Q159 / 160 · TS EAMCET 2015 · Physics
In a circuit $L, C$ and $R$ are connected in series with an alternating voltage source of frequency $f$. When current in the circuit leads the voltage by $45^{\circ}$, the value of $C$ is
- Correct answerA$\frac{1}{2 \pi f(2 \pi f L+R)}$
- B$\frac{1}{2 \pi f(2 \pi f R+L)}$
- C$\frac{2}{2 \pi f(R+L)}$
- D$\frac{2}{2 \pi f\left(R+\frac{1}{L}\right)}$
Solution
Current leads the voltage by $45^{\circ}$
$$
\begin{array}{rlrl}
& \therefore & \phi & =45^{\circ} \\
\text { Since, }, & \tan \phi & =\frac{X_C-X_L}{R} \\
\Rightarrow & 1 & =\frac{X_C-X_L}{R}
\end{array}
$$
$$
\begin{aligned}
\Rightarrow & \frac{1}{\omega C}-\omega L & =R \\
\Rightarrow & \quad \frac{1}{\omega C} & =R+\omega L \\
\Rightarrow & C & =\frac{1}{\omega[R+\omega L]} \\
\Rightarrow & C & =\frac{1}{2 \pi f[2 \pi f L+R]}
\end{aligned}
$$
Hence, the correct answer is option: 'A'.