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Multiple Choice
A 300 µF capacitor is connected to an AC source operating at an RMS voltage of 120 V. If the maximum current in the circuit is 1.5 A, what is the oscillation frequency of the AC source?
A
29.4 Hz
B
4.68 Hz
C
41.7 Hz
D
6.63 Hz
Verified step by step guidance
1
Start by understanding the relationship between the maximum current (I_max), the RMS voltage (V_rms), and the capacitive reactance (X_c) in an AC circuit. The formula for capacitive reactance is X_c = 1 / (2πfC), where f is the frequency and C is the capacitance.
Use the formula for the maximum current in an AC circuit: I_max = V_max / X_c. Note that V_max (maximum voltage) is related to V_rms by the equation V_max = V_rms * √2.
Substitute V_max = V_rms * √2 into the equation for I_max, giving I_max = (V_rms * √2) / X_c.
Rearrange the equation to solve for X_c: X_c = (V_rms * √2) / I_max.
Substitute X_c = 1 / (2πfC) into the equation and solve for the frequency f: f = 1 / (2πC * X_c). Use the given values: C = 300 µF (or 300 x 10^-6 F), V_rms = 120 V, and I_max = 1.5 A, to find the frequency.