Question Number 146233 by tabata last updated on 12/Jul/21
Commented by tabata last updated on 12/Jul/21
$${From}\:{the}\:{figure}\:,{calculate}\: \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:{the}\:{electric}\:{potential}\:{difference}\:{between}\:{A}\:{and}\:{B} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\:{the}\:{electromotive}\:{force} \\ $$
Answered by Olaf_Thorendsen last updated on 12/Jul/21
$$\left.\mathrm{1}\right) \\ $$$$\mathrm{U}_{\mathrm{AB}} \:=\:\mathrm{V}_{\mathrm{A}} −\mathrm{V}_{\mathrm{B}} \\ $$$$\mathrm{U}_{\mathrm{AB}} \:=\:\mathrm{4A}×\mathrm{2}\Omega+\mathrm{3A}×\mathrm{4}\Omega+\mathrm{14V}+\mathrm{4A}×\mathrm{1}\Omega \\ $$$$\mathrm{U}_{\mathrm{AB}} \:=\:\mathrm{8V}+\mathrm{12V}+\mathrm{14V}+\mathrm{4V} \\ $$$$\mathrm{U}_{\mathrm{AB}} \:=\:\mathrm{38V} \\ $$$$ \\ $$$$\left.\mathrm{2}\right) \\ $$$$\mathrm{3A}×\mathrm{4}\Omega\:=\:−\mathrm{V}_{\mathrm{B}} +\left(\mathrm{4A}−\mathrm{3A}\right)×\mathrm{1}\Omega \\ $$$$\mathrm{12V}\:=\:−\mathrm{V}_{\mathrm{B}} +\mathrm{1V} \\ $$$$\mathrm{V}_{\mathrm{B}} \:=\:−\mathrm{11V} \\ $$$$ \\ $$