Question

The charge flowing through a resistance $R$ varies with $q=\sqrt{a}t-\sqrt{b^3}{t}^2$. The total heat produced in $R$ before current become zero.

• A. $\frac{R{a}^{\frac{3}{2}}}{6b^{\frac{3}{2}}}$
• B. $\frac{R{a}^3}{6b}$
• C. $\frac{R^2\sqrt{a^2}}{6b}$
• D. $\frac{R{a}^3}{2b^{\frac{3}{2}}}$

Answer 1

The correct option is D $\frac{R{a}^3}{2b^{\frac{3}{2}}}$

Given R$=\sqrt{q}t-\sqrt{b^3}{t}^2$

$dH=I^{2}Rdt-------\left(1\right)andI=\frac{dq}{dt}=\sqrt{a}-2t\sqrt{b^3}atI=0;\sqrt{a}=2t\sqrt{b^3}andt=\frac{\sqrt{a}}{2\sqrt{b^3}}\text{Now integrating 1 with respect to time}\int dH={\int }_0^{\frac{\sqrt{a}}{2\sqrt{b^3}}}{I}^{2}Rdt=R{\int }_0^{\frac{\sqrt{a}}{2\sqrt{b^3}}}(\sqrt{a}-2t\sqrt{b^3}{)}^{2}dt=R{\int }_0^{\frac{\sqrt{a}}{2\sqrt{b^3}}}{(\sqrt{a})}^2-2\{2t\sqrt{b^3}\times \sqrt{a}\}+(2t\sqrt{b^3}{)}^{2}dt=R[\frac{a\sqrt{a}}{2\sqrt{b^3}}-\{\frac{2\sqrt{a{b}^3}}{{2\sqrt{b^3}}^2}{\sqrt{a}}^2\}+4{\left\{\frac{\sqrt{a}}{2\sqrt{b^3}}\right\}}^3{b}^3{]}_0^{\frac{\sqrt{a}}{2\sqrt{b^3}}}=R\{\frac{a^{\frac{3}{2}}}{2b^{\frac{3}{2}}}-\frac{a^{\frac{3}{2}}{b}^{\frac{3}{2}}}{{2b}^3}+\frac{a^{\frac{3}{2}}{b}^3}{2b^{\frac{9}{2}}}\}=\frac{a^{\frac{3}{2}}}{{2b}^{\frac{3}{2}}}-\frac{a^{\frac{3}{2}}}{2b^{\frac{3}{2}}}+\frac{R{a}^{\frac{3}{2}}}{2b^{\frac{3}{2}}}=\frac{R{a}^{\frac{3}{2}}}{{2b}^{\frac{3}{2}}}$