Question

An AC source of maximum voltage V0 = 30 V is connected to a resistor R = 50 Ω, an inductor L = 0.6 H, and a capacitor C = 20 µF. What is the maximum charge on the capacitor when the frequency of the source is set to the resonance frequency of the circuit?

Answer 1

Hello!

We can begin by solving for the resonance ANGULAR frequency of the circuit.

For an RCL circuit, the resonance angular frequency is given as:
$\omega_0^2 = \frac{1}{LC}\\\\w_0 = \sqrt{\frac{1}{LC}}$

ω₀ = resonance angular frequency (rad/s)

L = Inductance (0.6 H)
C = Capacitance (20 μF)

Plug in the values and solve.

$\omega_0 = \sqrt{\frac{1}{(0.6)(0.00002)} } = 288.675 \frac{rad}{s}$

For an AC power source, the output is usually expressed as:

$V(t) = V_{max}sin(\omega_0 t})$

So, using the appropriate values and setting the source angular frequency equivalent to the circuit's resonance angular frequency:

$V(t) = 30sin(288.675t)$

To find the maximum charge on the capacitor when the frequency of the source is equivalent to the resonance frequency of the circuit (or the angular frequencies are equal), we can begin by finding the maximum voltage across the capacitor.

To find this, however, we must solve for the maximum current across the circuit by finding the total impedance of the circuit. When the circuit is at resonance, the impedance is equivalent to the resistance of the RESISTOR.

So, solve for the maximum current in the circuit using Ohm's Law:

$i = \frac{V}{R}$

In this instance AT RESONANCE:

$I_{Max} = \frac{V_Max}{R}\\\\I_{Max} = \frac{30}{50} = 0.6 A$

Now, we must solve for the capacitive reactance in order to find the maximum voltage across the capacitor. Using the following equation for capacitive reactance:
$X_c = \frac{1}{\omega C}\\\\X_c = \frac{1}{(288.675)(0.00002)} = 173.205 \Omega$

Now that we found the maximum current and capacitive reactance, we can now solve for the maximum voltage across the capacitor:
$V_{C, max} = X_C I_{Max}\\\\V_{C, max} = 173.205 * 0.6 = 103.923 V$

Finally, we can easily solve for the maximum charge on the capacitor using the relationship:
$C = \frac{Q}{V}\\\\Q = CV$

Plug in the values solved for above.

$Q = (0.00002)(103.923) = 0.00208 C = \boxed{2.078 mC}$