\Omega

resistor is

0.5

A. (b) Steady-state voltage across capacitor:

V = 1

V.

U = \\frac{1}{2}CV^2 = \\frac{1}{2} \\times 4 \\times 10^{-6} \\times 1^2 = 2 \\times 10^{-6}

J [citation:2].

All Questions in this Chapter

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