Question

The resistances of 20 and 40 ohms respectively can be connected to a source of e.m.f. compare the ratio of heat produced in the two resistances when they are connected in (i) Series (ii) Parallel ​

Answer 1

$$\begin{gathered} \mathrm{Given}, \\ {\mathrm{R}}_1=20\mathrm{ohm} \\ {\mathrm{R}}_2=40\mathrm{ohm} \\ \left(\mathrm{i}\right)\mathrm{The}\mathrm{resistance}\mathrm{for}\mathrm{series}\mathrm{combination}\mathrm{is}, \\ {\mathrm{R}}_{\mathrm{S}}={\mathrm{R}}_1+{\mathrm{R}}_2=20+40=60\mathrm{ohm} \\ \mathrm{Here}\mathrm{current}\mathrm{will}\mathrm{be}\mathrm{same}\mathrm{through}\mathrm{both}\mathrm{the}\mathrm{resistances}.\mathrm{i}.\mathrm{e}.\mathrm{I}=\frac{\mathrm{V}}{60} \\ \mathrm{Heat}\mathrm{produced}\mathrm{in}{\mathrm{R}}_1, \\ {\mathrm{H}}_1={\mathrm{I}}^2{\mathrm{R}}_1=(\frac{\mathrm{V}}{60}{)}^2\times 20=\frac{{\mathrm{V}}^2}{180} \\ \mathrm{Heat}\mathrm{produced}\mathrm{in}{\mathrm{R}}_2, \\ {\mathrm{H}}_2={\mathrm{I}}^2{\mathrm{R}}_2=(\frac{\mathrm{V}}{60}{)}^2\times 40=\frac{{\mathrm{V}}^2}{90} \\ {\mathrm{H}}_1:{\mathrm{H}}_2=\frac{{\mathrm{V}}^2}{180}:=1:2 \\ \left(\mathrm{ii}\right)\mathrm{In}\mathrm{this}\mathrm{case},\mathrm{voltage}\mathrm{will}\mathrm{be}\mathrm{same}\mathrm{across}\mathrm{both}\mathrm{the}\mathrm{resistances}. \\ \mathrm{Heat}\mathrm{produced}\mathrm{in}{\mathrm{R}}_1 \\ {\mathrm{H}}_1=\frac{{\mathrm{V}}^2}{{\mathrm{R}}_1}=\frac{{\mathrm{V}}^2}{20} \\ \mathrm{Heat}\mathrm{produced}\mathrm{in}{\mathrm{R}}_2 \\ {\mathrm{H}}_2=\frac{{\mathrm{V}}^2}{{\mathrm{R}}_2}=\frac{{\mathrm{V}}^2}{40} \\ \mathrm{So}, \\ {\mathrm{H}}_1:{\mathrm{H}}_2=\frac{{\mathrm{V}}^2}{20}:\frac{{\mathrm{V}}^2}{40}=2:1 \end{gathered}$$