An AC voltage V = 100 sin(100t) is applied to a capacitor of 1 microfarad. What is the peak current in the circuit?

A. 0.01 A — The peak current is given by I_peak = V_peak / XC, where XC = 1 / (omega * C). Here, omega = 100 and C = 10^-6 F, so XC = 1 / (100 * 10^-6) = 10000 ohms. Thus, I_peak = 100 / 10000 = 0.01 A.

JEE Advanced · PhysicsMedium

$An AC voltage V = 100 sin (100 t) is applied to a capacitor of 1 μ F . What is the peak current in the circuit?$

A. $0.01 A$
B. $0.1 A$
C. $1 A$
D. $10 A$

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Correct answer: A — $0.01 A$

Solution

I_{0} = \frac{V _{0}}{X _{C}} = V_{0} ω C = 100 \times 100 \times 1 0^{- 6} = 0.01 A .

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An AC voltage V=100sin(100t) is applied to a capacitor of 1 μF. What is the peak current in the circuit?

Solution

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More AC Circuits practice questions

$An AC voltage V = 100 sin (100 t) is applied to a capacitor of 1 μ F . What is the peak current in the circuit?$