Good luck
Jawaban:
[tex]\bf\large{I=4\ Ampere} [/tex]
Penjelasan:
Diketahui:
[tex]\sf\large{R_{1}=4\ Ω}[/tex]
[tex]\sf\large{R₂=1\ Ω}[/tex]
[tex]\sf\large{R₃=2\ Ω}[/tex]
[tex]\sf\large{R₄=3\ Ω}[/tex]
[tex]\sf\large{R_{5}=8\ Ω}[/tex]
[tex]\sf\large{R_{6}=3\ Ω}[/tex]
[tex]\sf\large{V=56\ Volt }[/tex]
Ditanyakan kuat arus pada rangkaian
Jawab:
Tentukanlah dahulu nilai hambatan penggantinya:
[tex]\sf\large{Rp=R_{1}+}\Large{(\frac{(R₂+R₃+R₄)×R_{6}}{R₂+R₃+R₄+R_{6}})}\large{+R_{5}} [/tex]
[tex]=\sf\large4+\Large{(\frac{(1+2+3)×3}{1+2+3+3})}\large{+8} [/tex]
[tex]=\sf\large4+\Large{(\frac{6×3}{9})}\large{+8} [/tex]
[tex]=\sf\large4+\Large{(\frac{18}{9})}\large{+8} [/tex]
[tex]=\sf\large4+2+8 [/tex]
[tex]=\sf\large{14\ Ω} [/tex]
Lalu hitung kuat arusnya:
Gunakan rumus:
[tex]\sf\large{V=I×R}[/tex]
[tex]\sf\large{I}=\LARGE{\frac{V}{R}}[/tex]
[tex]\sf\large{I}=\LARGE{\frac{56}{14}}[/tex]
[tex]\sf\large{I=4\ A}[/tex]
Jadi kuat arusnya sebesar [tex]\sf\large4\ Ampere [/tex]
Jawaban:
I = 4 A
Penjelasan:
Tambahkan dulu =
1 ohm + 2 ohm + 3 ohm = 6 ohm
Paralelkan =
1/R = 1/6 + 1/3
1/R = 1/6 + 2/6
1/R = 3/6
R = 6/3 = 2 ohm
4 ohm + 2 ohm (R paralel tadi) + 8 ohm = 14 ohm
I = V / R
I = 56 / 14
I = 4 A
semoga membantu kak ^^
jadikan yang terbaik ya kak makasihh
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