Alternating Current

JEE Main level, numerical answer, medium

31 of 190

Question 31 of 190

An AC circuit has an inductor and a resistor of resistance RR in series, such that XL=3R{X}_{L}=3R. Now, a capacitor is added in series such that XC=2R{X}_{C}=2R. the ratio of the new power factor with the old power factor of the circuit is 5:x\sqrt{5}:x. The value of xx is

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Answer: 1

First Recall the formula of power factor, cos⁡ϕ=RZ\cos \phi =\dfrac{R}{Z}⇒cos⁡ϕinitial=RZ=R(XL)2+R2=R9R2+R2=110\Rightarrow \cos {\phi }_{initial}=\dfrac{R}{Z}=\dfrac{R}{\sqrt{{\left({X}_{L}\right)}^{2}+{R}^{2}}}=\dfrac{R}{\sqrt{9{R}^{2}+{R}^{2}}}=\dfrac{1}{\sqrt{10}}.
⇒cos⁡ϕfinal=RZ=R(XL−XC)2+R2=RR2+R2=12\Rightarrow \cos {\phi }_{final}=\dfrac{R}{Z}=\dfrac{R}{\sqrt{{\left({X}_{L}-{X}_{C}\right)}^{2}+{R}^{2}}}=\dfrac{R}{\sqrt{{R}^{2}+{R}^{2}}}=\dfrac{1}{\sqrt{2}}
⇒cos⁡ϕfinalcos⁡ϕinitial=5x⇒x=1\Rightarrow \dfrac{\cos {\phi }_{final}}{\cos {\phi }_{initial}}=\dfrac{\sqrt{5}}{x}\Rightarrow x=1

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