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Question Number 8550 by tawakalitu last updated on 15/Oct/16

Answered by ridwan balatif last updated on 16/Oct/16

Commented by tawakalitu last updated on 16/Oct/16

$$\mathrm{Wow}.\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Answered by sandy_suhendra last updated on 16/Oct/16

Commented by sandy_suhendra last updated on 16/Oct/16

$$\mathrm{answer}\mathrm{for}\mathrm{pic}\mathrm{A}$$$$\mathrm{loop}1\Rightarrow \mathrm{\Sigma }\mathrm{E}+\mathrm{\Sigma }\mathrm{I}.\mathrm{R}=0$$$$-{\mathrm{E}}_1+{\mathrm{I}}_1.{\mathrm{R}}_1+({\mathrm{I}}_1-{\mathrm{I}}_2).{\mathrm{R}}_2=0$$$$-2+{3\mathrm{I}}_1+({\mathrm{I}}_1-{\mathrm{I}}_2).3=0$$$${3\mathrm{I}}_1+{3\mathrm{I}}_1-{3\mathrm{I}}_2=2$$$${6\mathrm{I}}_1-{3\mathrm{I}}_2=2.....\left(1\right)$$$$$$$$\mathrm{loop}2\Rightarrow \mathrm{\Sigma }\mathrm{E}+\mathrm{\Sigma }\mathrm{I}.\mathrm{R}=0$$$$-{\mathrm{E}}_2+({\mathrm{I}}_1-{\mathrm{I}}_2).{\mathrm{R}}_2-{\mathrm{I}}_2.{\mathrm{R}}_3=0$$$$-1+({\mathrm{I}}_1-{\mathrm{I}}_2).3-{4\mathrm{I}}_2=0$$$${3\mathrm{I}}_1-{3\mathrm{I}}_2-{4\mathrm{I}}_2=1$$$${3\mathrm{I}}_1-{7\mathrm{I}}_2=1.....\left(2\right)$$$$\left(1\right)-2\times \left(2\right)\Rightarrow {11\mathrm{I}}_2=1$$$${\mathrm{I}}_2=\frac{1}{11}\mathrm{A}\mathrm{subs}\mathrm{to}\left(1\right)$$$$\mathrm{we}\mathrm{get}\Rightarrow {\mathrm{I}}_1=\frac{6}{11}\mathrm{A}$$$$\mathrm{\Delta }\mathrm{V}\mathrm{at}{\mathrm{R}}_1={\mathrm{I}}_1.{\mathrm{R}}_1=\frac{6}{11}\times 3=\frac{18}{11}\mathrm{volt}$$$$\mathrm{P}\mathrm{at}{\mathrm{R}}_1=\mathrm{\Delta }\mathrm{V}.{\mathrm{I}}_1=\frac{18}{11}\times \frac{6}{11}=\frac{108}{121}\mathrm{watt}$$$$\mathrm{\Delta }\mathrm{V}\mathrm{at}{\mathrm{R}}_2=({\mathrm{I}}_1-{\mathrm{I}}_2).{\mathrm{R}}_2=(\frac{6}{11}-\frac{1}{11})\times 3=\frac{15}{11}\mathrm{volt}$$$$\mathrm{P}\mathrm{at}{\mathrm{R}}_2=\mathrm{\Delta }\mathrm{V}.({\mathrm{I}}_1-{\mathrm{I}}_2)=\frac{15}{11}\times \frac{5}{11}=\frac{75}{121}\mathrm{watt}$$$$\mathrm{\Delta }\mathrm{V}\mathrm{at}{\mathrm{R}}_3={\mathrm{I}}_2.{\mathrm{R}}_3=\frac{1}{11}\times 4=\frac{4}{11}\mathrm{volt}$$$$\mathrm{P}\mathrm{at}{\mathrm{R}}_3=\mathrm{\Delta }\mathrm{V}.{\mathrm{I}}_2=\frac{4}{11}\times \frac{1}{11}=\frac{4}{121}\mathrm{watt}$$

Commented by tawakalitu last updated on 16/Oct/16

$$\mathrm{Wow}.\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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