Resonance: An ac source having a maximum voltage output of 30 V is connected in series with a 50-Ω resistor, a 0.60-H inductor, and a 20-µF capacitor. What is the maximum charge on the capacitor when the frequency of the source is adjusted so that the circuit is in resonance?

Question

Quizplus · Accepted Answer

At resonance, the capacitive reactance and inductive reactance are equal and opposite. Using the formula for the resonant frequency:

f = 1 / (2π√(LC))

The resonant frequency is found to be:

f = 1 / (2π√(0.60 H × 20 µF)) = 2.07 kHz

The maximum charge on the capacitor can be found using the formula:

Q = CV

where C is the capacitance and V is the voltage at resonance. The voltage at resonance can be found using the voltage divider rule:

V = (Vmax)(XC) / √(R^2 + XC^2)

where XC is the reactance of the capacitor and R is the resistance. At resonance XC = XL, so:

V = (Vmax)(XL) / √(R^2 + XL^2)

At resonance:

XL = XC = 1 / (2πfC) = 12.1 Ω

Plugging in the values:

V = (30 V)(12.1 Ω) / √(50 Ω^2 + (12.1 Ω)^2) = 6.36 V

Finally, the maximum charge on the capacitor is:

Q = CV = (20 µF)(6.36 V) = 2.1 mC

Resonance: an Ac Source Having a Maximum Voltage Output of 30