Question Number 5189 by sanusihammed last updated on 28/Apr/16
Commented by FilupSmith last updated on 28/Apr/16
$${Q}\mathrm{1} \\ $$$${V}={IR} \\ $$$${I}=\frac{{V}}{{R}}\:\Rightarrow\:{I}=\frac{\mathrm{12}}{\mathrm{16}+\mathrm{8}} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{Voltage}\:\mathrm{over}\:\mathrm{16ohm}: \\ $$$${V}={IR} \\ $$$${V}=\frac{\mathrm{1}}{\mathrm{2}}\centerdot\mathrm{16} \\ $$$${V}_{\mathrm{1}} =\mathrm{8}{v} \\ $$$$ \\ $$$$\mathrm{Voltage}\:\mathrm{over}\:\mathrm{8ohm}: \\ $$$${V}={IR} \\ $$$${V}=\frac{\mathrm{1}}{\mathrm{2}}\centerdot\mathrm{8} \\ $$$${V}_{\mathrm{2}} =\mathrm{4} \\ $$$$ \\ $$$$\mathrm{Chect}\:\mathrm{voltages}: \\ $$$${V}={V}_{\mathrm{1}} +{V}_{\mathrm{2}} \\ $$$$\mathrm{12}=\mathrm{8}+\mathrm{4} \\ $$$$\mathrm{12}=\mathrm{12} \\ $$$$\therefore\mathrm{Correct} \\ $$