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Question

Physics Question on Resistance

The charge flowing through a resistance RR varies with time t as Q=atbt2 Q = at - bt^2 , where aa and bb are positive constants, The total heat produced in RR is :

A

a3R3b\frac{a^3 R}{3b}

B

a3R2b\frac{a^3 R}{2b}

C

a3Rb\frac{a^3 R}{b}

D

a3R6b\frac{a^3 R}{6b}

Answer

a3R6b\frac{a^3 R}{6b}

Explanation

Solution

Q=atbt2Q = at - bt^2
i=a2bti = a - 2bt { for i = 0 t=a2b\Rightarrow \, t = \frac{a}{2b} }
From joule's law of heating
dH=i2RdtdH = i^2Rdt
H=0a/2b(a2bt)2RdtH = \int\limits^{a/2b}_0 (a - 2bt)^2 Rdt
H=(a2bt)3R3×2b0a2b=a3R6bH = \frac{(a - 2bt)^3 R}{ - 3 \times 2b} \Bigg|^{\frac{a}{2b}}_{0} = \frac{a^3R}{6b}