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Diah tugas 3

# Diah tugas 3

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12/24/2014

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Nama : Diah fauziahProdi : pend.teknik elektronikaNo reg : 5215122642
Tugas 3 pengukur elektro teknik
1. Jika RῨ = R1 = R2 = R3 dan RΔ = Ra = Rb = Rc, Buktikan bahwa nilai RΔ adalah 3 kali nilai RῨ.
Pembahasan soal :
RΔ = (R1.R2+R2.R3+R3.R1)/R1

RΔ = (RῨ. RῨ + RῨ. RῨ + RῨ. RῨ)/ RῨ

RΔ = {(RῨ)2 + (RῨ)2 + (RῨ)2}/ RY

RΔ = 3.( RῨ)2/ RῨ

RΔ = 3 RῨ
2.

Pada sebuah rangkaian diketahui R1 = 100 Ω, R2 = 80 Ω, R3 = 50 Ω
Transformasikan dari Bintang ke Segitiga, Berapakah nilai Ra=?, Rb=?, Rc=?Jawab :Ra = (R1.R2+R2.R3+R3.R1)/R1Ra = (100.80+80.50+50.100)/100Ra = (8000+4000+5000)/100Ra = 17000/100
Ra = 170 Ω
Rb = (R1.R2+R2.R3+R3.R1)/R2Rb = (100.80+80.50+50.100)/80

Rb = (8000+4000+5000)/80Rb = 17000/80
Rb = 212,5 Ω
Rc = (R1.R2+R2.R3+R3.R1)/R3Rc = (100.80+80.50+50.100)/50Rc = (8000+4000+5000)/50Rc = 17000/50
Rc = 340 Ω
3.

Jika diketahui rangkaian Ra = 10 Ω, Rb = 5 Ω, Rc = 2 Ω
Transformasikan dari segitiga ke bintang, berapa nilai R1=?, R2=?, R3=?Jawab :R1 = (Rb.Rc)/(Ra+Rb+Rc)R1 = (5.2)/(10+5+2)R1 = 10/17
R1 = 0,588 Ω
R2 = (Ra.Rc)/(Ra+Rb+Rc)R2 = (10.2)/(10+5+2)R2 = 20/17
R2 = 1,176 Ω

R3 = (Ra.Rb)/(Ra+Rb+Rc)R3 = (10.5)/(10+5+2)
R3 = 50/17 R3 = 2,94 Ω
4.

Transformasikan dari Segitiga ke bintang jika diketahui Ra = 20 Ω, Rb = 30 Ω, Rc = 40 Ω
tranformasikan dari segitiga bintang diatasJawab :R1 = (Rb.Rc)/(Ra+Rb+Rc)R1 = (30.40)/(20+30+40)R1 = 1200/90
R1 = 13,333 Ω
R2 = (Ra.Rc)/(Ra+Rb+Rc)R2 = (20.40)/(20+30+40)R2 = 800/90
R2 = 8,888 Ω
R3 = (Ra.Rb)/(Ra+Rb+Rc)R3 = (20.30)/(20+30+40)R3 = 50/90
R3 = 0,555 Ω
5.

Transformasikan dari Bintang ke Segitiga jika diketahui R1=18 Ω, R2 = 20 Ω, R3 = 22 Ω
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