JEE MAIN - Physics (2002 - No. 13)
The power factor of $$AC$$ circuit having resistance $$(R)$$ and inductance $$(L)$$ connected in series and an angular velocity $$\omega $$ is
$$R/\omega L$$
$$R/{\left( {{R^2} + {\omega ^2}{L^2}} \right)^{1/2}}$$
$$\omega L/R$$
$$R/{\left( {{R^2} - {\omega ^2}{L^2}} \right)^{1/2}}$$
Explanation
The impedance triangle for resistance $$\left( R \right)$$ and inductor $$(L)$$ connected in series is shown in the figure.
Power factor $$\cos \phi = {R \over {\sqrt {{R^2} + {\omega ^2}{L^2}} }}$$
Power factor $$\cos \phi = {R \over {\sqrt {{R^2} + {\omega ^2}{L^2}} }}$$
Comments (0)
