The alternating emf of `e=e_(0)` sin `omegat` is applied across capacitor C. the current through the circuit is given by

Options

A. `I=I_(0)sinomegat`
B. `I=I_(0)sin(omegat+(pi)/(2))`
C. `I=I_(0)sin(omegat-(pi)/(2))`
D. `I=I_(0)sin(omegat-pi)`

Answer

To solve the problem, we need to analyze the relationship between the alternating emf applied across a capacitor and the resulting current through the circuit.

### Step-by-Step Solution:

1. **Understand the given emf**: The alternating emf is given by the equation:
\[ e = e_0 \sin(\omega t) \]
where \( e_0 \) is the maximum emf, \( \omega \) is the angular frequency, and \( t \) is time.

2. **Know the relationship between current and voltage in a capacitor**: In a capacitor, the current \( i \) leads the voltage \( v \) across it by a phase angle of \( \frac{\pi}{2} \) radians (or 90 degrees). This means that the current reaches its maximum value a quarter cycle before the voltage does.

3. **Express the voltage across the capacitor**: The voltage across the capacitor can be expressed in terms of the emf:
\[ v = e = e_0 \sin(\omega t) \]

4. **Determine the current through the capacitor**: Since the current leads the voltage by \( \frac{\pi}{2} \), we can express the current as:
\[ i = I_0 \sin(\omega t + \frac{\pi}{2}) \]
where \( I_0 \) is the maximum current.

5. **Use the trigonometric identity**: We know that:
\[ \sin(x + \frac{\pi}{2}) = \cos(x) \]
Therefore, we can rewrite the current as:
\[ i = I_0 \cos(\omega t) \]

6. **Relate the maximum current to the maximum voltage**: The maximum current \( I_0 \) can be related to the maximum voltage \( e_0 \) and the capacitance \( C \) using the formula:
\[ I_0 = C \frac{de}{dt} \]
The derivative of the voltage \( e \) with respect to time \( t \) is:
\[ \frac{de}{dt} = e_0 \omega \cos(\omega t) \]
Therefore, the maximum current becomes:
\[ I_0 = C e_0 \omega \]

7. **Final expression for the current**: Substituting \( I_0 \) back into the expression for current, we get:
\[ i = C e_0 \omega \cos(\omega t) \]

### Conclusion:
The current through the circuit when an alternating emf \( e = e_0 \sin(\omega t) \) is applied across a capacitor \( C \) is given by:
\[ i = C e_0 \omega \cos(\omega t) \]