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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

dear sir: if you just do your own exercises, i.e. when you solve your questions by yourself, you don′t need to use the forum function of this app. forum function is for exchanging with other people, i.e. other people help you to solve your questions or you help other people to solve their questions. in forum please use a language which other people may understand, generally english. thanks!

$${dear}\:{sir}: \\ $$$${if}\:{you}\:{just}\:{do}\:{your}\:{own}\:{exercises},\:{i}.{e}. \\ $$$${when}\:{you}\:{solve}\:{your}\:{questions}\:{by} \\ $$$${yourself},\:{you}\:{don}'{t}\:{need}\:{to}\:{use}\:{the} \\ $$$${forum}\:{function}\:{of}\:{this}\:{app}.\:{forum} \\ $$$${function}\:{is}\:{for}\:{exchanging}\:{with} \\ $$$${other}\:{people},\:{i}.{e}.\:{other}\:{people}\:{help} \\ $$$${you}\:{to}\:{solve}\:{your}\:{questions}\:{or}\:{you} \\ $$$${help}\:{other}\:{people}\:{to}\:{solve}\:{their} \\ $$$${questions}.\:\:{in}\:{forum}\:{please}\:{use}\:{a} \\ $$$${language}\:{which}\:{other}\:{people}\:{may} \\ $$$${understand},\:{generally}\:{english}.\:{thanks}! \\ $$

Answered by DELETED last updated on 05/Sep/21

Jawab •i_3 =i_1 +i_2 (1) •ΣE+Σi(R+r)=0 lihat loop 1: −6+i_1 (2+1)+3i_3 =0 3i_1 +3(i_1 +i_2 )=6 6i_1 +3i_2 =6→:3 2i_1 +i_2 =2 (2) lihat loop 2: −6+i_2 (5+1)+3i_3 =0 6i_2 +3i_3 =6 6i_2 +3(i_1 +i_2 )=6 3i_1 +9i_2 =6→:3 i_1 +3i_2 =2 (3) (2) →(3) 2i_1 +i_2 =2 ∣×1 i_1 +3i_2 =2 ∣×2 −−−−−−−− 2i_1 +i_2 =2 2i_1 +6i_2 =4 −−−−−−− −5i_2 =−2 →i_2 =0.4 A i_1 +3i_2 =2→i_1 +1,2=2 →i_1 =0,8 A i_3 =i_1 +i_2 =0,8+0,4=1,2 A

$$\mathrm{Jawab} \\ $$$$\bullet\mathrm{i}_{\mathrm{3}} =\mathrm{i}_{\mathrm{1}} +\mathrm{i}_{\mathrm{2}} \:\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\bullet\Sigma\mathrm{E}+\Sigma\mathrm{i}\left(\mathrm{R}+\mathrm{r}\right)=\mathrm{0} \\ $$$$\:\mathrm{lihat}\:\mathrm{loop}\:\mathrm{1}: \\ $$$$−\mathrm{6}+\mathrm{i}_{\mathrm{1}} \left(\mathrm{2}+\mathrm{1}\right)+\mathrm{3i}_{\mathrm{3}} =\mathrm{0} \\ $$$$\mathrm{3i}_{\mathrm{1}} +\mathrm{3}\left(\mathrm{i}_{\mathrm{1}} +\mathrm{i}_{\mathrm{2}} \right)=\mathrm{6}\: \\ $$$$\mathrm{6i}_{\mathrm{1}} +\mathrm{3i}_{\mathrm{2}} =\mathrm{6}\rightarrow:\mathrm{3}\: \\ $$$$\mathrm{2i}_{\mathrm{1}} +\mathrm{i}_{\mathrm{2}} =\mathrm{2}\:\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$$\mathrm{lihat}\:\mathrm{loop}\:\mathrm{2}: \\ $$$$−\mathrm{6}+\mathrm{i}_{\mathrm{2}} \left(\mathrm{5}+\mathrm{1}\right)+\mathrm{3i}_{\mathrm{3}} =\mathrm{0} \\ $$$$\mathrm{6i}_{\mathrm{2}} +\mathrm{3i}_{\mathrm{3}} =\mathrm{6} \\ $$$$\mathrm{6i}_{\mathrm{2}} +\mathrm{3}\left(\mathrm{i}_{\mathrm{1}} +\mathrm{i}_{\mathrm{2}} \right)=\mathrm{6} \\ $$$$\mathrm{3i}_{\mathrm{1}} +\mathrm{9i}_{\mathrm{2}} =\mathrm{6}\rightarrow:\mathrm{3} \\ $$$$\mathrm{i}_{\mathrm{1}} +\mathrm{3i}_{\mathrm{2}} =\mathrm{2}\:\:\:\:\left(\mathrm{3}\right) \\ $$$$\left(\mathrm{2}\right)\:\rightarrow\left(\mathrm{3}\right)\:\mathrm{2i}_{\mathrm{1}} +\mathrm{i}_{\mathrm{2}} =\mathrm{2}\:\mid×\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{i}_{\mathrm{1}} +\mathrm{3i}_{\mathrm{2}} =\mathrm{2}\:\mid×\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−−−−−−−− \\ $$$$\:\:\mathrm{2i}_{\mathrm{1}} +\mathrm{i}_{\mathrm{2}} =\mathrm{2} \\ $$$$\:\:\:\mathrm{2i}_{\mathrm{1}} +\mathrm{6i}_{\mathrm{2}} =\mathrm{4} \\ $$$$−−−−−−− \\ $$$$\:\:\:−\mathrm{5i}_{\mathrm{2}} =−\mathrm{2}\:\:\:\rightarrow\mathrm{i}_{\mathrm{2}} =\mathrm{0}.\mathrm{4}\:\mathrm{A} \\ $$$$\mathrm{i}_{\mathrm{1}} +\mathrm{3i}_{\mathrm{2}} =\mathrm{2}\rightarrow\mathrm{i}_{\mathrm{1}} +\mathrm{1},\mathrm{2}=\mathrm{2}\:\rightarrow\mathrm{i}_{\mathrm{1}} =\mathrm{0},\mathrm{8}\:\mathrm{A} \\ $$$$\mathrm{i}_{\mathrm{3}} =\mathrm{i}_{\mathrm{1}} +\mathrm{i}_{\mathrm{2}} =\mathrm{0},\mathrm{8}+\mathrm{0},\mathrm{4}=\mathrm{1},\mathrm{2}\:\mathrm{A} \\ $$

Answered by DELETED last updated on 05/Sep/21

jawab no 2 •V_(E→F) =i_3 ×R_3 =1,2×3=3,6 V •V_(E→C) =E+I_2 (R_2 +r_2 ) =6−0,4(5+1) =6−2,4=3,6 Volt

$$\mathrm{jawab}\:\mathrm{no}\:\mathrm{2} \\ $$$$\bullet\mathrm{V}_{\mathrm{E}\rightarrow\mathrm{F}} =\mathrm{i}_{\mathrm{3}} ×\mathrm{R}_{\mathrm{3}} =\mathrm{1},\mathrm{2}×\mathrm{3}=\mathrm{3},\mathrm{6}\:\mathrm{V} \\ $$$$\bullet\mathrm{V}_{\mathrm{E}\rightarrow\mathrm{C}} =\mathrm{E}+\mathrm{I}_{\mathrm{2}} \left(\mathrm{R}_{\mathrm{2}} +\mathrm{r}_{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{6}−\mathrm{0},\mathrm{4}\left(\mathrm{5}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{6}−\mathrm{2},\mathrm{4}=\mathrm{3},\mathrm{6}\:\mathrm{Volt} \\ $$

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