Question

Alternating e.m.f. of `E = 220 sin 100 pi t` is applied to a circuit containing an inductance of `(1//pi)` henry. Write equation for instantaneous current through the circuit. What will be the reading of a.c. galvanometer connected in the circuit ?

Answer 1

From `E = 220 sin 100 pi t`,
`E_(0) = 220 V, omega = 100 pi , L = (1//pi) H`
`X_(l) = omega L = 100 pi xx (1)/(pi) = 100 ohm`.
`:. I_(0) = (E _(0))/(X_(L)) = (220)/(100) = 2.2 A`
As current lags behind the e.m.f. by a phase angle `pi//2`
`:. I = I_(0) sin (omega t - pi//2)`
`I = 2.2 sin (100 pi t - pi//2)`
Reading of a.c. galvanometer,`I_(v) = (I_(0))/(sqrt2) = (2.2)/(sqrt2)`
`I_(v) = (2.2)/(sqrt2) (sqrt2)/(sqrt2) = 1.1 xx 1.414 = 1.555 A`