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Electric charge passing through a resistor changes with time t as Q = at -bt^(2) . Then total heat produce in resistor R = .... |
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Answer» `(a^(3) R)/(3B)` `Q= at - bt^(2)` `therefore I =(DQ)/(dt ) = (d)/(dt) "" [at - bt^(2) ]` `therefore I = a - 2 bt` `therefore 0 = a - 2 bt` `therefore t = (a)/(2b)` Now power P = `int_(0)^(t) I^(2) R dt = int_(0)^((a)/(2b)) ( a - 2bt)^(2) R dt` `thereforeP= int_(0)^((a)/(2b)) (a^(2) - 4abt + 4b^(2) t^(2) ) R dt ` = `[ a^(2) t - 4ab (t^(2))/(2) + 4b^(2) (t^(3))/(3) ]_(0)^((a)/(2b))R` = ` [ (a^(2) xx a)/(2b) - (4 ab xx a^(2))/(2 xx 4 b^(2)) + (4b^(2) xx a^(3))/(3 xx 8b^(3)) ]` R = `[ (a^(3))/(2b) - (a^(3))/(2b) + (a^(3))/(6b) ]R = (a^(3)R)/(6b)` |
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