Verified Solution • 2025–26 Session moderate

QPages 330-331: Q15: HC Verma: Concepts of Physics - Volume 2 Alternating Current Pages 330-331: Q15 Solution: Consider the situation of the previous problem. Find the…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often plug the peak voltage or current into $\frac12 C V^2$ or $\frac12 L I^2$ and call it the average stored energy, forgetting that the energy oscillates with $sin^2$ and the correct average uses RMS values: $\langle U_C\rangle=\frac12 C V_{\rm rms}^2$, $\langle U_L\rangle=\frac12 L I_{\rm rms}^2$. Using peak values overestimates the average by a factor of two.

Consider the situation of the previous problem. Find the average electric field energy stored in the capacitor and the average magnetic field energy stored in the coil.

Step-by-Step Verified Solution

**Answer:** Capacitor energy =25 mJ= 25 \, \text{mJ}, Inductor energy =5 mJ= 5 \, \text{mJ}. **Step-by-Step Explanation:** Using values from Q14: Irms=0.1 AI_{rms} = 0.1 \, \text{A}, VC,rms=50 VV_{C,rms} = 50 \, \text{V}. Energy in capacitor =12CVC,rms2=0.5×20×10−6×502=25 mJ= \frac{1}{2} C V_{C,rms}^2 = 0.5 \times 20 \times 10^{-6} \times 50^2 = 25 \, \text{mJ}. Energy in inductor =12LIrms2=0.5×1×0.12=5 mJ= \frac{1}{2} L I_{rms}^2 = 0.5 \times 1 \times 0.1^2 = 5 \, \text{mJ} [citation:4].

All Questions in this Chapter

Related Learning Resources & Exam Preparation