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QPages 330-331: Q14: HC Verma: Concepts of Physics - Volume 2 Alternating Current Pages 330-331: Q14 Solution: In a series LCR circuit with an AC source, , , , and . Find a)…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often add the individual rms voltages algebraically, getting a sum larger than the source voltage, and think the calculation is wrong. In a series LCR circuit the voltages are out of phase, so they must be added as phasors: $V_{ ext{total}}=V_R+V_L-V_C$ in the complex plane, not $V_R+V_L+V_C$. Only after vector addition does the magnitude equal the source rms voltage.

In a series LCR circuit with an AC source, R=300 ΩR = 300 \ \Omega, C=20 μFC = 20 \ \mu\text{F}, L=1.0 henryL = 1.0 \text{ henry}, εrms=50 V\varepsilon_{\text{rms}} = 50 \text{ V} and ν=50/π Hz\nu = 50/\pi \text{ Hz}. Find a) the rms current in the circuit and b) the rms potential difference across the capacitor, the resistor and the inductor. Note that the sum of the rms potential differences across the three elements is greater than the rms voltage of the source.

Step-by-Step Verified Solution

**Answer:** (a) Irms≈0.1 AI_{rms} \approx 0.1 \, \text{A} (b) VC=50 VV_C = 50 \, \text{V}, VR=30 VV_R = 30 \, \text{V}, VL=10 VV_L = 10 \, \text{V}. **Step-by-Step Explanation:** ω=2π×50π=100 rad/s\omega = 2\pi \times \frac{50}{\pi} = 100 \, \text{rad/s}. XL=100×1=100 ΩX_L = 100 \times 1 = 100 \, \Omega. XC=1100×20×10−6=500 ΩX_C = \frac{1}{100 \times 20 \times 10^{-6}} = 500 \, \Omega. Z=3002+(500−100)2=500 ΩZ = \sqrt{300^2 + (500 - 100)^2} = 500 \, \Omega. Irms=50500=0.1 AI_{rms} = \frac{50}{500} = 0.1 \, \text{A}. VC=IrmsXC=50 VV_C = I_{rms} X_C = 50 \, \text{V}, VR=30 VV_R = 30 \, \text{V}, VL=10 VV_L = 10 \, \text{V} [citation:4][citation:8].

All Questions in this Chapter

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