1.

In the circuit of fig, the source frequency is `omega=2000 rad s^(-1)`. The current in the will be A. `2A`B. `3.3 A`C. `2//sqrt(5)A`D. `sqrt(5)A`

Answer» Correct Answer - A
`L omega=5xx10^(-3)xx2000=10 Omega`
`1/(C omega)=1/(50xx10^(-6)xx2000)=100/10 Omega=10 Omega`
since `L omega=1/(L omega)`,
`:. Z=R=0.10+4=10.1 Omega`
`I_(0)=E/Z=20/10.1 A=2A`


Discussion

No Comment Found

Related InterviewSolutions