The current in a conductor varies with time $t$ as $I = 3 t + 4 t^{2}$ ,where $I$ in ampere and $t$ in seconds. The elelctric charge flows through the section of the conductor between $t = 1 s$ and $t = 3 s$ is $\frac{n}{3} C$ . Here value of $n$ is

140.00

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140.0

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140

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Solution

$I = \frac{d q}{d t} = 3 t + 4 t^{2}$

$\int d q = \int_{1}^{3} (3 t + 4 t^{2}) d t$

$Q = {[\frac{3 t^{2}}{2} + \frac{4 t^{2}}{2}]}_{1}^{3}$

$= [\frac{27}{2} + \frac{108}{3}] - [\frac{3}{2} + \frac{4}{3}]$

$= 13.5 + 36 - \frac{17}{6}$

$= 49.5 - \frac{17}{6} = \frac{297 - 17}{6}$

$= \frac{280}{6} = \frac{140}{3} C$

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