If emf and current in the circult is given by `e= E_(0) sin omega t`
`i=l_(0) sin (omega t- phi)` Then the average power in the circuit over one cycle of ac is

Options

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

Explanation

To find the average power in the circuit over one cycle of AC, we can follow these steps:

### Step 1: Identify the given equations
The electromotive force (emf) and current in the circuit are given by:
- \( e = E_0 \sin(\omega t) \)
- \( i = I_0 \sin(\omega t - \phi) \)

### Step 2: Determine the peak values
From the equations, we can identify:
- \( E_0 \) is the peak emf.
- \( I_0 \) is the peak current.

### Step 3: Calculate the root mean square (RMS) values
The RMS values of voltage and current can be calculated using the following formulas:
- \( V_{rms} = \frac{E_0}{\sqrt{2}} \)
- \( I_{rms} = \frac{I_0}{\sqrt{2}} \)

### Step 4: Use the formula for average power
The average power \( P \) in an AC circuit can be calculated using the formula:
\[ P = V_{rms} \cdot I_{rms} \cdot \cos(\phi) \]

### Step 5: Substitute the RMS values into the power formula
Substituting the RMS values we found in Step 3 into the average power formula:
\[ P = \left(\frac{E_0}{\sqrt{2}}\right) \cdot \left(\frac{I_0}{\sqrt{2}}\right) \cdot \cos(\phi) \]
This simplifies to:
\[ P = \frac{E_0 I_0}{2} \cdot \cos(\phi) \]

### Step 6: Final expression for average power
Thus, the average power in the circuit over one cycle of AC is given by:
\[ P = \frac{E_0 I_0}{2} \cos(\phi) \]