A voltage ν(t)=V0cos(ωt) is applied to the series combination of a capacitor and a resistor. Obtain expressions for the power factor and the average power dissipated in the load.

Question

A voltage [latex] u(t)=V_0 \cos (\omega t) [/latex] is applied to the series combination of a capacitor and a resistor. Obtain expressions for the power factor and the average power dissipated in the load.

Accepted Answer

The impedance of the load is

[latex] \begin{aligned} Z &=R+\frac{1}{j \omega C}=R-j \frac{1}{\omega C} \\ &=\sqrt{R^2+(\omega C)^{-2}} \angle an ^{-1}\left(-\frac{1}{\omega R C} ight) . \end{aligned} [/latex]

The power factor is

[latex] \begin{aligned} p f &=\cos (∡ Z)=\cos \left[ an ^{-1}\left(-\frac{1}{\omega R C} ight) ight] \\ &=\frac{\omega R C}{\sqrt{1+(\omega R C)^2}}, \end{aligned}[/latex]

as illustrated by Fig. 13.14. The load current (phasor) is given by

[latex] ilde{I}=\frac{ ilde{V}}{Z}=\frac{V_0}{\sqrt{R^2+(\omega C)^{-2}}} L- an ^{-1}\left(-\frac{1}{\omega R C} ight) . [/latex]

The peak load voltage equals [latex]V_0[/latex] and the peak load current (the magnitude of the phasor [latex] ilde{I}[/latex] obtained above) is given by

[latex] I_0=\frac{V_0}{\sqrt{R^2+(\omega C)^{-2}}}=\frac{\omega C V_0}{\sqrt{1+(\omega R C)^2}} . [/latex]

It follows from (13.11) that the average power dissipated is given by

[latex] \begin{gathered} P=\frac{1}{2}| ilde{V} ilde{I}| \cos \left( heta_Z ight)=V_{r m s} I_{r m s} \cos \left( heta_Z ight), \\ heta_Z=∡Z, (13.11) \end{gathered} [/latex]

[latex] \begin{aligned} P &=\frac{1}{2}\left(V_0 I_0 ight)(p f) \\ &=\frac{1}{2} \frac{\omega C V_0^2}{\sqrt{1+(\omega R C)^2}} \frac{\omega R C}{\sqrt{1+(\omega R C)^2}} \\ &=\frac{1}{2} \frac{R(\omega C)^2 V_0^2}{\left[1+(\omega R C)^2 ight]} . \end{aligned} [/latex]

Why is the average power dissipated equal to zero for [latex] f=0 [/latex] ? Does the expression above yield the correct average power dissipated for [latex] f ightarrow \infty [/latex] ?

Question 13.9: A voltage ν(t)=V0cos(ωt) is applied to the series combinati...

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