Question

The charge flowing through a resistance R varies with time t as $Q = at - b{t^2}$, where a and b are positive constants. The total heat produced in R is $\dfrac{{{b^3}R}}{{6a}}$.
A. True
B. False

Answer 1

Hint: From the given expression for charge, we can find out the expression of current. We know the expression for heat produced in a resistance by the flow of current. Solving for the total amount of heat produced in the resistance, we can check if the given value of total heat is correct or not.

Formula used:
The current due to flow of a charge is given as
$I = \dfrac{{dQ}}{{dt}}$
The heat produced due to flow of a current I through a resistance R is given as
$H = {I^2}Rt$

Complete step-by-step answer:
We are given the expression for the charge flowing through a resistance R. Its variation with time t is given as follows:
$Q = at - b{t^2}$
We know that the current due to the flow of this charge is given as the time rate of flow of the charge. Therefore, we can write
$I = \dfrac{{dQ}}{{dt}} = \dfrac{d}{{dt}}\left( {at - b{t^2}} \right) = a - 2bt$
Now if we put the current to be zero then we get
$
  a - 2bt = 0 \\
  t = \dfrac{a}{{2b}} \\
 $
This is the time at which the current stops flowing after having flown for a duration of $t = 0$ to $t = \dfrac{a}{{2b}}$.
Now the small amount of heat dH produced by the flow of a current I for a small duration dt is given as follows:
$dH = {I^2}Rdt$
The total amount of heat produced by the current can be calculated by integrating the both sides in the following way.
$
  H = \int\limits_0^{\dfrac{a}{{2b}}} {{I^2}Rdt} \\
   = R\int\limits_0^{\dfrac{a}{{2b}}} {{{\left( {a - 2bt} \right)}^2}dt} \\
   = R\int\limits_0^{\dfrac{a}{{2b}}} {\left( {{a^2} + 4{b^2}{t^2} - 4abt} \right)dt} \\
   = R\left[ {{a^2}\int\limits_0^{\dfrac{a}{{2b}}} {dt} + 4{b^2}\int\limits_0^{\dfrac{a}{{2b}}} {{t^2}dt} - 4ab\int\limits_0^{\dfrac{a}{{2b}}} {tdt} } \right] \\
   = R\left[ {{a^2}\left\{ t \right\}_0^{\dfrac{a}{{2b}}} + 4{b^2}\left\{ {\dfrac{{{t^3}}}{3}} \right\}_0^{\dfrac{a}{{2b}}} - 4ab\left\{ {\dfrac{{{t^2}}}{2}} \right\}_0^{\dfrac{a}{{2b}}}} \right] \\
   = R\left[ {{a^2} \times \dfrac{a}{{2b}} + \dfrac{4}{3}{b^2}{{\left( {\dfrac{a}{{2b}}} \right)}^3} - 2ab{{\left( {\dfrac{a}{{2b}}} \right)}^2}} \right] \\
   = R\left( {\dfrac{{{a^3}}}{{2b}} + \dfrac{{{a^3}}}{{6b}} - \dfrac{{{a^3}}}{{2b}}} \right) \\
   = \dfrac{{{a^3}R}}{{6b}} \\
 $
This is the actual value of the heat produced by the flow of the given charge. Therefore, the given statement is false.

So, the correct answer is “Option B”.

Note: It should be noted that the purpose of the resistance is to obstruct the flow of current across it. The energy is dissipated in the form of heat while offering this resistance to the flow of current. It is directly proportional to the amount of current flowing through it and also the amount of resistance being offered to the flow of current.