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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