Saved Bookmarks
| 1. |
A resistance of 300 Omega is connected in series with inductor of self-inductance 1 H and capacitor of capacitance 20 mF. This combination is connected to an AC source whose angular frequency is 100 rad/s. Impedance of the circuit is |
|
Answer» 200 `OMEGA`. `X_(L) = omegaL = 100 X 1 =100 Omega` `X_(C) = (1)/(omegaC) = (1)/(100xx20xx10^(-6)) Omega = 500 Omega` Net impedance of the circuit can be written as follows : = `sqrt(R^(2)+(X_C-X_(L))^(2))` `implies Z = sqrt(300^(2)+ (500-100)^(2)Omega)` Hence option ( c) is CORRECT. |
|