Question:medium
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 (\(V_L\)) across \(L\) is:
![Question Figure]()
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 (\(V_L\)) across \(L\) is:
![Question Figure]()
\(I = I_0 \sin \omega t\).
The voltage drop (\(V_L\)) across \(L\) is:
Updated On: Nov 26, 2025
- \(\omega L I_0 \sin \omega t\)
- \(\frac{I_0}{\omega L} \sin \omega t\)
- \(\frac{I_0}{\omega L} \cos \omega t\)
- \(\omega L I_0 \cos \omega t\)