Question

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

A

$22C$

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B

$20C$

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C

$18C$

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D

$5C$

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Solution

The correct option is A $22C$Step 1: Given dataCurrent-time variation, $I=3{t}^{2}+2t+5$Initial time, $t=0$Final time, $t=2s$Step 2: Formula used$I=\frac{dQ}{dt}$ …………………….(a)Where ($dQ$) is the small charge flowing in a small time duration ($dt$)Step 3: Calculation of charge flowing through the wireSubstituting the given values in equation (a), we get$3{t}^{2}+2t+5=\frac{dQ}{dt}$$dQ=\left(3{t}^{2}+2t+5\right)dt$ …………………(b)The total charge flowing through the wire can be calculated by integrating the equation (b) over limits $t=0$ to $t=2s$Integration equation (b), we get$\int dQ={\int }_{0}^{2}\left(3{t}^{2}+2t+5\right)dt\phantom{\rule{0ex}{0ex}}Q={\int }_{0}^{2}\frac{3{t}^{3}}{3}+\frac{2{t}^{2}}{2}+5t\phantom{\rule{0ex}{0ex}}Q={\left[{t}^{3}\right]}_{0}^{2}+{\left[{t}^{2}\right]}_{0}^{2}+5{\left[t\right]}_{0}^{2}\phantom{\rule{0ex}{0ex}}Q={\left(2\right)}^{3}+{\left(2\right)}^{2}+5\left(2\right)\phantom{\rule{0ex}{0ex}}Q=8+4+10\phantom{\rule{0ex}{0ex}}Q=22C$Hence, option (a) is correct.

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