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Question Number 146183 by tabata last updated on 11/Jul/21

Commented by tabata last updated on 11/Jul/21

$$intheadjacentfigure,calculate$$$$$$$$\left(1\right)theintensityofthecurrentpassingthrouphthecircuit$$$$$$$$\left(2\right)thepotentialdifferencebetweenthetwopointsx,y$$$$$$$$\left(3\right)thepotentialdifferencebetweenthetwopointsy,x$$

Answered by Olaf_Thorendsen last updated on 11/Jul/21

$${\mathrm{V}}_{\mathrm{X}}-{\mathrm{V}}_{\mathrm{Z}}={\mathrm{U}}_{\mathrm{XY}}-{\mathrm{U}}_{\mathrm{YZ}}={\mathrm{U}}_{\mathrm{R}}(\mathrm{R}=3\mathrm{\Omega })$$$$({\mathrm{E}}_1-r_1\mathrm{I})-({\mathrm{E}}_2+r_2\mathrm{I})=\mathrm{RI}$$$$\mathrm{I}=\frac{{\mathrm{E}}_1-{\mathrm{E}}_2}{\mathrm{R}+r_1+r_2}=\frac{8-6}{3+1+5}=\frac{2}{9}\mathrm{A}$$$${\mathrm{U}}_{\mathrm{XY}}={\mathrm{E}}_1-r_1\mathrm{I}=8-1\times \frac{2}{9}=\frac{70}{9}\mathrm{V}$$$${\mathrm{U}}_{\mathrm{YZ}}={\mathrm{E}}_2+r_2\mathrm{I}=6+5\times \frac{2}{9}=\frac{64}{9}\mathrm{V}$$$${\mathrm{U}}_{\mathrm{XZ}}={\mathrm{U}}_{\mathrm{R}}=\mathrm{RI}=3\times \frac{2}{9}=\frac{2}{3}\mathrm{V}(=\frac{70}{9}-\frac{64}{9})$$

Commented by Olaf_Thorendsen last updated on 11/Jul/21

$$1)\mathrm{I}=\frac{2}{9}\mathrm{A}$$$$2){\mathrm{U}}_{\mathrm{XY}}=\frac{70}{9}\mathrm{V}$$$$3){\mathrm{U}}_{\mathrm{YX}}=-{\mathrm{U}}_{\mathrm{XY}}=-\frac{70}{9}\mathrm{V}$$

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