Voltage input in a circuit is V = 300 `sin (omega t)` with current = 100 `cos(omega t)` . Average power loss in the circuit is

Options

A. 3000 units
B. 1500 units
C. 1000 units
D. Zero

Answer

To find the average power loss in the circuit given the voltage and current equations, we can follow these steps:

### Step 1: Identify the given voltage and current equations
The voltage input in the circuit is given as:
\[ V(t) = 300 \sin(\omega t) \]
The current is given as:
\[ I(t) = 100 \cos(\omega t) \]

### Step 2: Convert the voltage equation to cosine form
To find the phase difference, we can express the sine function in terms of cosine:
\[ V(t) = 300 \sin(\omega t) = 300 \cos\left(\omega t - \frac{\pi}{2}\right) \]

### Step 3: Identify the peak values of voltage and current
From the equations, we can identify:
- Peak voltage \( V_0 = 300 \)
- Peak current \( I_0 = 100 \)

### Step 4: Calculate the RMS values of voltage and current
The RMS (Root Mean Square) values are calculated as follows:
\[ V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{300}{\sqrt{2}} = 150\sqrt{2} \]
\[ I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{100}{\sqrt{2}} = 50\sqrt{2} \]

### Step 5: Determine the phase difference
The phase difference \( \phi \) between voltage and current can be determined from their forms:
- Voltage is \( 300 \cos(\omega t - \frac{\pi}{2}) \)
- Current is \( 100 \cos(\omega t) \)

Thus, the phase difference \( \phi \) is:
\[ \phi = \frac{\pi}{2} \]

### Step 6: Calculate the power factor
The power factor \( \cos(\phi) \) is:
\[ \cos\left(\frac{\pi}{2}\right) = 0 \]

### Step 7: Calculate the average power loss
Using the formula for average power:
\[ P = V_{rms} \cdot I_{rms} \cdot \cos(\phi) \]
Substituting the values we have:
\[ P = (150\sqrt{2}) \cdot (50\sqrt{2}) \cdot 0 \]
\[ P = 0 \]

### Conclusion
The average power loss in the circuit is:
\[ \boxed{0} \]