In an LCR series circuit, an inductor 30 mH and a resistor \(1 \Omega\) are connected to an AC source of angular frequency \(300 \mathrm{rad} / \mathrm{s}\). The value of capacitance for which, the current leads the voltage by \(45^{\circ}\) is \(\frac{1}{x} \times 10^{-3} \mathrm{~F}\). Then the value of x is \_\_\_\_.

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Alternating Current

In an LCR series circuit, an inductor 30 mH and a resistor

1Ω

are connected to an AC source of angular frequency

300 rad / s

. The value of capacitance for which, the current leads the voltage by

4 5^{\circ}

\frac{1}{x} \times 1 0^{- 3} F

. Then the value of x is ____.

Answer: 3

The current leads the voltage, so the capacitive reactance exceeds the inductive one and the phase angle satisfies

tan ϕ = \frac{X _{C} - X _{L}}{R}

. The inductive reactance is

X_{L} = ω L = 300 \times 30 \times 1 0^{- 3} = 9 Ω

. For

ϕ = 4 5^{\circ}

tan ϕ = 1

, so

X_{C} - X_{L} = R

and

X_{C} = 9 + 1 = 10 Ω

. From

X_{C} = \frac{1}{ω C}

C = \frac{1}{ω X _{C}} = \frac{1}{300 \times 10} = \frac{1}{3000} F = \frac{1}{3} \times 1 0^{- 3} F

Comparing with

\frac{1}{x} \times 1 0^{- 3}

F gives

x = 3

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Integer answer