# The proof of $PF_{p}=cos\theta_p=sin\theta_s$

There is a Series RC circuit ,which power factor is $$\\frac{\sqrt{3}}{2}\$$,now if i modify this "Series" RC circuit to "parallel" RC circuit,then what is the value of power factor of this circuit?

$$\PF_{p}=cos\theta_p=sin\theta_s=\frac{1}{2}=0.5\$$.does anyone know its proof?Why can this ,$$\cos\theta_p=sin\theta_s\$$,theorem hold?

$$\PF_{p}\$$:power factor of parallel RC circuit

$$\\theta_p\$$ : $$\\theta\$$ of parallel RC circuit, $$\\theta=\theta_v-\theta_i\$$

$$\\theta_s\$$ : $$\\theta\$$ of Series RC circuit, $$\\theta=\theta_v-\theta_i\$$

• Go through power triangle (vectors) and remember the formula: PF is the real (true) power over the apparent power. – Rohat Kılıç May 3 at 1:48
• i think you are explaining $PF_p=cos\theta_p$,i know this.the question i really want to ask is this :$cos\theta_p=sin\theta_s$,do you know why? – shineele May 3 at 1:56
• I just wanted to show you the way so that you can prove it by yourself. Anyway, I'm putting it into an answer. – Rohat Kılıç May 3 at 2:08

$$P=i^2\cdot R \\ Q=i^2\cdot X_C \\ S = \sqrt{P^2+Q^2}=i^2\sqrt{R^2+X_C^2} \\ \therefore PF_s=\frac{P}{S}=\frac {R}{\sqrt{R^2+X_C^2}}$$
$$P = \frac{V^2}{R} \\ Q=\frac{V^2}{X_C} \\ S = \sqrt{P^2+Q^2}=V^2\sqrt{\frac{1}{R^2}+\frac{1}{X_C^2}}=V^2\cdot \frac{\sqrt{R^2+X_C^2}}{R \cdot X_C} \\ \therefore PF_p=\frac{P}{S}=\frac {X_C}{\sqrt{R^2+X_C^2}}$$
So, finally we obtain $$PF_s^2 + PF_p^2 = 1$$