Question Number 36140 by Cheyboy last updated on 29/May/18
Commented by Cheyboy last updated on 29/May/18
$${Find}\:{the}\:{current}\:{and}\:{voltage}\:{of} \\ $$$${the}\:{letters}\:{A}\:{to}\:{L} \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 01/Jun/18
$$\frac{\mathrm{1}}{{R}}=\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${R}=\frac{\mathrm{15}}{\mathrm{8}} \\ $$$${totsl}\:{resistance}=\mathrm{8}+\frac{\mathrm{15}}{\mathrm{8}}+\mathrm{6} \\ $$$$=\frac{\mathrm{64}+\mathrm{15}+\mathrm{48}}{\mathrm{8}}=\frac{\mathrm{127}}{\mathrm{8}} \\ $$$${i}=\frac{\mathrm{12}}{\frac{\mathrm{127}}{\mathrm{8}}}=\frac{\mathrm{96}}{\mathrm{127}}{amp} \\ $$$${current}\:{through}\:{resistance}\:\mathrm{8}{ohm}=\frac{\mathrm{96}}{\mathrm{127}}{amp} \\ $$$${current}\:{through}\mathrm{6}\:{ohm}=\frac{\mathrm{96}}{\mathrm{127}}\:{amp} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}:\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$=\mathrm{5}:\mathrm{3} \\ $$$${current}\:{through}\:\mathrm{3}\:{ohm}=\frac{\mathrm{5}}{\mathrm{8}}×\frac{\mathrm{96}}{\mathrm{127}}=\frac{\mathrm{60}}{\mathrm{127}}{amp} \\ $$$${current}\:{through}\:\mathrm{5}\:{ohm}=\frac{\mathrm{3}}{\mathrm{8}}×\frac{\mathrm{96}}{\mathrm{127}}=\frac{\mathrm{36}}{\mathrm{127}}{amp} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$