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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

Asked in NEET-I 2016 · Heating and power in resistors

Answer: (3) (a³R)/(6b)

Step-by-step solution

Current I = (dQ)/(dt) = a - 2bt.

Current flows until I = 0, at t₀ = a/(2b).

H = ∫₀^t₀ I²R dt = R∫₀^a/2b(a-2bt)² dt.

Put u = a - 2bt, dt = -(du)/(2b): H = R/(2b)∫₀^au² du = R/(2b)·(a³)/3.

H = (a³R)/(6b).

Why the other options are wrong

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