An ac source is connected in given series LCR circuit. The rms potential difference across the capacitor of $$20\mu\text{F}$$ is ______ V.

Given:
L = 1 H
C = 20 μF = 20 × 10⁻⁶ F
R = 300 Ω
v(t) = 50√2 sin(100t)

So:
$$V_{rms}=50V$$
ω = 100 rad/s

step 1: calculate reactances

Inductive reactance:
$$X_L=ωL=100\times1=100Ω$$

Capacitive reactance:
$$X_C=1/(ωC)$$
= 1 / (100 × 20×10⁻⁶)
= 1 / (2×10⁻³)
= 500 Ω

step 2: impedance of circuit

$$Z=\sqrt{\ \left(R^2+(X_L−X_C)^2\right)}$$
$$=\sqrt{\ \left(300^2+(100−500)^2\right)}$$
$$=\sqrt{\ 250000}$$
$$=500Ω$$

step 3: current

$$I_{rms}=V_{rms}/Z=50/500=0.1A$$

step 4: voltage across capacitor

$$V_C=I\times X_C=0.1\times500=50V$$