An a.c. voltage is applied to a pure inductor L, drives a current in the inductor. The current in the inductor would be

Options

A. ahead voltage by π/2
B. lagging voltage by π/2
C. ahead voltage by 3π/2
D. lagging voltage by 3π/2

Answer

To solve the problem of the current in a pure inductor when an AC voltage is applied, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**:
We have a pure inductor (L) connected to an alternating current (AC) voltage source. The current flowing through the inductor will be sinusoidal due to the nature of AC voltage.

2. **Express the Current**:
The current (I) in the inductor can be expressed as:
\[ I(t) = I_0 \sin(\omega t) \]
where \(I_0\) is the maximum current and \(\omega\) is the angular frequency of the AC source.

3. **Magnetic Flux in the Inductor**:
The magnetic flux (\(\Phi\)) linked with the inductor is given by:
\[ \Phi = L \cdot I \]
Substituting the expression for current, we have:
\[ \Phi = L \cdot I_0 \sin(\omega t) \]

4. **Differentiate the Flux**:
To find the induced electromotive force (EMF), we differentiate the magnetic flux with respect to time:
\[ \frac{d\Phi}{dt} = L \cdot I_0 \cdot \frac{d}{dt}(\sin(\omega t)) = L \cdot I_0 \cdot \omega \cos(\omega t) \]

5. **Relate EMF to Voltage**:
The induced EMF (E) in the inductor is equal to the rate of change of magnetic flux:
\[ E = -\frac{d\Phi}{dt} = -L \cdot I_0 \cdot \omega \cos(\omega t) \]
Since we are interested in the magnitude, we can write:
\[ V = L \cdot I_0 \cdot \omega \cos(\omega t) \]

6. **Express Voltage in Sinusoidal Form**:
The voltage can also be expressed in terms of sine:
\[ V(t) = V_0 \sin(\omega t + \frac{\pi}{2}) \]
where \(V_0 = L \cdot I_0 \cdot \omega\). The \(\frac{\pi}{2}\) phase shift indicates that the voltage leads the current.

7. **Conclusion on Phase Relationship**:
From the above expressions, we can conclude that:
- The current \(I(t) = I_0 \sin(\omega t)\)
- The voltage \(V(t) = V_0 \sin(\omega t + \frac{\pi}{2})\)

This shows that the current lags behind the voltage by \(\frac{\pi}{2}\) radians (or 90 degrees).

### Final Answer:
The current in the inductor would be lagging the voltage by \(\frac{\pi}{2}\).