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NEET 2016 (Phase 1) · Electrical Power & Heating Effect

Electrical Power & Heating Effect question from NEET 2016 (Phase 1)

Q48NEET 2016 · PHASE 1Difficult

Electrical Power & Heating Effect

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

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Answer: (c)

  1. I=dQdt=a−2btI = \frac{dQ}{dt} = a - 2bt, which becomes zero at t=a2bt = \frac{a}{2b} (current flows only till then).
  2. H=∫0a/2b(a−2bt)2R dt=R[(a−2bt)3−6b]0a/2bH = \int_0^{a/2b} (a - 2bt)^2R\,dt = R\left[\frac{(a - 2bt)^3}{-6b}\right]_0^{a/2b}
  3. H=a3R6bH = \frac{a^3R}{6b}. So the answer is (c).
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