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The current in a conductor of 40 ohm resistance increases linearly from 5 A to 25 A in 10 s. How much heat is liberated in it during this time? Solve the problem using two methods: (a) numerical calculation, (b) integration. |
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Answer» `triangleQ=triangleQ_(1)+triangleQ_(2)+triangleQ_(9)=(i_(1 av)^(2)+i_(2av)^(2)+i_(9av)^(2))R trianglet= 2580 xx 40 x 1 =1.03 xx 10^(5)J` (b) Integration. The CURRENT according to the low i=5+2T. The quantity of heat is `Q=underset(0)overset(10)int i^2 R dt=40 underset(0)overset(10)int (5+2t)^(2) dt-20 underset(0)overset(10)int (5+2t)^(2) d (5+2t)=`  `=(20)/3 (5 2t)^(3) underset(0)overset(10)int =(20)/3 (25^(3)-5^(3))=(20 xx 775 xx 20)/(3)=1.03 xx 10^(5)J` We see that the result of the numerical calculation was ACCURATE. |
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