An ac source is connected in given series LCR circuit. The rms potential difference across the capacitor of 20 μ F is __________ V.

Solution

Physics sir · 4 steps

Find both reactances from ω = 100, get Z, then the rms current times X_C gives the capacitor voltage.

1. Read the source. From V = 50√2 sin 100t, the angular frequency is ω = 100 rad s⁻¹. The peak is 50√2 V, so Vᵣₘₛ = 50 V.
2. Reactances. X_L = ω L = 100 × 1 = 100 Ω. X_C = 1/(ω C) = 1/(100 × 20 × 10⁻⁶) = 500 Ω.

   Z = √(R² + (X_C - X_L)²) = √(300² + 400²) = 500 Ω

3. Current. Iᵣₘₛ = Vᵣₘₛ/Z = 50/500 = 0.1 A.
4. So. V_C = Iᵣₘₛ X_C = 0.1 × 500 = 50 V.

The common slip: 70.7. Using the peak voltage 50√2 instead of the rms value gives V_C = 50√2 ≈ 70.7, which is the peak capacitor voltage, not the rms one.

The idea to keep. In a series LCR circuit the same current flows through every element, so find I from V/Z first, then the voltage across any one part is I times its own reactance or resistance.

Answer from the NTA final answer key. Solution by PrepWiser.