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An a.c.source is connected to a capacitor of capacitance C. Find the expression for current flowing through it. |
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Answer» Solution : Let an alternating voltage V = `V_(m) sin omega t` be applied across a pure capacitor C. Then instantaneous value of CHARGE on a PLATE of the capacitor is: `q =CV = CV_(m) sin omega t` Current FLOWING through the capaictor `I = (dq)/(dt) = CV_(m) omega cos omega t = V_(m)/(1//Comega). sin (omega t + pi/2)`  `=I_(m) sin (omega t + pi/2)` Expression for current clearly shows that for a pure capacitive a.c. circuit the current leads the voltage by a PHASE angle `pi/2` |
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