NVAEasyJEE 2023LCR Circuits & Resonance

JEE Physics 2023 Question with Solution

A series LCR circuit is connected to an AC source of 220V220 \, \text{V}, 50Hz50 \, \text{Hz}. The circuit contains a resistance R=80ΩR = 80 \, \Omega, an inductor of inductive reactance XL=70ΩX_L = 70 \, \Omega, and a capacitor of capacitive reactance XC=130ΩX_C = 130 \, \Omega. The power factor of the circuit is x/10x/10. The value of xx is:

Answer

Correct answer:8

Step-by-step solution

Standard Method

Given: A series LCR circuit has R=80ΩR = 80 \, \Omega, XL=70ΩX_L = 70 \, \Omega, and XC=130ΩX_C = 130 \, \Omega.

Find: The value of xx if the power factor is x/10x/10.

For a series LCR circuit, the power factor is

cosϕ=RZ\cos \phi = \frac{R}{Z}

where the impedance is

Z=R2+(XLXC)2Z = \sqrt{R^2 + (X_L - X_C)^2}

Step-by-step Calculation

Substituting the given values,

cosϕ=80802+(70130)2\cos \phi = \frac{80}{\sqrt{80^2 + (70 - 130)^2}}

Now simplify the reactance difference,

70130=6070 - 130 = -60

So,

cosϕ=80802+(60)2\cos \phi = \frac{80}{\sqrt{80^2 + (-60)^2}}

Evaluating further,

cosϕ=806400+3600=8010000=80100=0.8\cos \phi = \frac{80}{\sqrt{6400 + 3600}} = \frac{80}{\sqrt{10000}} = \frac{80}{100} = 0.8

Relating to x/10

Since the power factor is given as x/10x/10, compare directly:

x10=0.8=810\frac{x}{10} = 0.8 = \frac{8}{10}

Therefore, x=8x = 8. The numerical answer is 8.

Common mistakes

  • Using XL+XCX_L + X_C instead of XLXCX_L - X_C for net reactance. In a series LCR circuit, inductive and capacitive reactances oppose each other. Use XLXCX_L - X_C or XCXLX_C - X_L inside the square term.

  • Writing the power factor as Z/RZ/R instead of R/ZR/Z. Power factor is cosϕ=R/Z\cos \phi = R/Z, so reversing the ratio gives a value greater than 11, which is not possible.

  • Ignoring the square on the reactance difference. The impedance formula is Z=R2+(XLXC)2Z = \sqrt{R^2 + (X_L - X_C)^2}. Do not substitute XLXCX_L - X_C without squaring it.

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