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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?

The answer shows me that

\$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\$

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  • \$\begingroup\$ Go through power triangle (vectors) and remember the formula: PF is the real (true) power over the apparent power. \$\endgroup\$ – Rohat Kılıç May 3 at 1:48
  • \$\begingroup\$ 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? \$\endgroup\$ – shineele May 3 at 1:56
  • \$\begingroup\$ I just wanted to show you the way so that you can prove it by yourself. Anyway, I'm putting it into an answer. \$\endgroup\$ – Rohat Kılıç May 3 at 2:08
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In series RC, current passes through both R and C.

$$ P=i^2\cdot R \\ Q=i^2\cdot X_C \\ S = \sqrt{P^2+Q^2}=i^2\sqrt{R^2+X_C^2} \\ \therefore PF_s=\frac{P}{S}=\frac {R}{\sqrt{R^2+X_C^2}} $$

In parallel RC (with the same R and C), the supply voltage is applied to both R and C.

$$ P = \frac{V^2}{R} \\ Q=\frac{V^2}{X_C} \\ S = \sqrt{P^2+Q^2}=V^2\sqrt{\frac{1}{R^2}+\frac{1}{X_C^2}}=V^2\cdot \frac{\sqrt{R^2+X_C^2}}{R \cdot X_C} \\ \therefore PF_p=\frac{P}{S}=\frac {X_C}{\sqrt{R^2+X_C^2}} $$

So, finally we obtain $$ PF_s^2 + PF_p^2 = 1 $$

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