The current I passed in any instruement in alternating current circuit is I = 2 `sin omegat` amp and potential difference applied is given by V = 5 `cos omegat` volt then power loss in instruement is

Options

A. 2.5 watt
B. 5 watt
C. 10 watt
D. zero

Explanation

To find the power loss in the instrument given the current and potential difference in an alternating current circuit, we can follow these steps:

### Step 1: Identify the expressions for current and voltage
The current \( I \) is given as:
\[ I = 2 \sin(\omega t) \quad \text{(in amperes)} \]
The voltage \( V \) is given as:
\[ V = 5 \cos(\omega t) \quad \text{(in volts)} \]

### Step 2: Convert voltage expression to sine form
To compare the current and voltage expressions, we can convert the cosine function into a sine function. We know that:
\[ \cos(\omega t) = \sin\left(\omega t + \frac{\pi}{2}\right) \]
Thus, we can rewrite the voltage as:
\[ V = 5 \sin\left(\omega t + \frac{\pi}{2}\right) \]

### Step 3: Determine the phase difference
From the expressions:
- Current \( I = 2 \sin(\omega t) \)
- Voltage \( V = 5 \sin\left(\omega t + \frac{\pi}{2}\right) \)

We can see that the voltage is leading the current by \( \frac{\pi}{2} \) radians (or 90 degrees). Therefore, the phase difference \( \phi \) is:
\[ \phi = \frac{\pi}{2} \]

### Step 4: Calculate the power loss
The average power \( P \) in an AC circuit can be calculated using the formula:
\[ P = \frac{V_0}{\sqrt{2}} \cdot \frac{I_0}{\sqrt{2}} \cdot \cos(\phi) \]
Where:
- \( V_0 = 5 \) (the peak voltage)
- \( I_0 = 2 \) (the peak current)

Substituting the values:
\[ P = \frac{5}{\sqrt{2}} \cdot \frac{2}{\sqrt{2}} \cdot \cos\left(\frac{\pi}{2}\right) \]

### Step 5: Evaluate the cosine term
Since \( \cos\left(\frac{\pi}{2}\right) = 0 \):
\[ P = \frac{5}{\sqrt{2}} \cdot \frac{2}{\sqrt{2}} \cdot 0 = 0 \]

### Conclusion
The power loss in the instrument is:
\[ \boxed{0} \text{ watts} \]