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A 0.30 μF capacitor is connected across an AC generator that produces a peak voltage of 10 V. What is the peak current to and from the capacitor if the emf frequency is (a) 100 Hz?
31. Alternating Current
Capacitors in AC Circuits
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The peak current to and from a capacitor is i when the emf source has a peak voltage V and frequency f. What is the peak current to and from the capacitor if the frequency and peak voltage are doubled to 2f and , respectively?
A
i/4
B
i/2
C
i
D
2i
E
4i
Verified step by step guidance1
Understand that the peak current in a capacitor circuit is given by the formula: I = V * ω * C, where I is the peak current, V is the peak voltage, ω is the angular frequency, and C is the capacitance.
Recall that the angular frequency ω is related to the frequency f by the equation: ω = 2πf.
When the frequency is doubled, the new frequency becomes 2f, and the new angular frequency becomes ω' = 2π(2f) = 4πf.
When the peak voltage is doubled, the new peak voltage becomes 2V.
Substitute the new values into the peak current formula: I' = (2V) * (4πf) * C. Simplify this to find that the new peak current I' is 4 times the original peak current I, which means I' = 4i.
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