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Current Electricity question

2016 · Q121
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Current Electricity question

2016 · Q121

NEETPhysicsCurrent ElectricityMCQ+4 / −1
The charge flowing through a resistance R varies with time t is Q = at −-− bt2, where aaa and bbb are positive constants. The total heat produced in R is
  1. A
    a3R2b{{{a^3}R} \over {2b}}2ba3R​
  2. B
    a3Rb{{{a^3}R} \over {b}}ba3R​
  3. C
    a3R6b{{{a^3}R} \over {6b}}6ba3R​
  4. D
    a3R3b{{{a^3}R} \over {3b}}3ba3R​
View written solutionFree

Correct answer: C

Given: Charge Q = at – bt2

∴\therefore∴ Current i=∂Q∂t=a−2bti = {{\partial Q} \over {\partial t}} = a - 2bti=∂t∂Q​=a−2bt
{for i=0⇒t=a2b}\left\{ {for\,i = 0 \Rightarrow t = {a \over {2b}}} \right\}{fori=0⇒t=2ba​}

From joule's law of heating, heat produced
dH = i2Rdt

H=∫0a/2b(a−2bt)2RdtH = \int\limits_0^{a/2b} {{{\left( {a - 2bt} \right)}^2}Rdt} H=0∫a/2b​(a−2bt)2Rdt

H=(a−2bt)2R−3×2b∣0a2b=a3R6bH = \left. {{{{{\left( {a - 2bt} \right)}^2}R} \over { - 3 \times 2b}}} \right|_0^{{a \over {2b}}} = {{{a^3}R} \over {6b}}H=−3×2b(a−2bt)2R​​02ba​​=6ba3R​

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