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

Answer: 1. a³R/6b Key idea: current I=dQ/dt ; heat H=∫ I²R dt . I=a-2bt , which becomes zero at t=a/2b . H=R∫₀^(a/2b)(a-2bt)²dt=R[-(a-2bt)³/6b]₀^(a/2b) . H=a³R/6b .

Topic: Electrical energy and power· Current Electricity

The charge flowing through a resistance R varies with time t as Q = at - bt^2, where a and b are positive constants. The total heat produced in R is :

1. \(\frac{a^3R}{6b}\)

2. \(\frac{a^3R}{3b}\)

3. \(\frac{a^3R}{2b}\)

4. \(\frac{a^3R}{b}\)

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Answer: 1. \(\frac{a^3R}{6b}\)

Key idea: current \(I=\dfrac{dQ}{dt}\); heat \(H=\int I^2R\,dt\).

\(I=a-2bt\), which becomes zero at \(t=\dfrac{a}{2b}\).
\(H=R\int_0^{a/2b}(a-2bt)^2dt=R\left[\dfrac{-(a-2bt)^3}{6b}\right]_0^{a/2b}\).
\(H=\dfrac{a^3R}{6b}\).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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The charge flowing through a resistance R varies with time t as Q = at - bt^2, where a and b are positive constants. The total heat produced in R is :

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