Current in an ac circuit is given by `i=2 sqrt(2) sin [(pi t+((pi)//4)]`. Then find the average value of current during time t=0 to t=1 s.

Explanation

To find the average value of the current in the given AC circuit over the time interval from \( t = 0 \) to \( t = 1 \) second, we will follow these steps:

### Step 1: Write down the expression for current
The current in the AC circuit is given by:
\[ i(t) = 2\sqrt{2} \sin\left(\pi t + \frac{\pi}{4}\right) \]

### Step 2: Set up the formula for average current
The average value of the current over a time interval from \( t_1 \) to \( t_2 \) is given by:
\[ I_{\text{avg}} = \frac{1}{t_2 - t_1} \int_{t_1}^{t_2} i(t) \, dt \]
In this case, \( t_1 = 0 \) and \( t_2 = 1 \), so:
\[ I_{\text{avg}} = \frac{1}{1 - 0} \int_{0}^{1} 2\sqrt{2} \sin\left(\pi t + \frac{\pi}{4}\right) \, dt \]

### Step 3: Factor out constants from the integral
Since \( 2\sqrt{2} \) is a constant, we can factor it out:
\[ I_{\text{avg}} = 2\sqrt{2} \int_{0}^{1} \sin\left(\pi t + \frac{\pi}{4}\right) \, dt \]

### Step 4: Evaluate the integral
To evaluate the integral, we use the substitution:
\[ u = \pi t + \frac{\pi}{4} \quad \Rightarrow \quad du = \pi \, dt \quad \Rightarrow \quad dt = \frac{du}{\pi} \]
When \( t = 0 \), \( u = \frac{\pi}{4} \) and when \( t = 1 \), \( u = \pi + \frac{\pi}{4} = \frac{5\pi}{4} \).

Now, we can rewrite the integral:
\[ \int_{0}^{1} \sin\left(\pi t + \frac{\pi}{4}\right) \, dt = \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \sin(u) \frac{du}{\pi} = \frac{1}{\pi} \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \sin(u) \, du \]

### Step 5: Calculate the definite integral
The integral of \( \sin(u) \) is:
\[ -\cos(u) \]
Thus:
\[ \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \sin(u) \, du = -\cos\left(\frac{5\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right) \]
Calculating the cosine values:
\[ \cos\left(\frac{5\pi}{4}\right) = -\frac{1}{\sqrt{2}} \quad \text{and} \quad \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]
So:
\[ -\left(-\frac{1}{\sqrt{2}}\right) + \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

### Step 6: Substitute back to find \( I_{\text{avg}} \)
Now substituting back:
\[ I_{\text{avg}} = 2\sqrt{2} \cdot \frac{1}{\pi} \cdot \sqrt{2} = \frac{4}{\pi} \]

### Final Answer
Thus, the average value of the current during the time interval from \( t = 0 \) to \( t = 1 \) second is:
\[ I_{\text{avg}} = \frac{4}{\pi} \]