Emf applied and current in an A.C. circuit are `E=10sqrt(2) sin omega t ` volt and `l=5sqrt(2)cos omega t` ampere respectively. Average power loss in the circuit is

Options

A. 50 W
B. 25 W
C. `(50)/(sqrt(2))W`
D. Zero

Answer

To find the average power loss in the given AC circuit, we will follow these steps:

### Step 1: Identify the given values
The given EMF and current in the AC circuit are:
- EMF: \( E = 10\sqrt{2} \sin(\omega t) \) volts
- Current: \( I = 5\sqrt{2} \cos(\omega t) \) amperes

### Step 2: Calculate the RMS values
The RMS (Root Mean Square) values for voltage and current can be calculated as follows:
- For voltage:
\[ E_{\text{rms}} = \frac{E_0}{\sqrt{2}} = \frac{10\sqrt{2}}{\sqrt{2}} = 10 \text{ volts} \]
- For current:
\[ I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = \frac{5\sqrt{2}}{\sqrt{2}} = 5 \text{ amperes} \]

### Step 3: Determine the phase difference
The phase difference \( \phi \) between the voltage and current can be determined from their sine and cosine forms:
- The voltage is in the form of \( \sin(\omega t) \) and the current is in the form of \( \cos(\omega t) \).
- We can express \( \sin(\omega t) \) as \( \cos(\omega t - \frac{\pi}{2}) \). This indicates that the current lags the voltage by \( \frac{\pi}{2} \).

Thus, the phase difference \( \phi = \frac{\pi}{2} \).

### Step 4: Calculate the power factor
The power factor \( \cos(\phi) \) can now be calculated:
\[ \cos\left(\frac{\pi}{2}\right) = 0 \]

### Step 5: Calculate the average power
The average power \( P \) in an AC circuit is given by the formula:
\[ P = I_{\text{rms}} \cdot E_{\text{rms}} \cdot \cos(\phi) \]
Substituting the values we have:
\[ P = 5 \cdot 10 \cdot 0 = 0 \text{ watts} \]

### Conclusion
The average power loss in the circuit is \( 0 \text{ watts} \).

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