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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` |
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