Electric Resistance and Simple Circuits | Physics JEE previous year questions with Solutions

(2)

Given that, current $I = I_{0} + β t$

Also, $I_{0} = 20 A$ and $β = 3 A / s$

Hence, $d q / d t = 20 + 3 t \Rightarrow d q = (20 + 3 t) d t$

$\int_{0}^{q} d q = \int_{0}^{20} (20 + 3 t) d t$

$\Rightarrow q = {[20 t + \frac{3 t^{2}}{2}]}_{0}^{20} = 1000 C$

(22) $I = 3 t^{2} + 4 t^{3}$

$\int d Q = \int I d t$

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

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

$= (8 + 16) - (2) = 22 C$

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