Step 1: {Finding the Area under the \( i-t \) Graph}
\[
{Total area} = \frac{1}{2} \times 1 \times 10 + \frac{1}{2} \times (2-1) \times 10
\]
\[
= 5 + 5 = 10 { A}
\]
Step 2: {Finding Average Current}
\[
i_{{avg}} = \frac{{Total area}}{{time interval}}
\]
\[
i_{{avg}} = \frac{10}{2} = 5 { A}
\]
Thus, the correct answer is \( 5 \) A.
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Approach Solution -2
Step 1: Understand the graph
The graph shows a triangular waveform of current \( i(t) \) versus time from \( t = 0 \) to \( t = 2 \) seconds.
- The current increases linearly from 0 A at \( t = 0 \) to 10 A at \( t = 1 \) s
- Then it decreases linearly from 10 A at \( t = 1 \) s to 0 A at \( t = 2 \) s
Step 2: Use the formula for average value of current
The average value of a function over a time interval \([0, T]\) is:
\[
I_{avg} = \frac{1}{T} \int_0^T i(t) \, dt
\]
Here, the area under the current-time curve represents the total charge, and dividing by time gives average current.
Step 3: Calculate area under the triangle
This is a triangle with:
- Base = 2 s
- Height = 10 A
Area = \( \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 10 = 10 \) amp-seconds
Step 4: Calculate average current
\[
I_{avg} = \frac{\text{Area}}{\text{Total Time}} = \frac{10}{2} = 5 \text{ A}
\]
Final Answer:
The average current from \( t = 0 \) to \( t = 2 \) s is: 5 A
