A complex current wave is given by `i=(5+5sin100omegat)A`. Its average value over one time period is given as

Options

A. `10A`
B. `5A`
C. `sqrt(50)A`
D. 0

Explanation

To find the average value of the given complex current wave \( i = (5 + 5 \sin(100 \omega t)) \, A \) over one time period, we can follow these steps:

### Step 1: Identify the components of the current wave
The current wave consists of a constant component and a sinusoidal component:
- Constant component: \( 5 \, A \)
- Sinusoidal component: \( 5 \sin(100 \omega t) \, A \)

### Step 2: Write the formula for average value
The average value of a function \( i(t) \) over one time period \( T \) is given by:
\[ I_{\text{avg}} = \frac{1}{T} \int_0^T i(t) \, dt \]

### Step 3: Determine the time period \( T \)
The time period \( T \) for the sinusoidal component \( \sin(100 \omega t) \) is:
\[ T = \frac{2\pi}{100 \omega} \]

### Step 4: Set up the integral
Now we can set up the integral for the average value:
\[ I_{\text{avg}} = \frac{1}{T} \int_0^T (5 + 5 \sin(100 \omega t)) \, dt \]

### Step 5: Split the integral
We can split the integral into two parts:
\[ I_{\text{avg}} = \frac{1}{T} \left( \int_0^T 5 \, dt + \int_0^T 5 \sin(100 \omega t) \, dt \right) \]

### Step 6: Evaluate the first integral
The first integral is straightforward:
\[ \int_0^T 5 \, dt = 5T \]

### Step 7: Evaluate the second integral
The second integral, \( \int_0^T 5 \sin(100 \omega t) \, dt \), evaluates to zero over one complete cycle because the positive and negative areas cancel each other out:
\[ \int_0^T 5 \sin(100 \omega t) \, dt = 0 \]

### Step 8: Combine results
Now, substituting back into the average value formula:
\[ I_{\text{avg}} = \frac{1}{T} \left( 5T + 0 \right) = 5 \]

### Final Answer
Thus, the average value of the current wave over one time period is:
\[ I_{\text{avg}} = 5 \, A \]