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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^2R)/(6b)`B. `(a^3R)/(3b)`C. `(a^3R)/(2b)`D. `(a^3R)/(b)` |
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Answer» Correct Answer - A `Q = a t - bt^(2)` `I = (dQ)/(dt) = a-2bt, i=0 at t = a/(2b)` `H = int_(0)^(t=(a)/(ab)) i^(2)Rdt = R int_(0)^(a/2b) (a-2bt)^(2) dt` `= R int_(0)^(a/2b) (a^(2)-4abt+4b^(2)t^(2))dt` `R|a^(2)t - 4ab(t^2)/(2)+(4b^2)t^3))/(3)|_(0)^(a/2b)` `=R[a^(2) . a/(2b) -(4ab)/(2) * (a^2)/(4b^2) +(4b^2)/(3) * (a^3)/(8b^3)]` `=(a^3R)/(b) [1/2 - 1/2 + 1/6] = (a^3R)/(6b)`. |
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