A source of e.m.f. E = 15,V and having neglig | vedclass

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(C) The relationship between current $i$ and charge $Q$ is given by $i = \frac{dQ}{dt}$.Therefore,the charge $dQ$ flowing in a small time interval $dt

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$30$

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(C) The relationship between current $i$ and charge $Q$ is given by $i = \frac{dQ}{dt}$.Therefore,the charge $dQ$ flowing in a small time interval $dt$ is $dQ = i \, dt$.To find the total charge $Q$ flowing in the first $5$ seconds,we integrate the current with respect to time from $t = 0$ to $t = 5$:$Q = \int_{0}^{5} i \, dt = \int_{0}^{5} (1.2t + 3) \, dt$$Q = \left[ \frac{1.2t^2}{2} + 3t ight]_{0}^{5}$$Q = [0.6t^2 + 3t]_{0}^{5}$Substituting the limits:$Q = (0.6 	imes 5^2 + 3 	imes 5) - (0.6 	imes 0^2 + 3 	imes 0)$$Q = (0.6 	imes 25 + 15) - 0$$Q = 15 + 15 = 30\,C$.

$A$ source of $e.m.f.$ $E = 15\,V$ and having negligible internal resistance is connected to a variable resistance so that the current in the circuit increases with time as $i = 1.2t + 3$. Then,the total charge that will flow in the first five seconds will be ............... $C$.

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