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Question Number 153057 by DELETED last updated on 04/Sep/21

Answered by DELETED last updated on 04/Sep/21

i_1 +i_2 =i_3  .....(1)  ΣE+Σi.R=0  hk kirchoff II  loop 1 (searah jarum jam)  −4+4i_1 +4i_3 =0  i_1 +i_3 =1.....(2)   loop 2 (berlawanan jarum jam)  −2+2i_2 +4i_3 =0  i_2 +2i_3 =1....(3)  (1)→(2)  i_1 +i_1 +i_2 =1       2i_1 +i_2 =1....(4)  (1)→(3)  i_2 +2(i_1 +i_2 )=1                 2i_1 +3i_2 =1....(5)  (4)→(5)    2i_1 +3i_2 =1     2i_1 +i_2 =1  −−−−−−−  2i_2 =0→i_2 =0  2i_1 +0=1→2i_1 =1  i_1 =(1/2)  →i_3 =(1/2)+0=(1/2)  Daya_(4Ω) =i^2 ×R=((1/2))^2 ×4         =(1/4)×4=1 A//

$$\mathrm{i}_{\mathrm{1}} +\mathrm{i}_{\mathrm{2}} =\mathrm{i}_{\mathrm{3}} \:.....\left(\mathrm{1}\right) \\ $$$$\Sigma\mathrm{E}+\Sigma\mathrm{i}.\mathrm{R}=\mathrm{0}\:\:\mathrm{hk}\:\mathrm{kirchoff}\:\mathrm{II} \\ $$$$\mathrm{loop}\:\mathrm{1}\:\left(\mathrm{searah}\:\mathrm{jarum}\:\mathrm{jam}\right) \\ $$$$−\mathrm{4}+\mathrm{4i}_{\mathrm{1}} +\mathrm{4i}_{\mathrm{3}} =\mathrm{0} \\ $$$$\mathrm{i}_{\mathrm{1}} +\mathrm{i}_{\mathrm{3}} =\mathrm{1}.....\left(\mathrm{2}\right)\: \\ $$$$\mathrm{loop}\:\mathrm{2}\:\left(\mathrm{berlawanan}\:\mathrm{jarum}\:\mathrm{jam}\right) \\ $$$$−\mathrm{2}+\mathrm{2i}_{\mathrm{2}} +\mathrm{4i}_{\mathrm{3}} =\mathrm{0} \\ $$$$\mathrm{i}_{\mathrm{2}} +\mathrm{2i}_{\mathrm{3}} =\mathrm{1}....\left(\mathrm{3}\right) \\ $$$$\left(\mathrm{1}\right)\rightarrow\left(\mathrm{2}\right)\:\:\mathrm{i}_{\mathrm{1}} +\mathrm{i}_{\mathrm{1}} +\mathrm{i}_{\mathrm{2}} =\mathrm{1} \\ $$$$\:\:\:\:\:\mathrm{2i}_{\mathrm{1}} +\mathrm{i}_{\mathrm{2}} =\mathrm{1}....\left(\mathrm{4}\right) \\ $$$$\left(\mathrm{1}\right)\rightarrow\left(\mathrm{3}\right)\:\:\mathrm{i}_{\mathrm{2}} +\mathrm{2}\left(\mathrm{i}_{\mathrm{1}} +\mathrm{i}_{\mathrm{2}} \right)=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2i}_{\mathrm{1}} +\mathrm{3i}_{\mathrm{2}} =\mathrm{1}....\left(\mathrm{5}\right) \\ $$$$\left(\mathrm{4}\right)\rightarrow\left(\mathrm{5}\right) \\ $$$$\:\:\mathrm{2i}_{\mathrm{1}} +\mathrm{3i}_{\mathrm{2}} =\mathrm{1} \\ $$$$\:\:\:\mathrm{2i}_{\mathrm{1}} +\mathrm{i}_{\mathrm{2}} =\mathrm{1} \\ $$$$−−−−−−− \\ $$$$\mathrm{2i}_{\mathrm{2}} =\mathrm{0}\rightarrow\mathrm{i}_{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{2i}_{\mathrm{1}} +\mathrm{0}=\mathrm{1}\rightarrow\mathrm{2i}_{\mathrm{1}} =\mathrm{1} \\ $$$$\mathrm{i}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\:\:\rightarrow\mathrm{i}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{0}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{Daya}_{\mathrm{4}\Omega} =\mathrm{i}^{\mathrm{2}} ×\mathrm{R}=\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} ×\mathrm{4} \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}×\mathrm{4}=\mathrm{1}\:\mathrm{A}// \\ $$

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