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Question Number 153173 by DELETED last updated on 05/Sep/21

Commented by mr W last updated on 05/Sep/21

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Answered by DELETED last updated on 05/Sep/21

$$\mathrm{Jawab}$$$$\bullet {\mathrm{i}}_3={\mathrm{i}}_1+{\mathrm{i}}_2\left(1\right)$$$$\bullet \mathrm{\Sigma }\mathrm{E}+\mathrm{\Sigma }\mathrm{i}(\mathrm{R}+\mathrm{r})=0$$$$\mathrm{lihat}\mathrm{loop}1:$$$$-6+{\mathrm{i}}_1(2+1)+{3\mathrm{i}}_3=0$$$${3\mathrm{i}}_1+3({\mathrm{i}}_1+{\mathrm{i}}_2)=6$$$${6\mathrm{i}}_1+{3\mathrm{i}}_2=6\to :3$$$${2\mathrm{i}}_1+{\mathrm{i}}_2=2\left(2\right)$$$$\mathrm{lihat}\mathrm{loop}2:$$$$-6+{\mathrm{i}}_2(5+1)+{3\mathrm{i}}_3=0$$$${6\mathrm{i}}_2+{3\mathrm{i}}_3=6$$$${6\mathrm{i}}_2+3({\mathrm{i}}_1+{\mathrm{i}}_2)=6$$$${3\mathrm{i}}_1+{9\mathrm{i}}_2=6\to :3$$$${\mathrm{i}}_1+{3\mathrm{i}}_2=2\left(3\right)$$$$\left(2\right)\to \left(3\right){2\mathrm{i}}_1+{\mathrm{i}}_2=2\mid \times 1$$$${\mathrm{i}}_1+{3\mathrm{i}}_2=2\mid \times 2$$$$--------$$$${2\mathrm{i}}_1+{\mathrm{i}}_2=2$$$${2\mathrm{i}}_1+{6\mathrm{i}}_2=4$$$$-------$$$$-{5\mathrm{i}}_2=-2\to {\mathrm{i}}_2=0.4\mathrm{A}$$$${\mathrm{i}}_1+{3\mathrm{i}}_2=2\to {\mathrm{i}}_1+1,2=2\to {\mathrm{i}}_1=0,8\mathrm{A}$$$${\mathrm{i}}_3={\mathrm{i}}_1+{\mathrm{i}}_2=0,8+0,4=1,2\mathrm{A}$$

Answered by DELETED last updated on 05/Sep/21

$$\mathrm{jawab}\mathrm{no}2$$$$\bullet {\mathrm{V}}_{\mathrm{E}\to \mathrm{F}}={\mathrm{i}}_3\times {\mathrm{R}}_3=1,2\times 3=3,6\mathrm{V}$$$$\bullet {\mathrm{V}}_{\mathrm{E}\to \mathrm{C}}=\mathrm{E}+{\mathrm{I}}_2({\mathrm{R}}_2+{\mathrm{r}}_2)$$$$=6-0,4(5+1)$$$$=6-2,4=3,6\mathrm{Volt}$$

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