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NEET 2016
Exam
NEET
Year
2016
Subject
Physics
Chapter
Current Electricity

NEET 2016 · Physics · Current Electricity

The charge flowing through a resistance R varies with time t as Q =...

The charge flowing through a resistance R\text{R} varies with time tt as Q=at−bt2Q = at - bt^{2} where a\text{a} and b\text{b} are positive constants. The total heat produced in R\text{R} is:

Options

  1. Aa3R6b\dfrac{{a}^{3}R}{6b}
  2. Ba3R3b\dfrac{{a}^{3}R}{3b}
  3. Ca3R2b\dfrac{{a}^{3}R}{2b}
  4. Da3Rb\dfrac{{a}^{3}R}{b}

Correct answer

A. a3R6b\dfrac{{a}^{3}R}{6b}

Step-by-step solution

given

Q=at−bt2Q = at - b{t}^{2}

i=dqdt=a−2bti = \dfrac{dq}{dt} = a - 2bt

i=0i = 0

at t=a2b\text{t} = \dfrac{\text{a}}{\text{2b}}

∴dH=i2Rdt∴ dH = {i}^{2}Rdt

=(a−2bt)2Rdt= {\left(a - 2bt\right)}^{2}Rdt

H=∫0a2b(a−2bt)2RdtH = \mathop{∫}_{0}^{\dfrac{a}{2b}}{\left(a - 2bt\right)}^{2}Rdt

{\left= \dfrac{{\left(a - 2bt\right)}^{3}R}{3\left(- 2b\right)}\right|}_{0}^{\dfrac{a}{2b}}

=1−6b[(a−2ba2b)3−(a)3]R= \dfrac{1}{-6 b}\left[{\left(a - 2b\dfrac{a}{2b}\right)}^{3} - {\left(a\right)}^{3}\right]R

=−16b[(0)3−a3]R= - \dfrac{1}{6b}\left[{\left(0\right)}^{3} - {a}^{3}\right]R

H=a3R6b\text{H} = \dfrac{{\text{a}}^{\text{3}} \text{R}}{6\text{b}}

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