The average current of a sinusoidally varrying alternating current of peak value `5A` with initial phase zero, between the instants `t = T//8` to `t = T//4` is (where `'T'` is time period)

Options

A. `(10)/(pi) sqrt(2) A`
B. `(5)/(pi) sqrt(2) A`
C. `(20 sqrt(2))/(pi)`
D. `(10)/(pi) A`

Explanation

To find the average current of a sinusoidally varying alternating current with a peak value of \(5A\) and an initial phase of zero between the instants \(t = \frac{T}{8}\) and \(t = \frac{T}{4}\), we can follow these steps:

### Step 1: Write the equation for the alternating current
The equation for the alternating current \(i(t)\) can be expressed as:
\[ i(t) = I_0 \sin(\omega t) \]
where \(I_0\) is the peak current, and \(\omega\) is the angular frequency.

Given that \(I_0 = 5A\) and the initial phase is zero, we have:
\[ i(t) = 5 \sin(\omega t) \]

### Step 2: Determine the angular frequency \(\omega\)
The angular frequency \(\omega\) is related to the time period \(T\) by:
\[ \omega = \frac{2\pi}{T} \]

### Step 3: Set the limits for the average current calculation
We need to calculate the average current between \(t = \frac{T}{8}\) and \(t = \frac{T}{4}\).

### Step 4: Use the formula for average current
The average current \(I_{\text{avg}}\) over a time interval from \(t_1\) to \(t_2\) is given by:
\[ I_{\text{avg}} = \frac{1}{t_2 - t_1} \int_{t_1}^{t_2} i(t) \, dt \]
Substituting our limits:
\[ I_{\text{avg}} = \frac{1}{\frac{T}{4} - \frac{T}{8}} \int_{\frac{T}{8}}^{\frac{T}{4}} 5 \sin(\omega t) \, dt \]

### Step 5: Calculate the time interval
The time interval is:
\[ \frac{T}{4} - \frac{T}{8} = \frac{2T}{8} - \frac{T}{8} = \frac{T}{8} \]

### Step 6: Substitute the time interval into the average current formula
Thus, we have:
\[ I_{\text{avg}} = \frac{1}{\frac{T}{8}} \int_{\frac{T}{8}}^{\frac{T}{4}} 5 \sin(\omega t) \, dt \]
\[ = 8 \int_{\frac{T}{8}}^{\frac{T}{4}} 5 \sin(\omega t) \, dt \]

### Step 7: Evaluate the integral
The integral of \(5 \sin(\omega t)\) is:
\[ \int 5 \sin(\omega t) \, dt = -\frac{5}{\omega} \cos(\omega t) \]
Now, we evaluate it from \(\frac{T}{8}\) to \(\frac{T}{4}\):
\[ = -\frac{5}{\omega} \left[ \cos\left(\omega \frac{T}{4}\right) - \cos\left(\omega \frac{T}{8}\right) \right] \]

### Step 8: Substitute \(\omega = \frac{2\pi}{T}\)
Substituting \(\omega\):
\[ = -\frac{5T}{2\pi} \left[ \cos\left(\frac{2\pi}{T} \cdot \frac{T}{4}\right) - \cos\left(\frac{2\pi}{T} \cdot \frac{T}{8}\right) \right] \]
\[ = -\frac{5T}{2\pi} \left[ \cos\left(\frac{\pi}{2}\right) - \cos\left(\frac{\pi}{4}\right) \right] \]
\[ = -\frac{5T}{2\pi} \left[ 0 - \frac{1}{\sqrt{2}} \right] \]
\[ = \frac{5T}{2\pi \sqrt{2}} \]

### Step 9: Calculate the average current
Now substituting back into the average current formula:
\[ I_{\text{avg}} = 8 \cdot \frac{5T}{2\pi \sqrt{2}} = \frac{40T}{2\pi \sqrt{2}} = \frac{20T}{\pi \sqrt{2}} \]

### Final Result
The average current between \(t = \frac{T}{8}\) and \(t = \frac{T}{4}\) is:
\[ I_{\text{avg}} = \frac{20}{\pi \sqrt{2}} \, A \]