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

Answered by Zainalarifin 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

Answered by Zainalarifin 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

Answered by Zainalarifin 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

Answered by Zainalarifin 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

Answered by Zainalarifin 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

Answered by yeti123 last updated on 09/Aug/21

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