Alternating Current

JEE Main level, numerical answer, medium

53 of 190

Question 53 of 190

A series LCR circuit with R=20 ΩR = 20\ \Omega, L=1.6 HL = 1.6\text{ H} and C=40 μFC = 40\ \mu\text{F} is connected to a variable frequency a.c. source. The inductive reactance at resonant frequency is _______ Ω\Omega.

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Answer: 200

Given L=1.6 HL = 1.6\text{ H} and C=40 μF=40×10−6 FC = 40\ \mu\text{F} = 40 \times 10^{-6}\text{ F}.

At resonance, the angular frequency is ωr=1LC\omega_r = \dfrac{1}{\sqrt{LC}}.

The inductive reactance at resonance is XL=ωrL=LLC=LCX_L = \omega_r L = \dfrac{L}{\sqrt{LC}} = \sqrt{\dfrac{L}{C}}.

Substituting the given values:

XL=1.640×10−6X_L = \sqrt{\dfrac{1.6}{40 \times 10^{-6}}}

XL=1.64×10−5X_L = \sqrt{\dfrac{1.6}{4 \times 10^{-5}}}

XL=4×104X_L = \sqrt{4 \times 10^4}

XL=200 ΩX_L = 200\ \Omega

Answer: 200200

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