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MCQEasyJEE 2023LCR Circuits & Resonance

JEE Physics 2023 Question with Solution

In a series LRLR circuit with XL=RX_L = R, the power factor is P1P_1. If a capacitor of capacitance CC with XC=XLX_C = X_L is added to the circuit, the power factor becomes P2P_2. The ratio of P1P_1 to P2P_2 will be:

  • A

    3:13:1

  • B

    1:21:\sqrt{2}

  • C

    1:11:1

  • D

    1:21:2

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: A series LRLR circuit has XL=RX_L = R. After adding a capacitor in series, XC=XLX_C = X_L.

Find: The ratio P1:P2P_1 : P_2 of the initial and final power factors.

For the initial LRLR circuit, the impedance is

Z=R2+R2=2RZ = \sqrt{R^2 + R^2} = \sqrt{2}R

So the initial power factor is

P1=cos⁡ϕ=RZ=R2R=12P_1 = \cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{2}R} = \frac{1}{\sqrt{2}}

When the capacitor is added with XC=XLX_C = X_L, the net reactance becomes zero, so the circuit is in resonance and

Z=RZ = R

Hence the new power factor is

P2=cos⁡ϕ=RZ=RR=1P_2 = \cos \phi = \frac{R}{Z} = \frac{R}{R} = 1

Therefore,

P1P2=121=12\frac{P_1}{P_2} = \frac{\frac{1}{\sqrt{2}}}{1} = \frac{1}{\sqrt{2}}

So,

P1:P2=1:2P_1 : P_2 = 1 : \sqrt{2}

the solution marks B as correct. However, its final written line states 1:21:2, which is inconsistent with the derived result. Therefore, the correct option is B.

Common mistakes

  • Using the ratio 12:1\frac{1}{\sqrt{2}} : 1 as 1:21:2 is incorrect because rationalizing or squaring is not needed in a simple ratio comparison. Write it directly as 1:21:\sqrt{2}.

  • Forgetting that when XC=XLX_C = X_L in a series circuit, the net reactance is zero is wrong. At resonance, the impedance becomes only RR, so the power factor becomes 11.

  • Calculating impedance of the initial LRLR circuit as R+XLR+X_L is incorrect because resistance and reactance combine vectorially. Use Z=R2+XL2Z = \sqrt{R^2 + X_L^2}, and with XL=RX_L = R this gives 2R\sqrt{2}R.

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