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)`.

Answer

A

Explanation

`20 A,(pi)/(4)`
Inductive reactance
`X_(L)=omegaL=(50)(2pi)(35xx10^(-3))gt gt 11 Omega`
Impedance
`Z=sqrt(R^(2)+X_(L)^(2))=sqrt((11)^(2)+(11)^(2))=11sqrt2Omega`
Given `v_(rms)=220` Volt
Hence, amplitude of voltage
`v_(0)=sqrt(2) v_(rms)=220 sqrt(2)` volt
`therefore` Amplitude of current
`i_(0)=(v_(0))/(Z)=(220sqrt2)/(11sqrt2)=20 A`
Phase difference
`f=tan^(-1)((X_(L))/(R ))=tan^(-1)((11)/(11))=(pi)/(4)`
In `L-R` circuit voltage leads the current. Hence, instantaneous current inthe circuit is,
`i=(20A) sin (omegat-pi//4)`
Corresponding `i-t` graph is shown in figure.