CUET Preparation Today
A series L.C.R circuit with $R=3 \Omega, X_{L}=10 \Omega$ and $X_{C}=6 \Omega$ is connected to an a.c. source. The power factor of the circuit is: |
0 0.6 0.75 1 |
0.6 |
The correct answer is Option (2) → 0.6 Impedance, $Z=\sqrt{R^2+(X_L-X_C)^2}$ $=\sqrt{3^2+(10-6)^2}$ $=5Ω$ Power factor = $\frac{R}{Z}=\frac{3}{5}=0.6$ |
