A 60 μF capacitor is connected to a 110 V, 60 Hz AC supply. What is the root mean square value of current in the circuit?
Answer & explanation
Correct answer: option 3
The correct answer is Option 3:2.49A
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Capacitance ($C$): $60\ \mu\text{F} = 60 \times 10^{-6}\ \text{F}$
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$rms$ Voltage ($V_{rms}$): $110\ \text{V}$
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Frequency ($f$): $60\ \text{Hz}$
The opposition offered by the capacitor to the AC flow is given by:
$X_C = \frac{1}{2\pi fC}$
$X_C = \frac{1}{2 \times 3.141 \times 60 \times 60 \times 10^{-6}}$
$X_C = \frac{10^6}{22619.4} \approx 44.21\ \Omega$
Using Ohm's law for AC circuits:
$I_{rms} = \frac{V_{rms}}{X_C}$
$I_{rms} = \frac{110\ \text{V}}{44.21\ \Omega}$
$I_{rms} \approx 2.488\ \text{A}$
≈2.49 A