Q82 / 90 · JEE Main 2021 (20 Jul Shift 1) · Physics

In an $L C R$ series circuit, an inductor $30 mH$ and a resistor $1 Ω$ are connected to an $AC$ source of angular frequency $300 rad s^{- 1} .$ The value of capacitance for which the current leads the voltage by $45 °$ is $\frac{1}{x} \times 10^{- 3} F$ . Then the value of $x$ is

Correct answer: 3

Solution

\tan ϕ = \frac{x_{C} - x_{L}}{R}

\tan 45 = \frac{x_{C} - x_{L}}{R}

x_{C} - x_{L} = R

\frac{1}{ω C} - ω L = R

\frac{1}{ω C} - 300 \times 0.03 = 1

\frac{1}{ω C} = 10

C = \frac{1}{10 ω} = \frac{1}{10 \times 300}

C = \frac{1}{3} \times 10^{- 3}

X = 3

Hence, the correct answer is 3.

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