Question 439: AC fundamentals

Question

In the circuit shown below, the inductance $$L$$ is connected to an ac source. The current flowing in the circuit is $$I=I_0 \sin \omega t$$. The voltage drop $$\left(V_L\right)$$ across $$L$$ is

Accepted Answer

$$V_L$$ leads current $$I$$ by $$\frac{\pi}{2}$$
$$\begin{aligned}
& 	herefore V_L=V_0 \sin \left(\omega t+\frac{\pi}{2}ight) \quad\left(\because I=I_0 \sin \omega tight) \
& V_0=I_0 X_L \
& \Rightarrow V_L=I_0 X_L \cos (\omega t)=I_0 \omega L \cos (\omega t)
\end{aligned}$$

Answer

$$\omega L I_0 \sin \omega t$$

Answer

$$\frac{I_0}{\omega L} \sin \omega t$$

Answer

$$\frac{I_0}{\omega L} \cos \omega t$$

Answer

$$\omega L I_0 \cos \omega t$$

AC fundamentals

In the circuit shown below, the inductance $L is connected to an ac source. The current flowing in the circuit is I=I_0 \sin \omega t. The voltage drop \left(V_L\right) across L$ is

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