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

In the circuit of fig, the source freqency is `omega=2000 rad s^(-1)`.

1.	In the circuit of fig, the source freqency is `omega=2000 rad s^(-1)`. The current in the will be A. (A) `2A`B. (B) `3.3A`C. (C) `2 //sqrt(5)A`D. (D) `sqrt(5) A`
Answer» Correct Answer - A `(X_L) = L omega = 5 xx 10 ^(-3) xx 2000 = 10 Omega` `(X_C) = (1)/(C omega)=(1)/(50 xx 10^(-6) xx 2000) =(100)/(10) Omega = 10 Omega` since `(X_L)=(X_C)` therefore, `Z=R = 5.9 + 0.10 +4 = 10 Omega` `(I_V) =(E_V)/(Z)=(20)/(10) A =2A`.

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

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