Question

In a series LCR circuit with an AC source R = 300 W, C = 20 mF, L = 1 H, E_rms = 50 V and n = $\frac{50}{\pi}$Hz . The potential difference across the capacitor is–

• A. 50 V
• B. $\frac{50}{\sqrt{2}}$V
• C. 40 V
• D. $\frac{40}{\sqrt{2}}$V

Answer 1

Answer: (A)

50 V

Answer 2

X_C = $\frac{1}{C\omega}$ = $\frac{1}{20 \times 10^{–6} \times 2\pi \times \frac{50}{\pi}}$ = 500 W

X_L = Lw = L × 2p × $\frac{50}{\pi}$ = 100 W

I_rms = $\frac{E_{rms}}{2}$ = $\frac{50}{\sqrt{R^{2} + (X_{C}–X_{L})^{2}}}$

= $\frac{50}{\sqrt{(300)^{2} + (500–100)^{2}}}$= $\frac{50}{500} = \frac{1}{10}A$

V_C = I_rms × X_C

= $\frac{1}{10}$ × 500 V

= 50 V