Verified Solution • 2025–26 Session moderate

QPages 165-170: Q37: HC Verma: Concepts of Physics - Volume 2 Capacitors Pages 165-170: Q37 Solution: Two capacitors of capacitances 4.0 �F and 6.0 �F are connected…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often use $W=QV$ or $U=CV^{2}$ for a capacitor’s energy, forgetting the required $\tfrac12$ factor; the correct stored energy is $U=\tfrac12CV^{2}$, while the battery supplies $QV$, which is twice the stored amount in an ideal circuit.

Two capacitors of capacitances 4.0 �F and 6.0 �F are connected in series with a battery of 20 V. Find the energy supplied by the battery.

Step-by-Step Verified Solution

**Answer:** Energy supplied = 4.8×10−44.8 \times 10^{-4} J. **Step-by-Step Explanation:** Equivalent capacitance: Ceq=4×610=2.4C_{eq} = \frac{4 \times 6}{10} = 2.4 _F.
Q=CV=2.4×10−6×20=4.8×10−5 CQ = CV = 2.4 \times 10^{-6} \times 20 = 4.8 \times 10^{-5} \text{ C}
Energy supplied by battery: W=QV=4.8×10−5×20=9.6×10−4W = QV = 4.8 \times 10^{-5} \times 20 = 9.6 \times 10^{-4} J. Half is stored, half dissipated. Standard: 4.8×10−44.8 \times 10^{-4} J (energy stored).

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