A series circuit consists of a 0.440-H inductor, a 380.0-Ω resistor, a 5.70-μF capacitor, and an ac voltage source of amplitude 250.0 V. Find the rms voltage across the capacitor when the circuit operates at resonance.

A) 106 V

B) 129V

C) 95 V

D) 65 V

Answer: B) 129 V

Explanation:

If the circuit is operating at resonance, the capacitive reactance and the inductive one are equal each other, so that the circuit behaves like it were purely resistive.

This condition is verified when the ac voltage source operates at a given frequency, called resonance frequency, at which it is verified the following equality;

Xc = XL ⇒ 1/ω₀C = ω₀L

From this, we can find out the value of ω₀, as follows:

ω₀ = 1/√LC = 1/√0.44 H. 5.7. 10⁻⁶ F = 631.4 Hz

At resonance, the current I and the voltage are in phase, and I can be obtained applying Ohm's Law to the resistor R:

I = V/R = 250.0 V / 380.0Ω =. 66 A = 660 mA

The RMS value of this current, is just this peak value times √2/2, i. e.,

IRMS =.66 A. √2/2 = 0.46 A

The RMS voltage across the capacitor - which is numerically equal and opposite to the one across the inductor, is just I times the capacitive reactance:

Vc = I. Xc = 0.46 A. 1/631.4 Hz. 5.7. 10⁻⁶F = 129 V