Hitunglah arus yang mengalir pada tiap hambatan r1, r2, r3, r4, dan r5 yang masing-masing nilainya 2 ohm, 2 ohm, 4 ohm, 2 ohm, 4 ohm pada rangkaian berikut jika E1=8 V dan E2=10 V
Answer (4)
you would first do 2*2 inside the parentheses to get 12.3 time {4+6}-2 cubed That is 12.3x10, which is 123. 2 cubed is 8, and 123-8 is 115 .
To find the approximate answer for the expression 12.3 \( \times \) {(2 \( \times \) 1.7)+6}-2 cubed , we follow these steps:
First, address the parentheses: 2 \( \times \) 1.7 = 3.4 . So, the expression inside the curly bracket becomes 3.4 + 6, which equals 9.4.
Next, multiply this result by 12.3: 2 \( \times \) 1.7 = 3.4 .
Lastly, calculate 2 cubed, which is 2\(^3\) = 8 .
Subtract the cubed number from the product: 115.62 - 8 = 107.62.
Therefore, the approximate answer for the expression is 107.62.
To estimate the expression 12.3 × { ( 2 × 1.7 ) + 6 } − 2 3 , we calculate the inner expression's result to be approximately 9, multiply it by 12 (roughly 108), and then subtract 8 (for 2 3 ), resulting in an estimated final answer of about 100.
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Soal:“Hitunglah arus yang mengalir pada tiap hambatan R₁, R₂, R₃, R₄, dan R₅ yang masing-masing nilainya 2 Ω, 2 Ω, 4 Ω, 2 Ω, dan 4 Ω.”Rangkaian terdiri dari lima resistor yang disusun paralel, dihubungkan dengan sumber tegangan (V). Arus yang melalui masing-masing hambatan dapat dihitung dengan hukum Ohm:I_i = \frac{V}{R_i}Dengan demikian, untuk setiap hambatan: • I₁ = V / 2 Ω • I₂ = V / 2 Ω • I₃ = V / 4 Ω • I₄ = V / 2 Ω • I₅ = V / 4 ΩJika diketahui nilai tegangan sumber (misalnya 12 V), maka: • I₁ = I₂ = I₄ = 12 / 2 = 6 A • I₃ = I₅ = 12 / 4 = 3 ATotal arus = jumlah seluruh arus = 6 + 6 + 3 + 6 + 3 = 24 A    Hambatan (Ω)Arus (A)R₁ = 2I₁ = V / 2R₂ = 2I₂ = V / 2R₃ = 4I₃ = V / 4R₄ = 2I₄ = V / 2R₅ = 4I₅ = V / 4Jika V = 12 V: • I₁, I₂, I₄ = 6 A • I₃, I₅ = 3 A⸻✳️ Catatan: • Arus di tiap resistor Paralel berbeda-beda karena hambatan berbeda, namun semua tegangan sama. • Hukum Kirchoff menyatakan total arus = jumlah arus cabang.
