A current through a wire depends on time as
i = $\alpha$0t + $\beta$t2
where $\alpha$0 = 20 A/s and $\beta$ = 8 As$-$2. Find the charge crossed through a section of the wire in 15 s.
Solution
Given, $i = {\alpha _0}t + \beta {t^2}$<br/><br/>where, $\alpha$<sub>0</sub> = 20 A/s, $\beta$ = 8 A/s<sup>2</sup><br/><br/>We know that, $i = {{dq} \over {dt}}$<br/><br/>$\Rightarrow {{dq} \over {dt}} = i = {\alpha _0}t + \beta {t^2} = 20t + 8{t^2}$<br/><br/>$\Rightarrow dq = (20t + 8{t^2})dt$<br/><br/>On integrating both sides, we get<br/><br/>$\int\limits_0^q {dq = \int\limits_0^{15} {(2t + 8{t^2})dt} }$<br/><br/>$$q = \left[ {{{20{t^2}} \over 2} + {{8{t^3}} \over 3}} \right]_0^{15} = 10 \times {(15)^2} + {8 \over 3} \times {(15)^3}$$<br/><br/>$\therefore$ q = 11250 C
About this question
Subject: Physics · Chapter: Current Electricity · Topic: Ohm's Law and Resistance
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