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Question: The charge flowing through a resistance \( R \) varies with the time \( t \) as \( Q = at - b{t^2} \...

The charge flowing through a resistance RR varies with the time tt as Q=at−bt2Q = at - b{t^2} . The total heat produced in RR is
(A) a2R6b\dfrac{{{a^2}R}}{{6b}}
(B) a3R3b\dfrac{{{a^3}R}}{{3b}}
(C) a3R2b\dfrac{{{a^3}R}}{{2b}}
(D) a3Rb\dfrac{{{a^3}R}}{b}

Explanation

Solution

Based on the joule's law when the current travels through the conductor free electrons in the material collide with the atom and generate heat.
If we heated the metals then the lattice vibrations increased.
When we heat the semiconductors the charge carriers are increased and hence the conductivity of the semiconductor also gets increased.

Complete answer:
When we heat the conductor the atoms in the conductor absorb the heat and vibrate then it transfers the heat energy into the near atoms.
The transfer of heat occurs the two ways they are, molecular vibration and free-electron transfer.
The electricity flows through the copper and aluminum easily because the electrons jump from atom to atoms easily.
The first step is current as a function of time,
i=dQdt=d(at−bt2)dti = \dfrac{{dQ}}{{dt}} = \dfrac{{d(at - b{t^2})}}{{dt}}
Now differentiate the above equations,
i=a−2bti = a - 2bt
Let, when i=0i = 0 the time t=t0t = {t_0}
∴0=a−2bt0\therefore 0 = a - 2b{t_0}
Now rearranging the above equation and we get,
t0=a2b{t_0} = \dfrac{a}{{2b}}
The from the Joule’s heating formula the heat generated at a time, t=0t = 0 and t=t0t = {t_0} is as follows,
H=∫0t0i2RdtH = \int_0^{{t_0}} {{i^2}} Rdt
Now substitute the values in the above equations we get,
H=∫0t0(a−2bt)2RdtH = \int_0^{{t_0}} {{{\left( {a - 2bt} \right)}^2}} Rdt
Now expand the above equation,
H=R∫0t0(a2−4abt+4b2t2)2dtH = R\int_0^{{t_0}} {{{\left( {{a^2} - 4abt + 4{b^2}{t^2}} \right)}^2}} dt
Now integrate the above equation we get,
=R[a2t−2abt2+4b23t3]0t0=a2b= R{\left[ {{a^2}t - 2ab{t^2} + \dfrac{{4{b^2}}}{3}{t^3}} \right]_0}^{{t_0} = \dfrac{a}{{2b}}}
Now substitute the upper limit and lower limit values and solve the equation,
=R[a2(a2b)−2ab(a2b)2+4b23(a2b)3]= R\left[ {{a^2}\left( {\dfrac{a}{{2b}}} \right) - 2ab{{\left( {\dfrac{a}{{2b}}} \right)}^2} + \dfrac{{4{b^2}}}{3}{{\left( {\dfrac{a}{{2b}}} \right)}^3}} \right]
After solving the equation we can get the Heat,
H=a3R6bH = \dfrac{{{a^3}R}}{{6b}}
Hence the heat produced is H=a3R6bH = \dfrac{{{a^3}R}}{{6b}}
Finally, the correct answer is, option (A).

Note:
The electrons are facing the trouble of switching the orbits in iron wire
The energy cannot be created or destroyed according to the law of thermodynamics.
When the amount of collision of electrons gets increased then the heat produced also gets increased.