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

Here, `E = 220 sin 100 pi t, L = (1)/(pi) H`
`E_(0) = 220 V`,
`I_(0) = (E_(0))/(X_(L)) - (E_(0))/(2 pi v L) = (220)/((100 pi) (1)/(pi)) = 2.2 A`
As current through L lags behind the applied e.m.f. through a phase angle of `90^(@)`, therefore instantaneous current through the circuit is
`I = I_(0) sin (omega t - pi // 2) = 2.2 sin (100 pi t - pi // 2)`
Reading of a.c. ammeter `= I_(v) = (I_(0))/(sqrt 2) = (2.2)/(sqrt 2)`
`= 1.556 A`