A capactior of `1 mu F` is connected ot an a.c. source of e.m.f `E = 250 sin 100 pi t`. Write an equation for instantaneous current through the circuit and given reading of a.c. ammeter connected in the circuit.

Explanation

Here, `C = 1 mu F = 10^(-6) F`
As `E = 250 sin 100 pi t`
`:. E_(0) = 250 V, omega = 100 pi`
The instantaneous current through the circuit is
`I = I_(0) sin (omega t + pi//2) = (E_(0))/(1 // omega C) sin (omega t + pi // 2)`
` = 250 xx 100 pi xx 10^(-6) sin (100 pi t + pi // 2)`
`I = 0.0786 sin (100 pi t + pi // 2)`
Reading of a.c. ammeter `= I_(v)`
`= (I_(0))/(sqrt 2) = (0.0786)/(sqrt 2) = 0.06 A`