Sebuah kawat memiliki panjang 90 m, berdiameter 3 mm dan hambatan jenisnya 6.28 * 10 ^ - 6 ohm.m Tentukan: a) Hambatan kawat, b) Hambatan kawat kedua dari bahan dan berat yang sama, tetapi memiliki diameter 3 x kawat pertama.
Answer (3)
2 3 ; 4 3 ; 8 3 ; 16 3 ; ... a 1 = 2 3 a 2 = 4 3 = 2 3 ⋅ 2 1 a 3 = 8 3 = 4 3 ⋅ 2 1 = 2 3 ⋅ ( 2 1 ) 2 a 4 = 16 3 = 8 3 ⋅ 2 1 = 2 3 ⋅ ( 2 1 ) 3 ⋮ a n = 2 3 ⋅ ( 2 1 ) n − 1 = 2 3 ⋅ ( 2 1 ) − 1 ⋅ ( 2 1 ) n = 2 3 ⋅ 2 ⋅ ( 2 1 ) n = 3 ⋅ ( 2 1 ) n
The explicit formula for the sequence 2 3 , 4 3 , 8 3 , 16 3 , … is a n = 2 n 3 . This formula allows us to find any term in the sequence by plugging in the value of n. The sequence represents a geometric pattern where each term is half of the one before it, multiplied by 3 in the numerator.
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Jawaban:Diketahui:Panjang kawat (L) = 90 mDiameter kawat pertama (d₁) = 3 mm = 3 × 10⁻³ mρ (hambatan jenis) = 6,28 × 10⁻⁶ Ω·m---a) Hambatan kawat pertamaRumus: R = ρ × L / AA = luas penampang lingkaran = π × (d/2)²A₁ = π × (1,5 × 10⁻³)²= π × (2,25 × 10⁻⁶)≈ 7,07 × 10⁻⁶ m²R₁ = (6,28 × 10⁻⁶ × 90) / (7,07 × 10⁻⁶)≈ (5,65 × 10⁻⁴) / (7,07 × 10⁻⁶)≈ 79,9 Ω Hambatan kawat pertama = ≈ 80 Ω---b) Hambatan kawat keduaDiameter kawat kedua: d₂ = 3 × d₁ = 9 mm = 9 × 10⁻³ mKawat dari bahan dan berat sama → berarti volumenya sama.Vol = A × LL₂ = (A₁ × L₁) / A₂Hitung A₂:A₂ = π × (4,5 × 10⁻³)²= π × (20,25 × 10⁻⁶)≈ 6,36 × 10⁻⁵ m²L₂ = (7,07 × 10⁻⁶ × 90) / (6,36 × 10⁻⁵)≈ (6,36 × 10⁻⁴) / (6,36 × 10⁻⁵)≈ 10 mSekarang hambatan kawat kedua:R₂ = ρ × L₂ / A₂= (6,28 × 10⁻⁶ × 10) / (6,36 × 10⁻⁵)≈ (6,28 × 10⁻⁵) / (6,36 × 10⁻⁵)≈ 0,99 Ω Hambatan kawat kedua = ≈ 1 Ω---✅ Kesimpulan:a) Hambatan kawat pertama = ± 80 Ωb) Hambatan kawat kedua = ± 1 ΩBantu kasih likenya ya, apabila jawaban ini membantu. Terimakasih
