Question

Three resistors 3 ohms, 6 ohms and 9 ohms are connected in parallel. This combination is connected in series with 4 ohm resistor. A cell of emf 20Vand internal resistance 2 ohms is connected across the final combination. Find the current through the 6 om resistor.

Answer 1

$$\begin{gathered} \mathrm{Parallel}\mathrm{combination}\mathrm{of}3,6,9\mathrm{ohms}: \\ \frac{1}{{\mathrm{R}}_{\mathrm{p}}}=\frac{1}{3}+\frac{1}{6}+\frac{1}{9} \\ =\frac{6}{18}+\frac{3}{18}+\frac{2}{18} \\ =\frac{11}{18} \\ {\mathrm{R}}_{\mathrm{p}}=\frac{18}{11}\mathrm{\Omega } \\ \frac{18}{11}\mathrm{and}4\mathrm{in}\mathrm{series}: \\ {\mathrm{R}}_{\mathrm{s}}=\frac{18}{11}+4 \\ =\frac{62}{11}\mathrm{\Omega } \\ \mathrm{Total}\mathrm{resistance}: \\ \mathrm{R}=\frac{62}{11}+2 \\ =\frac{84}{11}\mathrm{\Omega } \\ \mathrm{Total}\mathrm{current}: \\ \mathrm{I}=\frac{\mathrm{E}}{\mathrm{R}} \\ =\frac{20}{84/11} \\ =\frac{220}{84}\mathrm{A} \\ \mathrm{Potential}\mathrm{difference}\mathrm{across}6\mathrm{ohm}\mathrm{resistor}: \\ \mathrm{V}\text{'}={\mathrm{IR}}_{\mathrm{p}} \\ =\frac{220}{84}\times \frac{18}{11} \\ =4.29\mathrm{V} \\ \mathbf{Current}\mathbf{}\mathbf{through}\mathbf{}\mathbf{6}\mathbf{}\mathbf{ohm}\mathbf{}\mathbf{resistor}\mathbf{:} \\ \mathrm{I}\text{'}=\frac{\mathrm{V}\text{'}}{6} \\ =\frac{4.29}{6} \\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{71}\mathbf{}\mathbf{A} \end{gathered}$$