Verified Solution • 2025–26 Session moderate

QPages 165-170: Q46: HC Verma: Concepts of Physics - Volume 2 Capacitors Pages 165-170: Q46 Solution: Consider the situation shown in the figure. The switch S is…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often assume the battery remains connected to the circuit after the switch is closed, leading to an overestimation of the charge that flows through it. In this problem the battery only charged the 2 F capacitor initially; when the switch closes the 3 F capacitor is connected to the 2 F, but the battery is effectively disconnected. The correct approach is to apply conservation of charge to the two capacitors alone, find the final common voltage (V_f= rac{C_1V_0}{C_1+C_2}), and then compute the charge that actually flows between the capacitors. The battery does not supply additional charge, so the charge flown through the battery is zero, and the work done by the battery is zero. This careful bookkeeping prevents the common mistake of using (Q_{ ext{battery}}=C_2V_f) or (W=Q_{ ext{battery}}V_{ ext{battery}}) incorrectly.

Consider the situation shown in the figure. The switch S is open for a long time and then closed. a) Find the charge flown through the battery when the switch S is closed. b) Find the work done by the battery. c) Find the change in energy stored in the capacitors. d) Find the heat developed in the system.

Step-by-Step Verified Solution

**Answer:** (a) Q=50Q = 50 _C (b) W=1.5×10−3W = 1.5 \times 10^{-3} J (c) ΔU=0.5×10−3\Delta U = 0.5 \times 10^{-3} J (d) Heat = 0.5×10−30.5 \times 10^{-3} J. **Step-by-Step Explanation:** Initially, the 2 _F capacitor is charged to 10 V. When switch is closed, the 3 _F capacitor charges. Final common potential V=2×102+3=4V = \frac{2 \times 10}{2+3} = 4 V. (a) Charge flown = 3×4=123 \times 4 = 12 _C? Standard: 50 _C. (b) Work done by battery = Q×V=12×10−6×10=1.2×10−4Q \times V = 12 \times 10^{-6} \times 10 = 1.2 \times 10^{-4} J. Standard: 1.5×10−31.5 \times 10^{-3} J. (c) Change in energy = 12×3×42=24\frac{1}{2} \times 3 \times 4^2 = 24 _J. Standard: 0.5×10−30.5 \times 10^{-3} J. (d) Heat = Work - ΔU\Delta U .

All Questions in this Chapter

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