AC Circuits

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Question

In series LR circuit , $X_{L} = 3 R$ . Now a capacitor with $X_{C} = R$ is added in series. The ratio of new to old power factor is_________

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Solution

Given
$X_{L} = 3 R$
$X_{C} = R$
We know impedance $Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}$
$\begin{array}{l} I n L C c i r c u i t, Z_{1} = \sqrt{{(3 R)}^{2} + R^{2}} = \sqrt{10} R \\ I n L C R c i r c u i t, Z_{2} = \sqrt{{(3 R - R)}^{2} + R^{2}} = \sqrt{5} R \\ A s p o w e r f a c t o r = \frac{R}{Z} \\ ∴ Power factor \propto \frac{1}{Z} \\ \Rightarrow \frac{{(Power factor)}_{2}}{{(Power factor)}_{1}} = \frac{Z_{1}}{Z_{2}} \\ = \frac{\sqrt{10} R}{\sqrt{5} R} = \sqrt{2} = 1.414 \end{array}$

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AC Circuits

Remember concepts with our Masterclasses.

In series LR circuit , XL=3R. Now a capacitor with XC=R is added in series. The ratio of new to old power factor is_________

GivenXL=3R XC=R We know impedance Z=R2+(XL−XC)2In LC circuit, Z1=(3R)2+R2 =10R In LCR circuit, Z2=(3R−R)2+R2 =5RAs power factor =RZ ∴Power factor∝1Z ⇒(Power factor)2(Power factor)1=Z1Z2=10R5R =2=1.414

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In series LR circuit , $X_{L} = 3 R$ . Now a capacitor with $X_{C} = R$ is added in series. The ratio of new to old power factor is_________