A capacitor of capacity `C` is connected in `A.C` circuit. If the applied emf is `V = V_(0) sin omega t`, then the current is

Options

A. `I = (V_(0))/(L omega) sin omega t`
B. `I = (V_(0))/(omega C) sin (omega t + (pi)/(2))`
C. `I = V_(0) C omega sin omega t`
D. `t = V_(0) C omega sin (omega t + (pi)/(2))`

Answer

To find the current in an AC circuit with a capacitor of capacitance \( C \) when the applied emf is given by \( V = V_0 \sin(\omega t) \), we can follow these steps:

### Step 1: Understand the relationship between voltage and current in a capacitor
In a capacitor, the current \( I \) leads the voltage \( V \) by \( 90^\circ \) (or \( \frac{\pi}{2} \) radians). This means that the current can be expressed as:
\[ I = I_0 \sin(\omega t + \frac{\pi}{2}) \]

### Step 2: Relate the peak current \( I_0 \) to the peak voltage \( V_0 \)
The relationship between the peak voltage \( V_0 \) and the peak current \( I_0 \) in a capacitor is given by:
\[ V_0 = I_0 \cdot X_C \]
where \( X_C \) is the capacitive reactance, defined as:
\[ X_C = \frac{1}{\omega C} \]

### Step 3: Substitute \( X_C \) into the voltage-current relationship
From the equation \( V_0 = I_0 \cdot X_C \), we can express \( I_0 \) as:
\[ I_0 = \frac{V_0}{X_C} = V_0 \cdot \omega C \]

### Step 4: Substitute \( I_0 \) back into the current equation
Now, substituting \( I_0 \) into the equation for current:
\[ I = I_0 \sin(\omega t + \frac{\pi}{2}) = (V_0 \cdot \omega C) \sin(\omega t + \frac{\pi}{2}) \]

### Step 5: Simplify the current equation
Using the property of sine, we can express the current as:
\[ I = V_0 \cdot \omega C \cdot \sin(\omega t + \frac{\pi}{2}) \]

### Final Result
Thus, the current in the AC circuit with a capacitor is:
\[ I = V_0 \cdot \omega C \cdot \sin(\omega t + \frac{\pi}{2}) \]