Q.In a circuit L, C and R are connected in series with an alternating voltage of frequency f. The current leads the voltage by $45^\circ$. The value of C is (A) $\dfrac{1}{2\pi f(2\pi fL - R)}$ (B) $\dfrac{1}{2\pi f(2\pi fL + R)}$ (C) $\dfrac{1}{2\pi f\cdot R}$ (D) $\dfrac{1}{(2\pi f)^2 L}$

[!TLDR] The current LEADS the voltage, so the circuit is capacitance-dominated: $\tan45^\circ=1=\dfrac{X_C-X_L}{R}\Rightarrow X_C=2\pi fL+R\Rightarrow C=\dfrac …

Since the current leads the voltage, the circuit is capacitance-dominated, so $X_C>X_L$ and the phase angle satisfies $\tan\phi=\dfrac{X_C-X_L}{R}$ (current-leading convention). With $\phi=45^\circ$, $\tan45^\circ=1$, so $X_C-X_L=R$, giving $X_C=X_L+R=2\pi fL+R$. Since $X_C=\dfrac{1}{2\pi fC}$, in …

Since the current leads, use $\tan\phi=(X_C-X_L)/R$ with $\phi=45^\circ$ to solve for $X_C$ in terms of R and $X …

Using the voltage-leading sign convention $\tan\phi=(X_L-X_C)/R$ instead of the current-leading one, which flips the sign in …