Physics - Current Electricity Question with Solution | TestHub

Given here, $Q=\left(\alpha t-\beta t^2+\gamma t^3\right)$Current flowing through conductor is given by

$i=\frac{dQ}{dt}=\left(\alpha -2\beta t+3\gamma t^2\right)...\left(1\right)$

For current to be minimum, $\frac{di}{dt}=0$

So, $\frac{di}{dt}=\left(6\gamma t-2\beta \right)=0$

$\Rightarrow t=\frac{\beta }{3\gamma }$Substituting in equation (1),

$$\begin{gathered} i=\left(\alpha -2\beta t+3\gamma t^2\right)=\alpha -2\beta \left(\frac{\beta }{3\gamma }\right)+3\gamma {\left(\frac{\beta }{3\gamma }\right)}^2 \\ =\alpha -2\frac{{\beta }^2}{3\gamma }+3\frac{{\beta }^2}{9\gamma } \\ =\left(\alpha -\frac{{\beta }^2}{3\gamma }\right) \end{gathered}$$