Question 6: AC fundamentals

Question

In a circuit $L, C$ and $R$ are connected in series with an alternating voltage source of frequency $f$. The current leads the voltage by $45^{\circ}$. The value of $C$ is

Accepted Answer

$	herefore 	an \phi=\frac{\omega L-\frac{1}{\omega C}}{R}$
$\phi$ being the angle by which the current leads the voltage.
Given, $\phi=45^{\circ}$
$\begin{aligned}
	herefore \quad & 	an 45^{\circ} =\frac{\omega L-\frac{1}{\omega C}}{R} \Rightarrow 1=\frac{\omega L-\frac{1}{\omega C}}{R} \
& R =\omega L-\frac{1}{\omega C} \Rightarrow \omega C=\frac{1}{\omega L-R} \
\Rightarrow \quad & C =\frac{1}{\omega(\omega L-R)}=\frac{1}{2 \pi f(2 \pi f L-R)}
\end{aligned}$

Answer

$\frac{1}{2 \pi f(2 \pi f L+R)}$

Answer

$\frac{1}{\pi f(2 \pi f L+R)}$

Answer

$\frac{1}{2 \pi f(2 \pi f L-R)}$

Answer

$\frac{1}{\pi f(2 \pi f L-R)}$

AC fundamentals

In a circuit L, C and R are connected in series with an alternating voltage source of frequency f. The current leads the voltage by 45^{\circ}. The value of C is

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