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The charge flowing through a resistance `R` varies with time `t as Q = at - bt^(2)`. The total heat produced in `R` isA. `(a^(3)R)/(6b)`B. `(a^(2)R)/(27b)`C. `(a^(3)R)/(3b)`D. None of these |
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Answer» Correct Answer - A `I=(dQ)/(dt)=a-2bt` `I=0at t=(a)/(2b)` `therefore"Heat"=underset(0)overset(a/(2b))intI^(2)Rdt=underset(0)overset(a/(2b))int(a-2bt)^(2)Rdt` `=R[a^(2)t+(4b^(2)t^(3))/(3)-(4abt^(2))/(2)]^(a/(2b))` `=R[(a^(2)xxa)/(2b)+(4b^(2)xxa^(3))/(24b^(3))-(4aba^(2))/(8b^(2))]` `=R[(a^(3))/(2b)+(a^(3))/(6b)-(a^(3))/(2b)]=(Ra^(3))/(6b)` |
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