TOLONG KERJAKANPada rangkaian berikut, gunakan prinsip seri-paralel untuk menghitung besar kuat arus l₁ dan 12! 6 ΘΩ ΤΩ ΘΩ ΤΩ ΘΩ 62 - 18 V ΘΩ ΤΩ ΘΩ ΘΩ​

A very small video that is continuous is called a GIF, which stands for Graphics Interchange Format.

Answered by TacoGod | 2024-06-10

It's called Gif, after its usual file extenstion. GIF stands for Graphics Interchange Format.

Answered by mackiemesser | 2024-06-10

A very small video that repeats indefinitely is called a GIF, which stands for Graphics Interchange Format. GIFs compile multiple frames into one file, creating a loop of animation that is popular across the internet. They are widely used for sharing emotions or reactions in a fun and engaging way.
;

Answered by TacoGod | 2024-12-24

Jawaban:Diketahui:- Tegangan sumber (V) = 18 V- Hambatan: 6 Ω dan 9 Ωpenjelasan:- Rangkaian simetris, sehingga memudahkan perhitungan.- Arus I₁ dan I₂ adalah arus yang melewati bagian tengah rangkaian.caranya:cara untuk menghitung I₁ dan I₂:1. Hitung Resistansi Paralel Identik:- Dua resistor 6Ω seri: 6Ω + 6Ω = 12Ω- Paralel dengan 9Ω: 1 / ((1/12) + (1/9)) = 5.14Ω2. Tiga Bagian Paralel:5. - 14Ω / 3 = 1.71Ω3. Resistor Cabang Luar:- 6Ω + 6Ω = 12Ω4. Resistansi Total:- 12Ω + 1.71Ω = 13.71Ω5. Arus Total (I₁):- 18V / 13.71Ω = 1.31A6. Arus Bagian (I₂):7. - 31A / 3 × (12 / (12+9)) = 0.25AJadi, I₁ ≈ 1.31A dan I₂ ≈ 0.25A. maka jumlah l¹ yaitu 1.31A dan I² yaitu 0.25A

Answered by ara1412 | 2025-08-24

Diketahui:dari kiri atas ke kanan bawah[tex]R_1=R_1=R_3=R_4=6 \text { Ohm}\\R_5=R_6=9 \text { Ohm}, R_7=6 \text { Ohm}\\R_8=R_9=R_{10}=R_{11}= 6 \text { Ohm}\\V=18\text V[/tex]Ditanyakan:[tex]I_1, I_2=?[/tex]Alternatif penyelesaian:R₄ dan R₁₁ diabaikan karena rangkaian terbuka dan semua satuan hambatan dalam Ohm (Ω)[tex]R_{sa}=R_3+R_7+R_{10}\\R_{sa}=6+6+6\\R_{sa}=18\\R_a=R_{sa}\parallel R_6\\R_a=18\parallel 9\\R_a=\frac{18\times9}{18+9} \\R_a=6[/tex][tex]R_{sb}=R_3+R_a+R_9\\R_{sb}=6+6 +6\\R_{sb}=18\\R_b=R_{sb}\parallel R_5\\R_b=18\parallel9\\R_b=\frac{18\times9}{18+9} \\R_b=6[/tex][tex]R_t=R_1+R_b+R_8\\R_t=6+6+6\\R_t=18[/tex][tex]I_1=\frac{V}{R_t} \\I_1=\frac{18}{18} \\I_1=1 \text {A}[/tex][tex]I_2=\frac{V_b}{R_{sb}}\\I_2=\frac{\frac{R_b\times V}{R_1+R_b+R_8}}{R_{sb}}\\I_2=\frac{\frac{6\times18}{6+6+6}}{18}\\I_2=\frac{\frac{6\times18}{18}}{18}\\I_2=\frac{6}{18} \\I_2=\frac{1}{3} \text A[/tex]Jadi I₁ dan I₂ berturut-turut 1A dan ⅓ A

Answered by mastreeada | 2025-08-25