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In the value circuit, C=(sqrt(3))/(2)muF...


In the value circuit, `C=(sqrt(3))/(2)muF,R_2=20omega, L=sqrt(3)/(10)H`, and `R_1=10omega` .Current in `L-R_1` pathsis `I_1` and in `C-R_2` path it is `I_2`. The voltage of A.C source is given by, `V=200sqrt(2)sin(100t)` volts. The phase difference between `I_2` and `I_2` is :

A

`60^@`

B

`150^@`

C

`30^@`

D

`90^@`

Text Solution

Verified by Experts

The correct Answer is:
B
`X_L=omegaL`
`=100 xxsqrt3/10 =10sqrt3Omega`
`tan phi_1=(10 sqrt3)/10 =sqrt3 ,phi_1=60^@` (Current is lagging)

`tan phi_2=1/(omegaCR)=10^2/sqrt3 ,phi_2~~90^@` (Current s leading)
Total phase difference `= 150^@`
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