In an a.c. circuit the e.m.f. € and the current (i) al any instant are-given respectively by
`e = E_(0)sin omegat`
`I = l_(0)sin (omegat-phi)`
The average power in the circuit over one cycle of a.c. is

Options

A. `E_(0)l_(0)`
B. `(E_(0)l_(0))/(2)`
C. `(E_(0)l_(0))/(2)sinphi`
D. `(E_(0)l_(0))/(2)cosphi`

Answer

To find the average power in an AC circuit given the equations for e.m.f. and current, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given equations**:
- The e.m.f. is given by:
\[ e = E_0 \sin(\omega t) \]
- The current is given by:
\[ I = I_0 \sin(\omega t - \phi) \]

2. **Determine the RMS values**:
- The RMS (Root Mean Square) value of the e.m.f. is:
\[ E_{rms} = \frac{E_0}{\sqrt{2}} \]
- The RMS value of the current is:
\[ I_{rms} = \frac{I_0}{\sqrt{2}} \]

3. **Calculate the average power**:
- The average power \( P \) in an AC circuit is given by the formula:
\[ P = E_{rms} \cdot I_{rms} \cdot \cos(\phi) \]
- Substituting the RMS values we found:
\[ P = \left(\frac{E_0}{\sqrt{2}}\right) \cdot \left(\frac{I_0}{\sqrt{2}}\right) \cdot \cos(\phi) \]

4. **Simplify the expression**:
- This simplifies to:
\[ P = \frac{E_0 I_0 \cos(\phi)}{2} \]

5. **Final result**:
- The average power in the circuit over one cycle of AC is:
\[ P = \frac{E_0 I_0 \cos(\phi)}{2} \]