The e.m.f. and the current of an A.C. circuit are `e=5cos omega t` volt and `I=2sin omega t` ampere respectively. The power consumed in the circuit is

Options

A. zero
B. `10W`
C. `5W`
D. `2.5W`

Answer

To solve the problem, we need to find the power consumed in the AC circuit given the expressions for e.m.f. (e) and current (I).

### Step-by-Step Solution:

1. **Identify the given expressions**:
- The e.m.f. is given by \( e = 5 \cos(\omega t) \) volts.
- The current is given by \( I = 2 \sin(\omega t) \) amperes.

2. **Convert the current expression**:
- We can express the sine function in terms of cosine:
\[ I = 2 \sin(\omega t) = 2 \cos\left(\frac{\pi}{2} - \omega t\right) \]
This shows that the current lags the e.m.f. by \( \frac{\pi}{2} \) radians.

3. **Calculate the root mean square (RMS) values**:
- The peak value of e.m.f. \( V_0 = 5 \) volts, so:
\[ V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{5}{\sqrt{2}} \text{ volts} \]
- The peak value of current \( I_0 = 2 \) amperes, so:
\[ I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \text{ amperes} \]

4. **Determine the phase difference**:
- The phase difference \( \phi \) between e.m.f. and current is given by:
\[ \phi = \frac{\pi}{2} \]

5. **Calculate the power consumed**:
- The average power \( P \) consumed in an AC circuit is given by:
\[ P = V_{rms} \cdot I_{rms} \cdot \cos(\phi) \]
- Substituting the values we found:
\[ P = \left(\frac{5}{\sqrt{2}}\right) \cdot \left(\sqrt{2}\right) \cdot \cos\left(\frac{\pi}{2}\right) \]
- We know that \( \cos\left(\frac{\pi}{2}\right) = 0 \):
\[ P = \left(\frac{5}{\sqrt{2}}\right) \cdot \left(\sqrt{2}\right) \cdot 0 = 0 \]

6. **Final result**:
- Therefore, the power consumed in the circuit is:
\[ P = 0 \text{ watts} \]