JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Alternating Current

An inductor of inductance

2 μ H

is connected in series with a resistance, a variable capacitor and an AC source of frequency 7 kHz. The value of capacitance for which maximum current is drawn into the circuit is

\frac{1}{x} F

, where the value of

x

is ______.
(Take

π = \frac{22}{7}

)

Answer: 3872

Solution

The current is maximum at resonance, where the inductive and capacitive reactances cancel,

2 π f L = \frac{1}{2 π f C}

, so

C = \frac{1}{4 π ^{2} f ^{2} L}

. Substituting

L = 2 \times 1 0^{- 6}

f = 7 \times 1 0^{3}

Hz and

π = \frac{22}{7}

C = \frac{1}{4 ( \frac{22}{7} ) ^{2} ( 7 \times 1 0 ^{3} ) ^{2} ( 2 \times 1 0 ^{- 6} )} = \frac{1}{4 \times \frac{484}{49} \times 49 \times 1 0 ^{6} \times 2 \times 1 0 ^{- 6}} = \frac{1}{3872}

F. The capacitance for maximum current is

\frac{1}{3872}

F, so

x = 3872

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Integer answer

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JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Alternating Current

Solution

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