Question

In series R–L–C circuit, the rms voltage across the resistor and inductor are respectively 400 V and 700 V. If the equation for the applied voltage is E = 500$\sqrt{2}$sin wt, then the peak voltage across the capacitor is -

• A. 1200 V
• B. $1200\sqrt{2}V$
• C. 400 V
• D. $400\sqrt{2}V$

Answer 1

Answer: (D)

$400\sqrt{2}V$

Answer 2

$V_{R}^{2} + (V_{L} - V_{C})^{2} = E_{rms}^{2}$ (V_L – V_C)² = (500)² – (400)²

V_L – V_C = $\sqrt{(500)^{2} - (400)^{2}}$= 300

V_C = V_L – 300 = 700 – 300 = 400

\ V_C (peak) = $\sqrt{2}$V_C =$\sqrt{2}$× 400 volts