In any `AC` circuit the emf `(e)` and the current `(i)` at any instant are given respectively by `e= E_(0)sin omega t`
`i=I_(0) sin (omegat-phi)`
The average power in the circuit over one cycle of `AC` is

Options

A. `(E_(0)I_(0))/2`
B. `(E_(0)I_(0))/2 sin phi`
C. `(E_(0)I_(0))/2 cos phi`
D. `E_(0)I_(0)`

Explanation

To find the average power in an AC circuit where the emf \( e \) and the current \( i \) are given by:

\[ e = E_0 \sin(\omega t) \]
\[ i = I_0 \sin(\omega t - \phi) \]

we can follow these steps:

### Step 1: Understand the Power Formula
The average power \( P \) in an AC circuit can be expressed as:
\[ P = V_{\text{rms}} \cdot I_{\text{rms}} \cdot \cos(\phi) \]
where \( V_{\text{rms}} \) is the root mean square (RMS) voltage, \( I_{\text{rms}} \) is the root mean square (RMS) current, and \( \phi \) is the phase difference between the voltage and the current.

### Step 2: Calculate RMS Values
The RMS values of the peak voltage \( E_0 \) and peak current \( I_0 \) are given by:
\[ V_{\text{rms}} = \frac{E_0}{\sqrt{2}} \]
\[ I_{\text{rms}} = \frac{I_0}{\sqrt{2}} \]

### Step 3: Substitute RMS Values into Power Formula
Substituting the RMS values into the power formula gives:
\[ P = \left(\frac{E_0}{\sqrt{2}}\right) \cdot \left(\frac{I_0}{\sqrt{2}}\right) \cdot \cos(\phi) \]

### Step 4: Simplify the Expression
Now, simplify the expression:
\[ P = \frac{E_0 I_0}{2} \cdot \cos(\phi) \]

### Final Result
Thus, the average power in the circuit over one cycle of AC is:
\[ P = \frac{E_0 I_0}{2} \cos(\phi) \]