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A current (i) through a wire depends on time (t) as $i = α t^{2} + β t^{3}$ where $α = 3 A / s^{2}$ and $β = 4 A / s^{3}$ . Then the charge crossed through a section of the wire in 5s is

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2250 C

11250 C

50 C

750 C

answer is D.

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Detailed Solution

$i = \frac{d q}{d t} \Rightarrow \int d q = \int i d t \Rightarrow q = \int_{0}^{5} (α t^{2} + β t^{3}) d t = \int_{0}^{5} (3 t^{2} + 4 t^{3}) d t = {[3 (\frac{t^{3}}{3}) + 4 (\frac{t^{4}}{4})]}_{0}^{5} = 5^{3} + 5^{4} = 125 + 625 = 750 C$

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