Problem 3: The A.C voltage across a 0.5 μF capacitor is 16sin(2×10^{3}t). Find (a) the capacitive reactance (b) the peak value of current through the capacitor.

Solution

Given Capacitance = C = 0.5 μF = 0.5×10^{-6} F, V = 16sin(2×10^{3}t)

Required Capacitive Inductance, X_{C} = ?
I_{m} = ? Formula

Now we know that for a capacitor, the instantaneous voltage V is

V = V_{m}sinωt … (1)

Comparing this equation with the equation in the given data, we have

Peak voltage = V_{m} = 16 V, and ω = 2×10^{3} rad/sec. So put in (A) to find X_{C}.

Similarly, the peak value of current, I_{m} = V_{m}/X_{C}. Put the value of V_{m} and X_{C} calculated above,

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