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 quantities
The EMF (E) and current (I) in the circuit are given as:
- \( E = 10\sqrt{2} \sin(\omega t) \) volts
- \( I = 5\sqrt{2} \cos(\omega t) \) amperes

### Step 2: Convert the current to sine form
We can express the current in terms of sine function to match the form of the EMF. The cosine function can be converted to sine using the identity:
\[ \cos(\omega t) = \sin\left(\omega t + \frac{\pi}{2}\right) \]
Thus, we can rewrite the current as:
\[ I = 5\sqrt{2} \cos(\omega t) = 5\sqrt{2} \sin\left(\omega t + \frac{\pi}{2}\right) \]

### Step 3: Determine the phase difference
From the expressions for EMF and current:
- The EMF \( E \) has a phase angle of \( 0 \) (since it is \( \sin(\omega t) \)).
- The current \( I \) has a phase angle of \( \frac{\pi}{2} \).

The phase difference \( \phi \) between the EMF and the current is:
\[ \phi = \frac{\pi}{2} - 0 = \frac{\pi}{2} \]

### Step 4: Calculate the RMS values
The RMS (Root Mean Square) values for voltage and current are 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 5: Calculate the average power loss
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:
- \( I_{\text{rms}} = 5 \)
- \( E_{\text{rms}} = 10 \)
- \( \cos\left(\frac{\pi}{2}\right) = 0 \)

Thus, the average power loss is:
\[ P = 5 \cdot 10 \cdot 0 = 0 \text{ watts} \]

### Final Answer
The average power loss in the circuit is \( 0 \) watts.

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