A small signal voltage `V(t)=V_(0)sin omegat` is applied across an ideal capacitor `C`:

Options

A. over a full cycle the capacitor C does not consume any energy from the voltage source.
B. current I(t) is in phase with voltage V(t)
C. current `I(t)` leads voltage V(t) by `180^(@)`
D. current I(t), lags voltage V(t) by `90^(@)`.

Explanation

To solve the problem, we need to analyze the behavior of a small signal voltage applied across an ideal capacitor. The voltage is given by:

\[ V(t) = V_0 \sin(\omega t) \]

### Step 1: Understand the relationship between voltage and current in a capacitor
For an ideal capacitor, the current \( I(t) \) is related to the voltage \( V(t) \) by the following equation:

\[ I(t) = C \frac{dV(t)}{dt} \]

where \( C \) is the capacitance of the capacitor.

### Step 2: Differentiate the voltage function
Now, we need to differentiate the voltage function \( V(t) \):

\[ V(t) = V_0 \sin(\omega t) \]

Taking the derivative with respect to time \( t \):

\[ \frac{dV(t)}{dt} = V_0 \omega \cos(\omega t) \]

### Step 3: Substitute the derivative into the current equation
Now we can substitute this derivative back into the equation for current:

\[ I(t) = C \frac{dV(t)}{dt} = C \cdot V_0 \omega \cos(\omega t) \]

### Step 4: Express the current in terms of sine function
We can express the cosine function in terms of sine:

\[ I(t) = C \cdot V_0 \omega \cos(\omega t) = C \cdot V_0 \omega \sin\left(\omega t + \frac{\pi}{2}\right) \]

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

### Step 5: Analyze the power consumed by the capacitor
The average power \( P \) consumed by the capacitor over a full cycle can be calculated using the formula:

\[ P = \frac{1}{T} \int_0^T V(t) I(t) dt \]

However, since the current leads the voltage by \( \frac{\pi}{2} \), the average power over a full cycle is zero:

\[ P = V_{rms} I_{rms} \cos(\phi) \]

where \( \phi = \frac{\pi}{2} \). Since \( \cos\left(\frac{\pi}{2}\right) = 0 \):

\[ P = 0 \]

### Conclusion
Thus, the capacitor does not consume any energy from the voltage source over a full cycle.

### Final Answer
The current through the capacitor is given by:

\[ I(t) = C V_0 \omega \cos(\omega t) \]

And the average power consumed by the capacitor is zero.

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