Question 319: AC fundamentals

Question

With an alternating voltage source frequency ' f ', inductor ' $L$ ', capacitor ' $C$ ' and resistance ' $R$ ' are connected in series. The voltage leads the current by $45^{\circ}$. The value of ' $L$ ' is $\left(	an 45^{\circ}=1ight)$

Accepted Answer

The phase difference between the current and
the voltage is given by $	an \phi=\frac{\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}}{\mathrm{R}}$
$$\begin{array}{ll}
	herefore & \omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}=\mathrm{R} \quad \ldots\left(\because 	an \phi=	an 45^{\circ}=1ight) \
	herefore & \omega \mathrm{L}=\mathrm{R}+\frac{1}{\omega \mathrm{C}} \
	herefore & \mathrm{~L}=\frac{\mathrm{R}}{\omega}+\frac{1}{\omega^2 \mathrm{C}}=\frac{\mathrm{R} \omega \mathrm{C}+1}{\omega^2 \mathrm{C}} \
	herefore & \mathrm{~L}=\frac{1+2 \pi \mathrm{fCR}}{4 \pi^2 \mathrm{f}^2 \mathrm{C}} \quad \ldots .(\because \omega=2 \pi \mathrm{f})
\end{array}$$

Answer

$\left(\frac{1+2 \pi \mathrm{fCR}}{4 \pi^2 \mathrm{f}^2 \mathrm{C}}\right)$

Answer

$\left(\frac{1-2 \pi f C R}{4 \pi^2 \mathrm{f}^2 \mathrm{C}}\right)$

Answer

$\left(\frac{4 \pi^2 \mathrm{f}^2 \mathrm{C}}{1+2 \pi \mathrm{fCR}}\right)$

Answer

$\left(\frac{4 \pi^2 \mathrm{f}^2 \mathrm{C}}{1-2 \pi \mathrm{fCR}}\right)$

AC fundamentals

With an alternating voltage source frequency ' f ', inductor ' L ', capacitor ' C ' and resistance ' R ' are connected in series. The voltage leads the current by 45^{\circ}. The value of ' L ' is \left(\tan 45^{\circ}=1\right)

Choose the correct answer: