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The current in a conductor varies with time $' t'$ as $I = 3 t + 4 t^{2}$ . Where I in amp and t in sec. The electric charge flows through the section of the conductor between t = 1s and t = 3s (in coulombs)

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$\frac{100}{3}$

$\frac{140}{3}$

$\frac{280}{3}$

$\frac{140}{6}$

answer is B.

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Detailed Solution

$q = \int d q = \int_{1}^{3} i d t,$

$q = \int_{1}^{3} (3 t + 4 t^{2}) d t = \frac{140}{3} C = 46.67$

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