If an alternating current i = `i_(m )sinomegat` is flowing through an inductor then voltage drop `DeltaV_(L)` across inductor L will be-

Options

A. `i_(m)omega L sin omegat`
B. `i_(m) omegaL cos omegat`
C. `i_(m) omegaL sin (omegaL +(pi)/(4))`
D. `i_(m) omegaL cos (omegaL -(pi)/(4))`

Answer

To find the voltage drop \( \Delta V_L \) across an inductor \( L \) when an alternating current \( i = I_m \sin(\omega t) \) is flowing through it, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship Between Current and Voltage in an Inductor**:
The voltage across an inductor is related to the rate of change of current through it. The formula for the voltage across an inductor is given by:
\[ V_L = L \frac{di}{dt} \]

2. **Differentiate the Current**:
Given the current \( i = I_m \sin(\omega t) \), we need to differentiate this with respect to time \( t \):
\[ \frac{di}{dt} = I_m \omega \cos(\omega t) \]

3. **Substitute into the Voltage Formula**:
Now, substitute \( \frac{di}{dt} \) back into the voltage formula:
\[ V_L = L \cdot I_m \omega \cos(\omega t) \]

4. **Express the Voltage in Terms of Sine**:
Since \( \cos(\omega t) \) can be expressed in terms of sine as:
\[ \cos(\omega t) = \sin\left(\omega t + \frac{\pi}{2}\right) \]
We can rewrite the voltage across the inductor:
\[ V_L = L \cdot I_m \omega \sin\left(\omega t + \frac{\pi}{2}\right) \]

5. **Final Expression**:
Thus, the voltage drop across the inductor \( \Delta V_L \) is:
\[ \Delta V_L = L I_m \omega \sin\left(\omega t + \frac{\pi}{2}\right) \]

### Conclusion:
The voltage drop across the inductor is given by:
\[ \Delta V_L = L I_m \omega \sin\left(\omega t + \frac{\pi}{2}\right) \]