Question of The Day

The charge flowing through a resistance $$R$$ varies with time $$t$$ as $$Q=at-b{t}^{2}$$. The total heat produced in $$R$$ by the time current ceases is



1) $${a}^{3}R/6b$$

2) $${a}^{3}R/3b$$

3) $$\cfrac{{a}^{3}R}{2b}$$

4) $$\cfrac{{a}^{3}R}{b}$$



 


Correct Answer :1

Solution

Given, 

$$Q= at- bt^{2}$$
$$i=\cfrac{dQ}{dt}=\cfrac{d}{dt} \cdot (at-bt^{2})$$
$$\Rightarrow i= \cfrac{d}{dt} \cdot at -\cfrac{d}{dt} \cdot bt^{2}$$
$$\Rightarrow i= a\cdot \cfrac{d}{dt} - b\cdot \cfrac{d}{dt}t^{2}$$
$$\Rightarrow i= a- 2bt$$
we know, $$H=\int { { i }^{ 2 }\cdot  } Rdt$$
Now, for the limits of integration,
when $$i=0, \quad a=2bt$$
$$\therefore t=\cfrac { a }{ 2b } $$
$$H= \int _{ a }^{ a/2b}{ { { i }^{ 2 }\cdot  }Rdt }$$
$$\Rightarrow H=\int _{ a }^{a/2b} \left( { a }^{ 2 }+4{ b }^{ 2 }t^{ 2 }-\quad 4bt \right) Rdt=\quad R\left[ { a }^{ 2 }t+\cfrac { 4{ b }^{ 2 }t^{ 3 } }{ 3 } -\cfrac { 4ab{ t }^{ 2 } }{ 2 }  \right] _{ 0 }^{ a/2b }$$
$$\Rightarrow H=R\left[ { a }^{ 2 }\left( \cfrac { a }{ 2b }  \right) +\cfrac { 4{ b }^{ 2 } }{ 3 } \left( \cfrac { { a }^{ 3 } }{ 8{ b }^{ 3 } }  \right) -\cfrac { 4ab }{ 2 } \left( \cfrac { { a }^{ 2 } }{ 4{ b }^{ 2 } }  \right)  \right] =\quad R\left[ \cfrac { { a }^{ 3 } }{ 2b } +\cfrac { { a }^{ 3 }{ b }^{ 3 } }{ 6{ b }^{ 3 } } -\cfrac { { a }^{ 3 } }{ 6b }  \right] =\quad R\left[ \cfrac { 3{ a }^{ 3 }{ b }^{ 3 }+\quad { a }^{ 3 }{ b }^{ 2 }-{ 3a }^{ 3 }{ b }^{ 3 } }{ 6{ b }^{ 3 } }  \right] $$
$$\Rightarrow H=\quad R\left[ \cfrac { { a }^{ 3 } }{ 6b }  \right] =\quad \cfrac { R{ a }^{ 3 } }{ 6b } $$