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Problem 3: The A.C voltage across a 0.5 μF capacitor is 16sin(2×103t). 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(2×103t)

Required             Capacitive Inductance, XC = ?                     Im = ?
Formula              

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

V = Vmsinωt     …     (1)

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

Peak voltage = Vm = 16 V, and ω = 2×103 rad/sec. So put in (A) to find XC.

Similarly, the peak value of current, Im = Vm/XC. Put the value of Vm and XC calculated above,

3 Comments

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