Question Number 174974 by sciencestudent last updated on 15/Aug/22
Answered by MikeH last updated on 15/Aug/22
$$\mathrm{total}\:\mathrm{resistance}\:=\:\mathrm{2}.\mathrm{0}\:+\:\mathrm{14}.\mathrm{0}\:+\:\mathrm{9}.\mathrm{5}\:=\:\mathrm{25}.\mathrm{5}\:\Omega \\ $$$$\mathrm{but}\:{V}\:=\:{IR}\:\Rightarrow\:{I}\:=\:\frac{{V}}{{R}}\:=\:\frac{\mathrm{9}.\mathrm{0}}{\mathrm{25}.\mathrm{5}}\:=\:\mathrm{0}.\mathrm{35}\:\mathrm{A} \\ $$$$\mathrm{now}\:\mathrm{transversing}\:\mathrm{the}\:\mathrm{circuit}, \\ $$$$\mathrm{9}.\mathrm{5}\left(\mathrm{0}.\mathrm{35}\right)\:+\:\mathrm{14}\left(\mathrm{0}.\mathrm{35}\right)+\mathrm{2}.\mathrm{0}\left(\mathrm{0}.\mathrm{35}\right)−\mathrm{9}.\mathrm{0}\:\approx\:\mathrm{0}.\mathrm{0}{V} \\ $$