Salwa R

26 Maret 2024 21:27

Iklan

Salwa R

26 Maret 2024 21:27

Pertanyaan

tolongg yya

tolongg yya

alt

Ikuti Tryout SNBT & Menangkan E-Wallet 100rb

Habis dalam

00

:

09

:

42

:

52

Klaim

1

2

Jawaban terverifikasi

Iklan

Sumber W

Community

27 Maret 2024 03:52

Jawaban terverifikasi

<p>1. Hambatan paralel</p><p>&nbsp; &nbsp; &nbsp;1/Rp = 1/x + 1/x + 1/x</p><p>&nbsp; &nbsp; &nbsp;1/Rp = 3/x</p><p>&nbsp; &nbsp; &nbsp; Rp = x/3 ohm</p><p>&nbsp; &nbsp; &nbsp;</p><p>&nbsp; &nbsp; &nbsp; Hambatan total (R<sub>total</sub> )</p><p>&nbsp; &nbsp; &nbsp; R<sub>total</sub> &nbsp;= Rp + x</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;= x/3 + x</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;= (1/3 + 3/3) x</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;= 4/3 x</p><p>&nbsp;</p><p>2. Hambatan paralel</p><p>&nbsp; &nbsp; &nbsp;1/Rp = 1/(x + x) + 1/x</p><p>&nbsp; &nbsp; &nbsp;1/Rp = 1/(2x) + 1/x</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; = 1/(2x) + 2/(2x)</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; = 3/(2x)</p><p>&nbsp; &nbsp; &nbsp; Rp = 2x/3 ohm</p><p>&nbsp; &nbsp; &nbsp;</p><p>&nbsp; &nbsp; &nbsp; Hambatan total (R<sub>total</sub> )</p><p>&nbsp; &nbsp; &nbsp; R<sub>total</sub> &nbsp;= Rp + x</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;= 2x/3 + x</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;= (2/3 + 3/3) x</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;= 5/3 x</p><p>&nbsp;</p><p>3. Hambatan paralel</p><p>&nbsp; &nbsp; &nbsp;1/Rp = 1/(x + x) + 1/(x + x)</p><p>&nbsp; &nbsp; &nbsp;1/Rp = 1/(2x) + 1/(2x)</p><p>&nbsp; &nbsp; &nbsp; 1/Rp = 2/(2x)</p><p>&nbsp; &nbsp; &nbsp; Rp = 2x/2 = x ohm</p><p>&nbsp; &nbsp;&nbsp;</p><p>4. Hambatan paralel</p><p>&nbsp; &nbsp; &nbsp;1/Rp = 1/x + 1/x&nbsp;</p><p>&nbsp; &nbsp; &nbsp;1/Rp = 2/x</p><p>&nbsp; &nbsp; &nbsp; Rp = x/2 ohm</p><p>&nbsp; &nbsp; &nbsp;</p><p>&nbsp; &nbsp; &nbsp; Hambatan total (R<sub>total</sub> )</p><p>&nbsp; &nbsp; &nbsp; R<sub>total</sub> &nbsp;= Rp + x</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;= x/2 + x + x</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;= (1/2 + 2/2 + 2/2) x</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;= 5/2 x</p><p>&nbsp; &nbsp; &nbsp;&nbsp;</p><p>Nilai hambatan pengganti :</p><ul><li>Paling besar adalah nomor 4</li><li>Paling kecil adalah nomor 3</li></ul><p>&nbsp;</p>

1. Hambatan paralel

     1/Rp = 1/x + 1/x + 1/x

     1/Rp = 3/x

      Rp = x/3 ohm

     

      Hambatan total (Rtotal )

      Rtotal  = Rp + x

                   = x/3 + x

                   = (1/3 + 3/3) x

                   = 4/3 x

 

2. Hambatan paralel

     1/Rp = 1/(x + x) + 1/x

     1/Rp = 1/(2x) + 1/x

                = 1/(2x) + 2/(2x)

                = 3/(2x)

      Rp = 2x/3 ohm

     

      Hambatan total (Rtotal )

      Rtotal  = Rp + x

                   = 2x/3 + x

                   = (2/3 + 3/3) x

                   = 5/3 x

 

3. Hambatan paralel

     1/Rp = 1/(x + x) + 1/(x + x)

     1/Rp = 1/(2x) + 1/(2x)

      1/Rp = 2/(2x)

      Rp = 2x/2 = x ohm

    

4. Hambatan paralel

     1/Rp = 1/x + 1/x 

     1/Rp = 2/x

      Rp = x/2 ohm

     

      Hambatan total (Rtotal )

      Rtotal  = Rp + x

                   = x/2 + x + x

                   = (1/2 + 2/2 + 2/2) x

                   = 5/2 x

      

Nilai hambatan pengganti :

  • Paling besar adalah nomor 4
  • Paling kecil adalah nomor 3

 


Iklan

Raissa F

27 Maret 2024 04:44

Jawaban terverifikasi

<p>anggap nilai R = 2</p><p>&nbsp;</p><p>1. R total = 1/(1/2 + 1/2 + 1/2) + 2</p><p>R total = 2/3+6/3&nbsp;</p><p>R total = 8/3 = 2,67</p><p>&nbsp;</p><p>2. R total = 1/(1/4+1/2) + 2</p><p>R total = 4/3 + 6/3</p><p>R total = 10/3 = 3,33</p><p>&nbsp;</p><p>3. R total = 1/(1/4+1/4)</p><p>R total = 4/2</p><p>R total = 2</p><p>&nbsp;</p><p>4. R total = 2 + 1/(1/2+1/2) + 2</p><p>R total = 2 + 1 + 2 = 5</p><p>&nbsp;</p><p>R terbesar adalah 4</p><p>R terkecil adalah 3</p>

anggap nilai R = 2

 

1. R total = 1/(1/2 + 1/2 + 1/2) + 2

R total = 2/3+6/3 

R total = 8/3 = 2,67

 

2. R total = 1/(1/4+1/2) + 2

R total = 4/3 + 6/3

R total = 10/3 = 3,33

 

3. R total = 1/(1/4+1/4)

R total = 4/2

R total = 2

 

4. R total = 2 + 1/(1/2+1/2) + 2

R total = 2 + 1 + 2 = 5

 

R terbesar adalah 4

R terkecil adalah 3


Mau pemahaman lebih dalam untuk soal ini?

Tanya ke AiRIS

Yuk, cobain chat dan belajar bareng AiRIS, teman pintarmu!

Chat AiRIS

LATIHAN SOAL GRATIS!

Drill Soal

Latihan soal sesuai topik yang kamu mau untuk persiapan ujian

Cobain Drill Soal

Perdalam pemahamanmu bersama Master Teacher
di sesi Live Teaching, GRATIS!

Pertanyaan serupa

Ada berapa banyak kemungkinan menyusun 3 buku matematika, 2 Kimia dan 2 buku fisika pada sebuah rak yang disusun secara berderet horizontal, .Jika ada syarat buku Matematika tidak boleh dipisahkan?

54

0.0

Jawaban terverifikasi