Electromagnetic Induction

JEE Main level, single correct, medium

149 of 153

Question 149 of 153

In a series LRLR circuit XL=R{X}_{L}=R and power factor of the circuit is P1{P}_{1}. When capacitor with capacitance CC such that XL=XC{X}_{L}={X}_{C} is put in series, the power factor becomes P2{P}_{2}. The ratio P1P2\dfrac{{P}_{1}}{{P}_{2}} is

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Answer: (B) 12\dfrac{1}{\sqrt{2}}

In case of L−RL-R circuit
Z=XL2+R2Z=\sqrt{{{X}_{L}}^{2}+{R}^{2}} Therefore, power factor
P1=cos⁡ϕ=RZ{P}_{1}=\cos \phi =\dfrac{R}{Z}
As XL=R{X}_{L}=R ⇒Z=2R\Rightarrow Z=\sqrt{2}R
⇒P1=R2R⇒P1=12\Rightarrow {P}_{1}=\dfrac{R}{\sqrt{2}R}\Rightarrow {P}_{1}=\dfrac{1}{\sqrt{2}}
In case of L−C−RL-C-R circuit
Z=R2+(XL−XC)2Z=\sqrt{{R}^{2}+{\left({X}_{L}-{X}_{C}\right)}^{2}}
According to the question XL=XC{X}_{L}={X}_{C} ⇒Z=R\Rightarrow Z=R ⇒P2=cos⁡ϕ=RR=1\Rightarrow {P}_{2}=\cos \phi =\dfrac{R}{R}=1
⇒P1P2=12\Rightarrow \dfrac{{P}_{1}}{{P}_{2}}=\dfrac{1}{\sqrt{2}}

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