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Equipotential Surfaces

Current I v...

Question

Current $I$ versus time $t$ graph through a conductor is shown in the figure. Average current through the conductor in the interval $0$ to $15 s$ is:

1 A

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10 A

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7.5 A

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5 A

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Solution

The correct option is D $5 A$
Area under the $i, t = q$
$\begin{matrix} T_{a v g} = \frac{q_{1} + q_{2}}{t} A r e a q_{1} = \frac{1}{2} \times 5 \times 10 = 25 c A r e a q_{2} = \frac{1}{2} (15 - 5) \times 10 = 50 c \Rightarrow I a v g = \frac{25 c + 50 c}{15 sec} = 5 A \end{matrix}$

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i = i_{0} + 3 α t^{2}

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