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Question Number 90632 by awlia last updated on 25/Apr/20
Answered by ajfour last updated on 25/Apr/20
$$−{i}\left(\mathrm{10}\Omega\right)+\mathrm{10}{V}−\mathrm{2}{i}\left(\mathrm{5}\Omega\right)+\mathrm{5}{V}=\mathrm{0} \\ $$$$\left({for}\:{the}\:{left}\:{loop}\:{or}\:{right}\:{loop}\right) \\ $$$$\Rightarrow\:\:\mathrm{3}{i}=\frac{\mathrm{15}{V}}{\mathrm{15}\Omega}\:=\:\mathrm{1}{A} \\ $$$${current}\:{in}\:\mathrm{5}\Omega\:{is}\:=\:\mathrm{2}{i}\:=\:\frac{\mathrm{2}}{\mathrm{3}}{A}. \\ $$