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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:

1. a³R / 6b
2. a³R / 3b
3. a³R / 2b
4. a³R / b

Correct answer: a³R / 6b

Solution

The current is the time derivative of charge, I = dQ/dt = a - 2bt. The heat produced is given by H = ∫(I²R) dt. Substituting I and integrating over the time interval where current flows (from t = 0 to t = a/2b, where I becomes zero), the result is H = a³R / 6b.

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