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QPages 330-331: Q12: HC Verma: Concepts of Physics - Volume 2 Alternating Current Pages 330-331: Q12 Solution: In a series RC circuit with an AC source, , , and . Find the…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently substitute the peak current $I_0$ directly into the power formula $P = I^2 R$, yielding an incorrect result of 3 W. You must convert to RMS current using $I_{rms} = I_0 / \sqrt{2}$ before calculating average power, or use the equivalent formula $P = \frac{1}{2} I_0^2 R$.

In a series RC circuit with an AC source, R=300 ΩR = 300 \ \Omega, C=25 μFC = 25 \ \mu\text{F}, ε0=50 V\varepsilon_0 = 50 \text{ V} and ν=50/π Hz\nu = 50/\pi \text{ Hz}. Find the peak current and the average power dissipated in the circuit.

Step-by-Step Verified Solution

**Answer:** Peak current ≈0.1 A\approx 0.1 \, \text{A}, average power =1.5 W= 1.5 \, \text{W}. **Step-by-Step Explanation:** XC=1ωC=12π×50π×25×10−6=12.5×10−3=400 ΩX_C = \frac{1}{\omega C} = \frac{1}{2\pi \times \frac{50}{\pi} \times 25 \times 10^{-6}} = \frac{1}{2.5 \times 10^{-3}} = 400 \, \Omega. Z=3002+4002=500 ΩZ = \sqrt{300^2 + 400^2} = 500 \, \Omega. I0=50500=0.1 AI_0 = \frac{50}{500} = 0.1 \, \text{A}. P=Irms2R=(0.12)2×300=1.5 WP = I_{rms}^2 R = \left(\frac{0.1}{\sqrt{2}}\right)^2 \times 300 = 1.5 \, \text{W} [citation:4][citation:9].

All Questions in this Chapter

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