Explanation

(b) $Q = at - bt^{2}$
$I = \frac{dQ}{dt} = a - 2bt\ \ \ \ \ \ldots(i)$
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Where, $t = t_{0},I = 0$
In loop $BCDEB\ I_{1}(2R) - \left( I - I_{1} \right)R = 0$
or $\ 3I_{1} = I$
$\therefore{\ I}_{1} = \frac{I}{3} = \frac{a - 2bt}{3}$
$\therefore H = \int_{0}^{t_{0}}{I_{1}^{2}(2R)dt}$
$\ = \frac{2R}{9}\int_{0}^{t_{0}}{(a - 2bt)^{2}dt}$
$= \frac{2R}{9}\left\lbrack \ \int_{0}^{t_{0}}{\left( a^{2} - 4b^{2}t^{2} - 4dt \right)dt\ } \right\rbrack$
$\ =$ $\frac{2R}{9}\left\lbrack \left\{ a^{2}t + \frac{4b^{2}t^{2}}{3} - \frac{4bat^{2}}{2} \right\}_{0}^{to} \right\rbrack\ $
$\ =$ $\frac{2R}{9}\ \left\lbrack a^{2}t_{0} + \frac{4b^{2}t_{0}^{3}}{3} - 2bat_{0}^{2} \right\rbrack$
$but{\ t}_{0} = \frac{a}{2b}$ [from Eqn.(i)]
$\ H =$ $\frac{2R}{9}\left\lbrack a^{2} \times \frac{a}{2b} + \frac{4b^{2}}{3}\frac{a^{3}}{8b^{3}} - 2ab\frac{a^{2}}{4b^{2}} \right\rbrack$
$= \frac{2R}{9}\left\lbrack \frac{a^{3}}{2b} + \frac{a^{3}}{6b} - \frac{a^{3}}{2b} \right\rbrack$=$\frac{a^{3}R}{27b}$