Question

The quantity of charge \(Q\) in coulombs (C) that has passed through a point in a wire up to time \(t\) (measured in seconds) is given by \(Q\left( t \right) = {t^3} - 2{t^2} + 6t + 2\). Find the current when (a) \(t = 0.5s\) and (b) \(t = 1\,s\). (See Example 3. The unit of current is an ampere \(t = 0.5s\left( {1\,A = 1\,C/s} \right)\).) At what time is the current lowest?

Short answer

(a). \({\rm{4}}{\rm{.75 A}}\)

(b). \({\rm{5 A}}\)

Step-by-step solution

Flow of Charge.

The Flow of a Chargecan be defined as the parameter named as Currentwhich determines the amount of charge that flows through a conductor after application of voltage.

Evaluating the derivative of the given function of Charge:

(a)

Thegiven function for the charge is:

\(Q\left( t \right) = {t^3} - 2{t^2} + 6t + 2\)

Differentiating with respect to t, we get:

\(\begin{array}{c}\frac{{dQ}}{{dt}} = \frac{d}{{dt}}\left( {{t^3} - 2{t^2} + 6t + 2} \right)\\ = 3{t^2} - 4t + 6\end{array}\)

Now, the current at \(t = 0.5\,s\)will be:

\(\begin{array}{c}\frac{{dQ}}{{dt}}\left| {_{t = 0.5}} \right. = 3{\left( {0.5} \right)^2} - 4\left( {0.5} \right) + 6\\ = 4.75{\rm{ A}}\end{array}\)

Calculating the current at 1 second:

(b)

The current at \(t = 1.0\,s\)will be:

\(\begin{array}{c}\frac{{dQ}}{{dt}}\left| {_{t = 1}} \right. = 3{\left( 1 \right)^2} - 4\left( 1 \right) + 6\\ = 5{\rm{ A}}\end{array}\)

Now, again differentiating the function of charge, we have:

\(\begin{array}{c}\frac{{{d^2}Q}}{{d{t^2}}} = \frac{d}{{dt}}\left( {3{t^2} - 4t + 6} \right)\\ = 6t - 4\end{array}\)

For lowest value of current:

\(\begin{array}{c}\frac{{{d^2}Q}}{{d{t^2}}} = 0\\6t - 4 = 0\\t = \frac{2}{3}\end{array}\)

Hence, the current will be lowest at \(t = \frac{2}{3}\,s\).