JEE MAIN - Physics (2021 - 18th March Evening Shift - No. 4)
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 :
$${1 \over 2}$$
$${{\sqrt 3 } \over 2}$$
$${1 \over {\sqrt 2 }}$$
$${1 \over {2\sqrt 2 }}$$
Explanation
Given :
XL = 10$$\Omega$$
XC = 4$$\Omega$$
R = 6$$\Omega$$
$$ \therefore $$ Power factor = cos$$\theta$$ = $${R \over Z}$$
$$ = {R \over {\sqrt {{R^2} + {{({X_L} - {X_C})}^2}} }}$$
$$ = {6 \over {\sqrt {{6^2} + {{(10 - 4)}^2}} }}$$
$$ = {6 \over {6\sqrt 2 }} = {1 \over {\sqrt 2 }}$$
XL = 10$$\Omega$$
XC = 4$$\Omega$$
R = 6$$\Omega$$
$$ \therefore $$ Power factor = cos$$\theta$$ = $${R \over Z}$$
$$ = {R \over {\sqrt {{R^2} + {{({X_L} - {X_C})}^2}} }}$$
$$ = {6 \over {\sqrt {{6^2} + {{(10 - 4)}^2}} }}$$
$$ = {6 \over {6\sqrt 2 }} = {1 \over {\sqrt 2 }}$$
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