In a series L-R circuit `(L=35 mH and R=11 Omega)`, a variable emf source `(V=V_(0) sin omega t)` of `V_(rms)=220V` and frequency 50 Hz is applied. Find the current amplitude in the circuit and phase of current with respect to voltage. Draw current-time graph on given graph `(pi=22/7)`.

Explanation

Inductive reactance,
`X_(L) = omega L=2pi xx 50xx 35 xx10^(-3)`
`=2 xx 22/7 xx50 xx 35 xx 10^(-3)=11 Omega`

Therefore, impedance of circuit
`Z + sqrt(R^(2)+X_(L)^(2))=sqrt(11^(2)+11^(2))=11sqrt(2) Omega`
Given, `V_(rms)=220 V`
Amplitude of voltage `V_(0)=sqrt(2) V_(rms)=220 sqrt(2)V`
Amplitude of current `i_(0)=(V_0)/(Z) =(220sqrt(2))/(11sqrt(2))=20A`
Phase lag of current over voltage
`phi = tan^(-1) ((omega L)/(R )) = tan^(-1) ((11)/(11)) = tan^(-1) (1)= (pi)/(4)`
Voltage and current variations are shown in fig.

The voltage and the current are expressed as `V=220 sqrt(2) sin omega t`
`I= 20 sin (omega t -(pi)/(4))`.