For a given series LCR circuit it is found that maximum current is drawn when value of variable capacitance is 25 nF . If resistance of 200 and 100 mH inductor is being used in the given circuit. The frequency of ac

Correct answer: 0 aligned & f_0= 1 2 π L × C & f_0= 1 2 π 100 × 10^ -3 × 25 × 10^ -9 & = 1 2 π 25 × 10^ -10 & =(1)/(2 π × 5) × 10^5 × 10 ~Hz & =(100)/(10) × 10^3 ~Hz & f_0=10 × 10^3 ~Hz aligned

JEE Main 20246 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 6 April
Shift: Evening Shift
Subject: Physics
Chapter: Alternating Current

JEE Main 2024 · 6 April · Evening Shift · Physics · Alternating Current

For a given series LCR circuit it is found that maximum current is...

For a given series LCR circuit it is found that maximum current is drawn when value of variable capacitance is

25 \mathrm{nF}

. If resistance of

200 \Omega

and

100 \mathrm{mH}

inductor is being used in the given circuit. The frequency of ac source is ______

\times 10^3 \mathrm{~Hz}

. (given

\pi^2=10

)

Correct answer

Step-by-step solution

\begin{aligned} & \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 25 \times 10^{-9}}} \\ & =\frac{1}{2 \pi \sqrt{25 \times 10^{-10}}} \\ & =\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}\end{aligned}

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