For a given series LCR circuit it is found that maximum current is drawn when value of variable capacitance is 2.5 nF . If resistance of 200 Ω and 100 mH inductor is being used in the given circuit. The frequency of ac source is × 10 3 Hz (given π 2 =10 )

Physics · Alternating Current

JEE Main 2024 — 6 April, Shift 2 — Question 52

For a given series LCR circuit it is found that maximum current is drawn when value of variable capacitance is 2.5 nF . If resistance of $200 \Omega$ and 100 mH inductor is being used in the given circuit. The frequency of ac source is $\qquad$ $\times 10^{3} \mathrm{~Hz}$ (given $\pi^{2}=10$ )

Answer: 10

Numerical answer — enter this value.

Step-by-step solution

for maximum current, circuit must be in resonance.

$\mathrm{f}_{0}=\frac{1}{2 \pi \sqrt{\mathrm{~L} \times \mathrm{C}}}$

$\mathrm{f}_{0}=\frac{1}{2 \pi \sqrt{100 \times 10^{-3} \times 2.5 \times 10^{-9}}}$

$=\frac{1}{2 \pi \sqrt{25 \times 10^{-11}}}$

$=\frac{1}{2 \pi \times 5} \times 10^{5} \times \sqrt{10} \mathrm{~Hz}$

$=\frac{100}{10} \times 10^{3} \mathrm{~Hz}$

$\mathrm{f}_{0}=10 \times 10^{3} \mathrm{~Hz}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 6 April, Shift 2
Subject: Physics
Chapter: Alternating Current
Topic: Series LCR Circuit and Power Factor