Question

The charge flowing through a resistance R varies with time t as Q = at – bt², where a and b are positive constants. The total heat produced in R is:
(a) $\frac{a^{3}R}{6b}$
(b) $\frac{a^{3}R}{3b}$
(c) $\frac{a^{3}R}{2b}$
(d) $\frac{a^{3}R}{b}$

Answer 1

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Answer 2

Answer:

Explanation:

Resistance = R (Given)

The charge starts from 0 and goes to positive peak at t= a/2b, after that it starts decreasing and crosses zero at t= a/b and then goes towards the negative infinity. Heat produced as a function of t will be -

Q = at - bt²

I = dQ/dt = a - 2bt

At t = 0 Q = 0 I = 0

Total heat produced in resistance R

H = ∫I²Rdt = R∫(a-2bt)²dt

R∫ (a²+4b²t²-4abt) dt

R[a²t+4b²t³/3 -4abt²/2]

R[a²×a/2b+4b²/3×a³/8b³ -4ab/2×a²/4b²]

= a³R/b[ 1/2+1/6-1/2]

= a³R/6b

Thus, the total heat produced in R is a³R/6b