Q 12-07-120JEE MainJEE Main 2021 (20 Jul, Shift 1)Easy
In an $LCR$ series circuit, an inductor 30 mH and a resistor $1\ \Omega$ are connected to an AC source of angular frequency $300\ \text{rad s}^{-1}$. The value of capacitance for which the current leads the voltage by $45^\circ$ is $\dfrac1x\times10^{-3}$ F. Then the value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 3
Current leading by $45^\circ$: $\tan45^\circ = \dfrac{X_C - X_L}{R} = 1 \Rightarrow X_C = X_L + R$.
$X_L = 300\times0.03 = 9\ \Omega$, so $X_C = 10\ \Omega$ and $C = \dfrac{1}{300\times10} = \dfrac13\times10^{-3}$ F. So $x = 3$.
Solution by Sreeraj P, M.Sc Physics