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In a series L-R circuit (L = 35 mH and R...

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)=220 V` and frequency 50 Hz is applied. The current amplitude in the circuit is

Text Solution

Verified by Experts

The correct Answer is:
`20A; (pi)//(4)`.
Inductive reactance
`X_(L)=omega L = (50)(2 pi) (35 xx 10^(-3))~~11 Omega`
Impedance `Z=sqrt(R^(2)+X_(L)^(2))=sqrt((11)^(2)+(11)^(2)) = 11sqrt(2) Omega`
Given, `V_(rms)=220 V`
Hence, amplitude of voltage
`V_(0)=sqrt(2)V_(rms)=220sqrt(2) V`
Amplitude of current `(i_0)=(V_0)/(Z)=(220sqrt(2))/(11 sqrt(2))=20 A`
phase difference `phi=tan^(-1)(X_L)/(R )=tan^(-1)((11)/(11))=(pi)/(4)`
In L-R circuit voltage leads the current. Hence, instantaneous current in the circuit is
`i=(20A)sin (omega t-(pi)/(4))`
Corresponding i-t graph is shown in the following figure.
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Knowledge Check

  • 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) .

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