Question

A WIRE OF RESISTANCE 1ohm connected to a cell of 4V .If another identical wire is strethced twice its length and connected in parallel to first wire .What will be the ratio of power developed across the wires ?

Answer 1

Given,
$$\begin{gathered} \mathrm{Resistance}\mathrm{of}\mathrm{first}\mathrm{wire},\mathrm{R}=1\mathrm{ohm} \\ \mathrm{When}\mathrm{this}\mathrm{wire}\mathrm{is}\mathrm{stretched}\mathrm{twice}\mathrm{its}\mathrm{length},\mathrm{then}\mathrm{according}\mathrm{to}\mathrm{R}=\frac{\mathrm{\rho L}}{\mathrm{A}}, \\ \mathrm{The}\mathrm{resistance}\mathrm{of}\mathrm{second}\mathrm{wire}\mathrm{will}\mathrm{be} \\ \mathrm{R}\text{'}=\frac{\mathrm{\rho L}\text{'}}{\mathrm{A}\text{'}} \\ =\frac{\mathrm{\rho }\left(2\mathrm{L}\right)}{\mathrm{A}/2}(\mathrm{Since}\mathrm{the}\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{wire}\mathrm{remains}\mathrm{same}) \\ =4\frac{\mathrm{\rho L}}{\mathrm{A}}=4\mathrm{R}=4\times 1=4\mathrm{ohm} \\ \mathrm{Power}\mathrm{developed}\mathrm{across}\mathrm{first}\mathrm{wire},\mathrm{P}=\frac{{\mathrm{V}}^2}{\mathrm{R}}=\frac{4^2}{1}=16\mathrm{W} \\ \mathrm{Power}\mathrm{developed}\mathrm{across}\mathrm{second}\mathrm{wire},\mathrm{P}\text{'}=\frac{{\mathrm{V}}^2}{{\mathrm{R}}^{\text{'}}}(\mathrm{Voltage}\mathrm{remains}\mathrm{same}\mathrm{in}\mathrm{parallel}\mathrm{combination}) \\ \mathrm{P}\text{'}=\frac{4^2}{4}=\frac{16}{4}=4\mathrm{W} \\ \mathrm{Thus},\mathrm{P}:\mathrm{P}\text{'}=16:4=4:1 \end{gathered}$$