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 ?

Explanation

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)`