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Question Number 149946 by DELETED last updated on 08/Aug/21

Answered by DELETED last updated on 08/Aug/21

Jawaban no 2 Gunakan Hk Kirchoff I ΣI_(masuk) =ΣI_(keluar) I_1 +I_2 +I_4 =I_3 +I_5 2+3+4=6+I_5 9=6+I_5 →I_5 =9−6=3 A

$$\boldsymbol{\mathrm{Jawaban}}\:\boldsymbol{\mathrm{no}}\:\mathrm{2} \\ $$$$\mathrm{Gunakan}\:\mathrm{Hk}\:\mathrm{Kirchoff}\:\mathrm{I} \\ $$$$\Sigma\mathrm{I}_{\mathrm{masuk}} =\Sigma\mathrm{I}_{\mathrm{keluar}} \\ $$$$\mathrm{I}_{\mathrm{1}} +\mathrm{I}_{\mathrm{2}} +\mathrm{I}_{\mathrm{4}} =\mathrm{I}_{\mathrm{3}} +\mathrm{I}_{\mathrm{5}} \\ $$$$\mathrm{2}+\mathrm{3}+\mathrm{4}=\mathrm{6}+\mathrm{I}_{\mathrm{5}} \: \\ $$$$\mathrm{9}=\mathrm{6}+\mathrm{I}_{\mathrm{5}} \:\rightarrow\mathrm{I}_{\mathrm{5}} =\mathrm{9}−\mathrm{6}=\mathrm{3}\:\mathrm{A} \\ $$$$ \\ $$

Answered by DELETED last updated on 08/Aug/21

Jawaban no 3 Tentang Jembatan Wheatston Syarat:R_1 ×R_3 =R_2 ×R_4 10×15 =15×10 R_1 dan R_4 →Hubungan seri 1 R_(s1) =10+10=20Ω R_2 dan R_3 →Hubungan seri 2 R_(s2) =15+15=30 (1/R_p )=(1/R_(s1) ) +(1/R_(s2) ) =(1/(20))+(1/(30)) =((30+20)/(600)) =((50)/(600)) R_p =R_(total) =((600)/(50)) =12Ω →R _(pengganti) =12Ω

$$\boldsymbol{\mathrm{Jawaban}}\:\boldsymbol{\mathrm{no}}\:\mathrm{3} \\ $$$$\mathrm{Tentang}\:\mathrm{Jembatan}\:\mathrm{Wheatston} \\ $$$$\mathrm{Syarat}:\mathrm{R}_{\mathrm{1}} ×\mathrm{R}_{\mathrm{3}} =\mathrm{R}_{\mathrm{2}} ×\mathrm{R}_{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{10}×\mathrm{15}\:=\mathrm{15}×\mathrm{10} \\ $$$$\mathrm{R}_{\mathrm{1}} \mathrm{dan}\:\mathrm{R}_{\mathrm{4}} \:\rightarrow\mathrm{Hubungan}\:\mathrm{seri}\:\mathrm{1} \\ $$$$\mathrm{R}_{\mathrm{s1}} =\mathrm{10}+\mathrm{10}=\mathrm{20}\Omega \\ $$$$\mathrm{R}_{\mathrm{2}} \:\mathrm{dan}\:\mathrm{R}_{\mathrm{3}} \rightarrow\mathrm{Hubungan}\:\mathrm{seri}\:\mathrm{2} \\ $$$$\mathrm{R}_{\mathrm{s2}} =\mathrm{15}+\mathrm{15}=\mathrm{30} \\ $$$$\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{p}} }=\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{s1}} }\:+\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{s2}} }\:=\frac{\mathrm{1}}{\mathrm{20}}+\frac{\mathrm{1}}{\mathrm{30}} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{30}+\mathrm{20}}{\mathrm{600}}\:=\frac{\mathrm{50}}{\mathrm{600}}\:\:\:\:\: \\ $$$$\mathrm{R}_{\mathrm{p}} =\mathrm{R}_{\mathrm{total}} =\frac{\mathrm{600}}{\mathrm{50}}\:=\mathrm{12}\Omega \\ $$$$\rightarrow\mathrm{R}\:_{\mathrm{pengganti}} =\mathrm{12}\Omega \\ $$

Answered by DELETED last updated on 08/Aug/21

R_(s1) =2+6=8Ω (1/R_(p1) )=(1/R_(8Ω) ) + (1/R_(s1) ) =(1/8) + (1/8) =(2/8)→R_(p1) =(8/2) =4Ω R_(s2) =R_(3Ω) +R_(1Ω) +R_(p1) =3+1+4=8 Ω R_(s3) =R_(10Ω) +R_(5Ω) +R_(9Ω) =10+5+9=24Ω (1/R_(p2) )=(1/R_(s2) )+(1/R_(s3) )=(1/8)+(1/(24)) =(3/(24))+(1/(24))=(4/(24)) R_(p2) =((24)/4) =6Ω →R_(p2) =R_(total) =6Ω I_(total) =(V/R_(total) )=((12)/6)=2 A

$$\mathrm{R}_{\mathrm{s1}} =\mathrm{2}+\mathrm{6}=\mathrm{8}\Omega \\ $$$$\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{p1}} }=\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{8}\Omega} }\:+\:\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{s1}} }\:=\frac{\mathrm{1}}{\mathrm{8}}\:+\:\frac{\mathrm{1}}{\mathrm{8}}\: \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{2}}{\mathrm{8}}\rightarrow\mathrm{R}_{\mathrm{p1}} =\frac{\mathrm{8}}{\mathrm{2}}\:=\mathrm{4}\Omega \\ $$$$\mathrm{R}_{\mathrm{s2}} =\mathrm{R}_{\mathrm{3}\Omega} +\mathrm{R}_{\mathrm{1}\Omega} +\mathrm{R}_{\mathrm{p1}} \\ $$$$\:\:\:\:\:\:\:=\mathrm{3}+\mathrm{1}+\mathrm{4}=\mathrm{8}\:\Omega \\ $$$$\mathrm{R}_{\mathrm{s3}} =\mathrm{R}_{\mathrm{10}\Omega} +\mathrm{R}_{\mathrm{5}\Omega} +\mathrm{R}_{\mathrm{9}\Omega} \\ $$$$\:\:\:\:\:\:\:=\mathrm{10}+\mathrm{5}+\mathrm{9}=\mathrm{24}\Omega \\ $$$$\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{p2}} }=\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{s2}} }+\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{s3}} }=\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{24}} \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{3}}{\mathrm{24}}+\frac{\mathrm{1}}{\mathrm{24}}=\frac{\mathrm{4}}{\mathrm{24}} \\ $$$$\mathrm{R}_{\mathrm{p2}} =\frac{\mathrm{24}}{\mathrm{4}}\:=\mathrm{6}\Omega\:\rightarrow\mathrm{R}_{\mathrm{p2}} =\mathrm{R}_{\mathrm{total}} =\mathrm{6}\Omega \\ $$$$\mathrm{I}_{\mathrm{total}} =\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{total}} }=\frac{\mathrm{12}}{\mathrm{6}}=\mathrm{2}\:\mathrm{A} \\ $$

Answered by DELETED last updated on 08/Aug/21

Jawaban no 4 R_(total) =5+15+30=50Ω Gunakan Hk Kirchoff 2 bunyinya: ΣI (R+r)=ΣE i (50+3/2)=8 i = (8/(51.5))=0.155 A V_(jepit) =i×R=0.155×50=7.75 Volt

$$\boldsymbol{\mathrm{Jawaban}}\:\boldsymbol{\mathrm{no}}\:\mathrm{4} \\ $$$$\mathrm{R}_{\mathrm{total}} =\mathrm{5}+\mathrm{15}+\mathrm{30}=\mathrm{50}\Omega \\ $$$$\mathrm{Gunakan}\:\mathrm{Hk}\:\mathrm{Kirchoff}\:\mathrm{2} \\ $$$$\mathrm{bunyinya}:\:\Sigma\mathrm{I}\:\left(\mathrm{R}+\mathrm{r}\right)=\Sigma\mathrm{E} \\ $$$$\mathrm{i}\:\left(\mathrm{50}+\mathrm{3}/\mathrm{2}\right)=\mathrm{8} \\ $$$$\mathrm{i}\:=\:\frac{\mathrm{8}}{\mathrm{51}.\mathrm{5}}=\mathrm{0}.\mathrm{155}\:\mathrm{A} \\ $$$$\mathrm{V}_{\mathrm{jepit}} =\mathrm{i}×\mathrm{R}=\mathrm{0}.\mathrm{155}×\mathrm{50}=\mathrm{7}.\mathrm{75}\:\mathrm{Volt} \\ $$

Answered by DELETED last updated on 08/Aug/21

Jawaban no 5 (1/R_p )=(1/R_(3Ω) )+(1/R_(6Ω) )=(1/3)+(1/6)=(3/6) R_p =(6/3)=2Ω R_s =R_(total) =R_p +R_(4Ω) =2+4=6Ω I_(total) =((12)/6)=2 A I_(total) =I_(4Ω) =2 A V_(4Ω) =I_(4Ω) ×R_(4Ω) =2×4=8 Volt

$$\boldsymbol{\mathrm{Jawaban}}\:\boldsymbol{\mathrm{no}}\:\mathrm{5} \\ $$$$\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{p}} }=\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{3}\Omega} }+\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{6}\Omega} }=\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{6}}=\frac{\mathrm{3}}{\mathrm{6}} \\ $$$$\mathrm{R}_{\mathrm{p}} =\frac{\mathrm{6}}{\mathrm{3}}=\mathrm{2}\Omega \\ $$$$\mathrm{R}_{\mathrm{s}} =\mathrm{R}_{\mathrm{total}} =\mathrm{R}_{\mathrm{p}} +\mathrm{R}_{\mathrm{4}\Omega} =\mathrm{2}+\mathrm{4}=\mathrm{6}\Omega \\ $$$$\mathrm{I}_{\mathrm{total}} =\frac{\mathrm{12}}{\mathrm{6}}=\mathrm{2}\:\mathrm{A} \\ $$$$\mathrm{I}_{\mathrm{total}} =\mathrm{I}_{\mathrm{4}\Omega} =\mathrm{2}\:\mathrm{A}\: \\ $$$$\mathrm{V}_{\mathrm{4}\Omega} =\mathrm{I}_{\mathrm{4}\Omega} ×\mathrm{R}_{\mathrm{4}\Omega} =\mathrm{2}×\mathrm{4}=\mathrm{8}\:\mathrm{Volt} \\ $$

Answered by yeti123 last updated on 09/Aug/21

it′s so easy to solve. more difficult problem, please...

$$\mathrm{it}'\mathrm{s}\:\mathrm{so}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solve}.\:\mathrm{more}\:\mathrm{difficult}\:\mathrm{problem},\:\mathrm{please}... \\ $$

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