PRECONCEPTOS ELECTROTECNIA

• 1.0×10⁻³ Ω/s
• 2.0×10⁻³ Ω/s
• 1.0×10⁻² Ω/s
• 2.0×10⁻² Ω/s

Answer explanation

dR/dt = k·dL/dt = 0.02·0.05 = 1e-3 Ω/s

• 15 C
• 30 C
• 45 C
• 60 C

Answer explanation

q = ∫0^3 5 t^2 dt = 5·(27/3)=45 C

• -30 V
• 30 V
• -25 V
• 25 V

Answer explanation

V(0.4)-V(0.1) = [-100·0.4+5] - [-100·0.1+5] = -35 - (-5) = -30 V

• Aumenta 1%
• Aumenta 2%
• Aumenta 3%
• Aumenta 2.02%

Answer explanation

R ∝ 1/r^2, 1/(0.99^2)-1 ≈ 0.0203 → ≈2.02%

• 10%
• 15%
• 21.05%
• 20%

Answer explanation

V' = 1.15/0.95 -1 ≈ 0.210526 → 21.05%

• 50 C
• 25 C
• 10 C
• 5 C

Answer explanation

q = ∫0^∞ 10 e^{-0.2 t} dt = 10/(0.2) = 50 C

• 1.92×10⁻⁸ Ω·m
• 3.20×10⁻⁸ Ω·m
• 2.00×10⁻⁸ Ω·m
• 8.00×10⁻⁸ Ω·m

Answer explanation

ρ = 1.6e-8·(1+0.004·50)=1.6e-8·1.2 = 1.92e-8 Ω·m

• (-20, 3) V/m
• (20, -3) V/m
• (-10, 3) V/m
• (-20, -3) V/m

Answer explanation

E = -∇V = (-10x, 3) ⇒ at (2,1): (-20,3) V/m

Un alambre tiene L = 3 m, A = 0.5 mm² y R = 0.10 Ω. Si se fabrica uno nuevo con el doble de área y la mitad de la longitud, ¿Cuál será su resistencia?

• 0.025 Ω
• 0.05 Ω
• 0.10 Ω
• 0.20 Ω

Answer explanation

R' = R·(L'/A')/(L/A) = 0.1·(1.5/1)/(3/0.5)=0.1·0.25=0.025 Ω

• 3.2×10⁻¹⁷ J
• -3.2×10⁻¹⁷ J
• 1.6×10⁻¹⁷ J
• -1.6×10⁻¹⁷ J

Answer explanation

Magnitud del trabajo = |q|·ΔV = 1.6e-19·200 = 3.2e-17 J