Question

The capacitor can withstand a peak voltage of 510 volts. If the voltage source operates at the resonance frequency, what maximum voltage amplitude Vmax can the source have if the maximum capacitor voltage is not exceeded

Answer 1

The complete question is Resistance is 440 Ω, inductance L = 0.380 H, capacitance, C = 1.4 ×10⁻² μF

Answer:

43.12 V

Explanation:

Find the current through the capacitor.

$I = \frac{V}{X_c}\\I = \frac{510 V}{X_c}\\I = (\omega C) (510V)\\I = \sqrt {\frac {C}{L}} (510V)\\I = \sqrt {\frac {1.4\times 10^{-8}}{0.380}} (510)= 0.098 A$

At resonance, impedance Z = R

V = IR

V = 0.098 × 440 = 43.12 V