Pada trafo 160 KVA tegangan 20kv,380/220 Hitung Arus pada sisi primer

2x - the first and the smallest even integer 2x+2 - the second even integer 2x+4 - the third even integer
2 x + 2 x + 2 + 2 x + 4 = 2 ⋅ 2 x − 48 6 x + 6 = 4 x − 48 2 x = − 54 2 x = − 54 2 x + 2 = − 54 + 2 = − 52 2 x + 4 = − 54 + 4 = − 50
These three consecutive even integers are -54, -52 and -50.

Answered by naǫ | 2024-06-10

The three consecutive even integers are -54, -52, and -50. These were determined by setting up an equation based on the problem's requirements. By solving for the smallest integer, we found the other two integers in the sequence.
;

Answered by naǫ | 2024-12-26

Jawaban:Penjelasan:Tidak diberikan efisiensi trafo maka dianggap efisiensi 100%[tex]P=160\text{kVA}\\V_p=20\text {kV}\\V_{sll}=380\text V\to \text{line to line}\\V_{sln}=220\text {V}\to \text{line to neutral}\\[/tex]Ditanyakan:[tex]I_p=?[/tex]Alternatif penyelesaian:Pada sisi primer hanya ada satu tegangan  sedangkan pada sisi sekunder ada dua tegangan sehingga trafo tersebut trafo tiga fasa dengan bagian primer dihubungkan Delta (Δ) dan bagian sekunder dihubungkan Bintang (Y) maka daya tiap fasa:[tex]P=\frac{1}{3} P_p\\VI_f=\frac{1}{3} P_p\\(20\text{kV})I_f=\frac{1}{3} (160\text{kVA})\\I_f=\frac{160}{3\times20} \text A\\I_f=\frac{8}{3} \text A[/tex][tex]I_p=I_f\sqrt{3} \to\text { antar fasa berbeda }120^{\circ}\\I_p=\frac{8}{3} \sqrt{3}\,\,\text A\\I_p\approx 4,62\text A[/tex]Jadi arus pada sisi primer adalah (8/3)√3 A atau mendekati 4,62 A.

Answered by mastreeada | 2025-07-12