Q.

The charge flowing through a resistance R varies with time t as Q=at-bt2 where a and bare positive constants. The total heat produced in R is

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a

a3R2b

b

a3Rb

c

a3R6b

d

a3R3b

answer is C.

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Detailed Solution

Given, Q=at-bt2∴I=dQdt=a-2bt At  t=0,Q=0, I=a Also,   I=0 at t=a/2b Total heat produced in resistance  R,H=∫0a/2bI2Rdt=R∫0a/2b(a-2bt)2dt=R∫0a/2ba2+4b2t2-4abtdt=Ra2t+4b2t33-4abt220a/2b=Ra2×a2b+4b23×a38b3-4ab2×a24b2=a3Rb12+16-12=a3R6b
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