If instantaneous value of current is I = 10 sin (314 t) A, then the average current for the half cycle will be –

Options

A. `10A`
B. 7.07 A
C. 6.37 A
D. 3.53 A

Answer

To find the average current for the half cycle given the instantaneous current \( I = 10 \sin(314t) \) A, we can follow these steps:

### Step 1: Identify the time period of the current
The given current can be expressed in the form \( I = I_0 \sin(\omega t) \), where:
- \( I_0 = 10 \) A (the amplitude)
- \( \omega = 314 \) rad/s (the angular frequency)

The time period \( T \) can be calculated using the formula:
\[ T = \frac{2\pi}{\omega} \]
Substituting the value of \( \omega \):
\[ T = \frac{2\pi}{314} \]

### Step 2: Calculate the average current over half a cycle
The average current \( I_{avg} \) over half a cycle (from \( t = 0 \) to \( t = \frac{T}{2} \)) can be calculated using the formula:
\[ I_{avg} = \frac{1}{\frac{T}{2}} \int_0^{\frac{T}{2}} I \, dt \]
Substituting \( I = 10 \sin(314t) \):
\[ I_{avg} = \frac{2}{T} \int_0^{\frac{T}{2}} 10 \sin(314t) \, dt \]

### Step 3: Evaluate the integral
The integral can be evaluated as follows:
\[ \int 10 \sin(314t) \, dt = -\frac{10}{314} \cos(314t) \]
Now, we need to evaluate this from \( 0 \) to \( \frac{T}{2} \):
\[ \int_0^{\frac{T}{2}} 10 \sin(314t) \, dt = \left[-\frac{10}{314} \cos(314t)\right]_0^{\frac{T}{2}} \]
Substituting the limits:
\[ = -\frac{10}{314} \left( \cos(314 \cdot \frac{T}{2}) - \cos(0) \right) \]
Since \( \frac{T}{2} \) corresponds to \( \cos(\pi) = -1 \):
\[ = -\frac{10}{314} \left( -1 - 1 \right) = -\frac{10}{314} \cdot (-2) = \frac{20}{314} \]

### Step 4: Substitute back to find \( I_{avg} \)
Now substituting back into the average current formula:
\[ I_{avg} = \frac{2}{T} \cdot \frac{20}{314} \]
Substituting \( T = \frac{2\pi}{314} \):
\[ I_{avg} = \frac{2}{\frac{2\pi}{314}} \cdot \frac{20}{314} = \frac{314}{\pi} \cdot \frac{20}{314} = \frac{20}{\pi} \]

### Step 5: Final answer
Thus, the average current for the half cycle is:
\[ I_{avg} = \frac{20}{\pi} \, \text{A} \]