Question

two identical resistors each of resistance 3 ohm are connected in turn 1) series 2) parallel to a battery of 120 V . calculate the ratio of power consumed in the two cases

Answer 1

Given,
$$\begin{gathered} {\mathrm{R}}_1=3\mathrm{ohm} \\ {\mathrm{R}}_2=3\mathrm{ohm} \end{gathered}$$
V=120 V
Series combination:
Resultant resistance is given by,
$$\begin{gathered} {\mathrm{R}}_{\mathrm{s}}={\mathrm{R}}_1+{\mathrm{R}}_2=3+3=6\mathrm{ohm} \\ \mathrm{Power},{\mathrm{P}}_{\mathrm{s}}=\frac{{\mathrm{V}}^2}{{\mathrm{R}}_{\mathrm{s}}}=\frac{{120}^2}{6}=2400\mathrm{W} \\ \mathrm{Parallel}\mathrm{combination}, \\ \mathrm{Resultant}\mathrm{resistance}\mathrm{is}\mathrm{given}\mathrm{by} \\ {\mathrm{R}}_{\mathrm{P}}=\frac{{\mathrm{R}}_1{\mathrm{R}}_2}{{\mathrm{R}}_1+{\mathrm{R}}_2}=\frac{3\times 3}{3+3}=\frac{9}{6}=1.5\mathrm{ohm} \\ \mathrm{Power},{\mathrm{P}}_{\mathrm{P}}=\frac{{\mathrm{V}}^2}{{\mathrm{R}}_{\mathrm{P}}}=\frac{{120}^2}{1.5}=9600\mathrm{W} \\ \mathrm{Now}, \\ {\mathrm{P}}_{\mathrm{s}}:{\mathrm{P}}_{\mathrm{P}}=2400:9600=1:4 \end{gathered}$$