If an alternating current i = `i_(m) sin omegat` is flowing through a capacitor then voltage drop `Delta V_(C)` across capacitor C will be

Options

A. `-(i_("m"))/(omegaC )sin omegat`
B. `-(i_("m"))/(omegaC )cos omegat`
C. `-(i_("m"))/(omegaC)(sinomegat+(pi)/(4))`
D. `(i_("m"))/(omegaC)(sinomegat-(pi)/(4))`

Answer

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

### Step-by-Step Solution:

1. **Identify the Current Expression**:
The alternating current flowing through the capacitor is given by:
\[ i = I_m \sin(\omega t) \]

2. **Understanding the Relationship Between Current and Voltage in a Capacitor**:
The voltage across a capacitor is related to the current through it. The relationship is given by:
\[ i = C \frac{dV_C}{dt} \]
where \( C \) is the capacitance and \( V_C \) is the voltage across the capacitor.

3. **Integrate the Current to Find Voltage**:
Rearranging the equation gives:
\[ dV_C = \frac{i}{C} dt \]
Substituting the expression for \( i \):
\[ dV_C = \frac{I_m \sin(\omega t)}{C} dt \]

4. **Integrate to Find Voltage**:
Integrating both sides:
\[ V_C = \int \frac{I_m \sin(\omega t)}{C} dt \]
The integral of \( \sin(\omega t) \) is:
\[ V_C = -\frac{I_m}{C \omega} \cos(\omega t) + V_0 \]
where \( V_0 \) is the constant of integration.

5. **Expressing the Voltage Drop**:
The voltage drop across the capacitor can be expressed as:
\[ \Delta V_C = -\frac{I_m}{C \omega} \cos(\omega t) \]
This indicates that the voltage across the capacitor lags the current by \( 90^\circ \) (or \( \frac{\pi}{2} \) radians).

6. **Final Expression**:
Thus, the voltage drop across the capacitor is:
\[ \Delta V_C = \frac{I_m}{\omega C} \cos(\omega t) \]

### Summary:
The voltage drop \( \Delta V_C \) across the capacitor when an alternating current \( i = I_m \sin(\omega t) \) flows through it is given by:
\[ \Delta V_C = \frac{I_m}{\omega C} \cos(\omega t) \]