"In a series LCR circuit, the inductive reactance (XL) is 10$\Omega$ and the capacitive reactance (XC) is 4$\Omega$. The resistance (R) in the circuit is 6$\Omega$. The power factor of the circuit is :"

Question

Accepted Answer

"Given :

X_L = 10$\Omega$

X_C = 4$\Omega$

R = 6$\Omega$

$\therefore$ Power factor = cos$\theta$ = ${R \over Z}$

$= {R \over {\sqrt {{R^2} + {{({X_L} - {X_C})}^2}} }}$

$= {6 \over {\sqrt {{6^2} + {{(10 - 4)}^2}} }}$

$= {6 \over {6\sqrt 2 }} = {1 \over {\sqrt 2 }}$"

In a series LCR circuit, the inductive reactance (X_L) is 10$\Omega$ and the capacitive reactance (X_C) is 4$\Omega$. The resistance (R) in the circuit is 6$\Omega$. The power factor of the circuit is :

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In a series LCR circuit, the inductive reactance (XL) is 10$\Omega$ and the capacitive reactance (XC) is 4$\Omega$. The resistance (R) in the circuit is 6$\Omega$. The power factor of the circuit is :

Solution

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In a series LCR circuit, the inductive reactance (X_L) is 10$\Omega$ and the capacitive reactance (X_C) is 4$\Omega$. The resistance (R) in the circuit is 6$\Omega$. The power factor of the circuit is :