The current flowing through a wire depends on time as $I = 3 t^{2} + 2 t + 5.$ The charge flowing through the cross-section of the wire in time from t = 0 to t = 2s is:

22C

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20C

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18C

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Solution

The correct option is A
22C

$I = 3 t^{2} + 2 t + 5 \frac{d q}{d t} = 3 t^{2} + 2 t + 5 d q = (3 t^{2} + 2 t + 5) d t \int d q = \int_{0}^{2} 3 t^{2} d t + \int_{0}^{2} 2 t d t + \int_{0}^{2} 4 d t q = [\frac{3 \times t^{2}}{3}]_{0}^{2} + [\frac{2 t^{2}}{2}]_{0}^{2} + 5 [t]_{0}^{2} = (4 - 0) + (4 - 0) + 5 (2 - 0) = 22 C$

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Q. The current flowing through a wire depends on time as

I = 3 t^{2} + 2 t + 5

. The charge flowing through the cross-section of the wire in time from

t = 0

t = 2 s

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