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A capacitor of `10 mu F` is connected to an a.c. source of e.m.f. `E = 220 sin 100 pi t`. Write the equaiton of instananeous current through the circuit. What will be the reading of a.c. ammeter connected in the circuit ? |
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Answer» Here, `C = 10 mu F = 10 xx 10^(-6) F` `E = 220 sin 100 pi t = E_(0) sin omega t` `:. E_(0) = 220 V, omega = 2 pi v = 100 pi, v = 50 hz`. `X_(C ) = (1)/(omega C) = (1)/(2 pi v C)` ` =(1)/(2 xx 3.14 xx 50 xx 10^(-5)) = 318.5 Omega` `I_(0) = (E_(0))/(X_(C )) = (220)/(318.5) = 0.691 A` Reading of a.c. ammeter, `I_(v) = (I_(0))/(sqrt2) = (0.691)/(1.414) = 0.489 A` Instantaneous current in the circuit, `I = I_(0) sin (omega t + pi//2)` `I = 0.691 sin (100 pi t + pi//2)` |
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