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MCQEasyJEE 2025LCR Circuits & Resonance

JEE Physics 2025 Question with Solution

In a series LCR circuit, a resistor of 300 Ω300 \, \Omega, a capacitor of 25 nF25 \, \text{nF}, and an inductor of 100 mH100 \, \text{mH} are used. For maximum current in the circuit, the angular frequency of the AC source is _____×104\times 10^4 radians s−1\text{s}^{-1}.

  • A

    22

  • B

    33

  • C

    44

  • D

    55

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: R=300 ΩR = 300 \, \Omega, C=25 nFC = 25 \, \text{nF}, L=100 mHL = 100 \, \text{mH}. Find: The angular frequency for maximum current in the series LCR circuit.

For maximum current in a series LCR circuit, resonance occurs. The resonant angular frequency is

ω0=1LC\omega_0 = \frac{1}{\sqrt{LC}}

Convert the given quantities to SI units:

L=100×10−3=0.1 HL = 100 \times 10^{-3} = 0.1 \, \text{H} C=25×10−9=2.5×10−8 FC = 25 \times 10^{-9} = 2.5 \times 10^{-8} \, \text{F}

Substitute these values into the resonance formula:

ω0=10.1×2.5×10−8\omega_0 = \frac{1}{\sqrt{0.1 \times 2.5 \times 10^{-8}}} =12.5×10−9= \frac{1}{\sqrt{2.5 \times 10^{-9}}}

Now simplify:

2.5×10−9=2.5×10−4.5\sqrt{2.5 \times 10^{-9}} = \sqrt{2.5} \times 10^{-4.5}

So,

ω0≈2×104 rad/s\omega_0 \approx 2 \times 10^4 \, \text{rad/s}

Therefore, the angular frequency is 2×104 rad/s2 \times 10^4 \, \text{rad/s}, so the correct option is A. The solution's lists answer key as option 2, but the extracted solution working clearly gives 2×1042 \times 10^4.

Working from the extracted steps

Given: L=100 mHL = 100 \, \text{mH} and C=25 nFC = 25 \, \text{nF}. Find: The coefficient multiplying 10410^4 in the resonant angular frequency.

Using the extracted approach,

ω0=1LC=10.1×2.5×10−8\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.1 \times 2.5 \times 10^{-8}}} =12.5×10−9= \frac{1}{\sqrt{2.5 \times 10^{-9}}}

Also,

2.5×10−9≈1.58×3.16×10−5≈5×10−5\sqrt{2.5 \times 10^{-9}} \approx 1.58 \times 3.16 \times 10^{-5} \approx 5 \times 10^{-5}

Hence,

ω0≈15×10−5=2×104 rad/s\omega_0 \approx \frac{1}{5 \times 10^{-5}} = 2 \times 10^4 \, \text{rad/s}

Thus the blank is 22. Hence the most defensible option is A.

Common mistakes

  • Using the resistor value RR in the resonance formula is incorrect because the resonant angular frequency of a series LCR circuit depends on LL and CC only. Use ω0=1LC\omega_0 = \frac{1}{\sqrt{LC}}.

  • Failing to convert 100 mH100 \, \text{mH} and 25 nF25 \, \text{nF} into SI units leads to an incorrect power of ten. Convert them to 0.1 H0.1 \, \text{H} and 2.5×10−8 F2.5 \times 10^{-8} \, \text{F} before substitution.

  • Reading the answer as 3×1043 \times 10^4 from the raw key is incorrect here because the extracted solution working gives 2×104 rad/s2 \times 10^4 \, \text{rad/s}. Verify the numerical result from the formula instead of trusting a mismatched key.

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