Resolución del Ejercicio de Corriente Alterna
a) La expresión temporal de la corriente que circula por el generador, tomando como origen de fases la tensión del generador. Primero, se calculan la frecuencia angular y las reactancias de cada componente. Datos:
R 1 = 40 Ω R_1=40 \ \Omega R 1 = 40 Ω R 2 = 16 Ω R_2=16 \ \Omega R 2 = 16 Ω L 1 = 80 mH = 0.08 H L_1= 80 \text{ mH} = 0.08 \text{ H} L 1 = 80 mH = 0.08 H L 2 = 150 mH = 0.15 H L_2=150 \text{ mH} = 0.15 \text{ H} L 2 = 150 mH = 0.15 H C 1 = 150 μ F = 150 × 10 − 6 F C_1=150 \ \mu\text{F} = 150 \times 10^{-6} \text{ F} C 1 = 150 μ F = 150 × 1 0 − 6 F C 2 = 280 μ F = 280 × 10 − 6 F C_2=280 \ \mu\text{F} = 280 \times 10^{-6} \text{ F} C 2 = 280 μ F = 280 × 1 0 − 6 F E eficaz = 230 V E_{\text{eficaz}}=230 \text{ V} E eficaz = 230 V f = 50 Hz f=50 \text{ Hz} f = 50 Hz Fórmulas:
ω = 2 π f \omega = 2\pi f ω = 2 π f X L = ω L X_L = \omega L X L = ω L X C = 1 ω C X_C = \frac{1}{\omega C} X C = ω C 1 Sustitución:
ω = 2 π ⋅ 50 Hz = 100 π rad/s ≈ 314.16 rad/s \omega = 2\pi \cdot 50 \text{ Hz} = 100\pi \text{ rad/s} \approx 314.16 \text{ rad/s} ω = 2 π ⋅ 50 Hz = 100 π rad/s ≈ 314.16 rad/s X L 1 = ( 100 π rad/s ) ⋅ ( 0.08 H ) = 8 π Ω ≈ 25.13 Ω X_{L1} = (100\pi \text{ rad/s}) \cdot (0.08 \text{ H}) = 8\pi \ \Omega \approx 25.13 \ \Omega X L 1 = ( 100 π rad/s ) ⋅ ( 0.08 H ) = 8 π Ω ≈ 25.13 Ω X L 2 = ( 100 π rad/s ) ⋅ ( 0.15 H ) = 15 π Ω ≈ 47.12 Ω X_{L2} = (100\pi \text{ rad/s}) \cdot (0.15 \text{ H}) = 15\pi \ \Omega \approx 47.12 \ \Omega X L 2 = ( 100 π rad/s ) ⋅ ( 0.15 H ) = 15 π Ω ≈ 47.12 Ω X C 1 = 1 ( 100 π rad/s ) ⋅ ( 150 × 10 − 6 F ) = 1 0.015 π = 200 3 π Ω ≈ 21.22 Ω X_{C1} = \frac{1}{(100\pi \text{ rad/s}) \cdot (150 \times 10^{-6} \text{ F})} = \frac{1}{0.015\pi} = \frac{200}{3\pi} \ \Omega \approx 21.22 \ \Omega X C 1 = ( 100 π rad/s ) ⋅ ( 150 × 1 0 − 6 F ) 1 = 0.015 π 1 = 3 π 200 Ω ≈ 21.22 Ω X C 2 = 1 ( 100 π rad/s ) ⋅ ( 280 × 10 − 6 F ) = 1 0.028 π = 250 7 π Ω ≈ 11.37 Ω X_{C2} = \frac{1}{(100\pi \text{ rad/s}) \cdot (280 \times 10^{-6} \text{ F})} = \frac{1}{0.028\pi} = \frac{250}{7\pi} \ \Omega \approx 11.37 \ \Omega X C 2 = ( 100 π rad/s ) ⋅ ( 280 × 1 0 − 6 F ) 1 = 0.028 π 1 = 7 π 250 Ω ≈ 11.37 Ω Ahora se calculan las impedancias equivalentes para cada rama en paralelo. Datos:
X L 1 = 8 π Ω X_{L1} = 8\pi \ \Omega X L 1 = 8 π Ω X L 2 = 15 π Ω X_{L2} = 15\pi \ \Omega X L 2 = 15 π Ω R 1 = 40 Ω R_1 = 40 \ \Omega R 1 = 40 Ω R 2 = 16 Ω R_2 = 16 \ \Omega R 2 = 16 Ω X C 1 = 200 3 π Ω X_{C1} = \frac{200}{3\pi} \ \Omega X C 1 = 3 π 200 Ω X C 2 = 250 7 π Ω X_{C2} = \frac{250}{7\pi} \ \Omega X C 2 = 7 π 250 Ω Fórmulas:
Z eq = Z 1 Z 2 Z 1 + Z 2 Z_{\text{eq}} = \frac{Z_1 Z_2}{Z_1 + Z_2} Z eq = Z 1 + Z 2 Z 1 Z 2 Z C = − j X C Z_C = -jX_C Z C = − j X C Sustitución:
Z L _ eq = ( j 8 π ) ( j 15 π ) j 8 π + j 15 π = − 120 π 2 j 23 π = j 120 π 23 Ω ≈ j 16.36 Ω Z_{L\_\text{eq}} = \frac{(j8\pi)(j15\pi)}{j8\pi + j15\pi} = \frac{-120\pi^2}{j23\pi} = j\frac{120\pi}{23} \ \Omega \approx j16.36 \ \Omega Z L _ eq = j 8 π + j 15 π ( j 8 π ) ( j 15 π ) = j 23 π − 120 π 2 = j 23 120 π Ω ≈ j 16.36 Ω Z R _ eq = 40 ⋅ 16 40 + 16 = 640 56 = 80 7 Ω ≈ 11.43 Ω Z_{R\_\text{eq}} = \frac{40 \cdot 16}{40 + 16} = \frac{640}{56} = \frac{80}{7} \ \Omega \approx 11.43 \ \Omega Z R _ eq = 40 + 16 40 ⋅ 16 = 56 640 = 7 80 Ω ≈ 11.43 Ω Z C _ eq = ( − j 200 3 π ) ( − j 250 7 π ) − j 200 3 π − j 250 7 π = − 50000 21 π 2 − j ( 1400 + 750 21 π ) = − 50000 21 π 2 − j 2150 21 π = − j 50000 2150 π = − j 1000 43 π Ω ≈ − j 7.42 Ω Z_{C\_\text{eq}} = \frac{(-j\frac{200}{3\pi})(-j\frac{250}{7\pi})}{-j\frac{200}{3\pi} - j\frac{250}{7\pi}} = \frac{-\frac{50000}{21\pi^2}}{-j\left(\frac{1400+750}{21\pi}\right)} = \frac{-\frac{50000}{21\pi^2}}{-j\frac{2150}{21\pi}} = -j\frac{50000}{2150\pi} = -j\frac{1000}{43\pi} \ \Omega \approx -j7.42 \ \Omega Z C _ eq = − j 3 π 200 − j 7 π 250 ( − j 3 π 200 ) ( − j 7 π 250 ) = − j ( 21 π 1400 + 750 ) − 21 π 2 50000 = − j 21 π 2150 − 21 π 2 50000 = − j 2150 π 50000 = − j 43 π 1000 Ω ≈ − j 7.42 Ω Se calcula la impedancia total del circuito, que es la suma de las impedancias equivalentes en serie. Datos:
Z R _ eq = 80 7 Ω Z_{R\_\text{eq}} = \frac{80}{7} \ \Omega Z R _ eq = 7 80 Ω Z L _ eq = j 120 π 23 Ω Z_{L\_\text{eq}} = j\frac{120\pi}{23} \ \Omega Z L _ eq = j 23 120 π Ω Z C _ eq = − j 1000 43 π Ω Z_{C\_\text{eq}} = -j\frac{1000}{43\pi} \ \Omega Z C _ eq = − j 43 π 1000 Ω Fórmulas:
Z total = Z R _ eq + Z L _ eq + Z C _ eq Z_{\text{total}} = Z_{R\_\text{eq}} + Z_{L\_\text{eq}} + Z_{C\_\text{eq}} Z total = Z R _ eq + Z L _ eq + Z C _ eq Sustitución:
Z total = 80 7 + j 120 π 23 − j 1000 43 π Z_{\text{total}} = \frac{80}{7} + j\frac{120\pi}{23} - j\frac{1000}{43\pi} Z total = 7 80 + j 23 120 π − j 43 π 1000 Z total = 80 7 + j ( 120 π 23 − 1000 43 π ) Ω Z_{\text{total}} = \frac{80}{7} + j\left(\frac{120\pi}{23} - \frac{1000}{43\pi}\right) \ \Omega Z total = 7 80 + j ( 23 120 π − 43 π 1000 ) Ω Z total ≈ 11.4286 + j ( 16.3638 − 7.4243 ) Ω Z_{\text{total}} \approx 11.4286 + j(16.3638 - 7.4243) \ \Omega Z total ≈ 11.4286 + j ( 16.3638 − 7.4243 ) Ω Z total ≈ 11.4286 + j 8.9395 Ω Z_{\text{total}} \approx 11.4286 + j8.9395 \ \Omega Z total ≈ 11.4286 + j 8.9395 Ω ∣ Z total ∣ = ( 11.4286 ) 2 + ( 8.9395 ) 2 = 130.612 + 79.914 = 210.526 ≈ 14.510 Ω |Z_{\text{total}}| = \sqrt{(11.4286)^2 + (8.9395)^2} = \sqrt{130.612 + 79.914} = \sqrt{210.526} \approx 14.510 \ \Omega ∣ Z total ∣ = ( 11.4286 ) 2 + ( 8.9395 ) 2 = 130.612 + 79.914 = 210.526 ≈ 14.510 Ω ϕ Z = arctan ( 8.9395 11.4286 ) ≈ 38.03 ∘ \phi_Z = \arctan\left(\frac{8.9395}{11.4286}\right) \approx 38.03^{\circ} ϕ Z = arctan ( 11.4286 8.9395 ) ≈ 38.0 3 ∘ Z total ≈ 14.510 ∠ 38.03 ∘ Ω Z_{\text{total}} \approx 14.510 \angle 38.03^{\circ} \ \Omega Z total ≈ 14.510∠38.0 3 ∘ Ω Finalmente, se calcula la corriente total eficaz y se convierte a su expresión temporal. Datos:
E eficaz = 230 ∠ 0 ∘ V E_{\text{eficaz}} = 230 \angle 0^{\circ} \text{ V} E eficaz = 230∠ 0 ∘ V Z total ≈ 14.510 ∠ 38.03 ∘ Ω Z_{\text{total}} \approx 14.510 \angle 38.03^{\circ} \ \Omega Z total ≈ 14.510∠38.0 3 ∘ Ω ω = 100 π rad/s \omega = 100\pi \text{ rad/s} ω = 100 π rad/s Fórmulas:
I eficaz = E eficaz Z total I_{\text{eficaz}} = \frac{E_{\text{eficaz}}}{Z_{\text{total}}} I eficaz = Z total E eficaz I m a ˊ ximo = I eficaz 2 I_{\text{máximo}} = I_{\text{eficaz}} \sqrt{2} I m a ˊ ximo = I eficaz 2 i ( t ) = I m a ˊ ximo sin ( ω t + ϕ I ) i(t) = I_{\text{máximo}} \sin(\omega t + \phi_I) i ( t ) = I m a ˊ ximo sin ( ω t + ϕ I ) Sustitución:
I eficaz = 230 ∠ 0 ∘ 14.510 ∠ 38.03 ∘ ≈ 15.85 ∠ − 38.03 ∘ A I_{\text{eficaz}} = \frac{230 \angle 0^{\circ}}{14.510 \angle 38.03^{\circ}} \approx 15.85 \angle -38.03^{\circ} \text{ A} I eficaz = 14.510∠38.0 3 ∘ 230∠ 0 ∘ ≈ 15.85∠ − 38.0 3 ∘ A I m a ˊ ximo = 15.85 A ⋅ 2 ≈ 22.42 A I_{\text{máximo}} = 15.85 \text{ A} \cdot \sqrt{2} \approx 22.42 \text{ A} I m a ˊ ximo = 15.85 A ⋅ 2 ≈ 22.42 A ϕ I = − 38.03 ∘ \phi_I = -38.03^{\circ} ϕ I = − 38.0 3 ∘ Resultado:
i ( t ) = 22.42 sin ( 100 π t − 38.03 ∘ ) A i(t) = 22.42 \sin(100\pi t - 38.03^{\circ}) \text{ A} i ( t ) = 22.42 sin ( 100 π t − 38.0 3 ∘ ) A b) El valor eficaz de la tensión en bornes de la resistencia R 1 R_1 R 1 . La tensión en bornes de la resistencia R 1 R_1 R 1 es la tensión eficaz a través de la rama paralela de las resistencias R 1 R_1 R 1 y R 2 R_2 R 2 . Datos:
I eficaz ≈ 15.85 ∠ − 38.03 ∘ A I_{\text{eficaz}} \approx 15.85 \angle -38.03^{\circ} \text{ A} I eficaz ≈ 15.85∠ − 38.0 3 ∘ A Z R _ eq = 80 7 Ω ≈ 11.43 Ω Z_{R\_\text{eq}} = \frac{80}{7} \ \Omega \approx 11.43 \ \Omega Z R _ eq = 7 80 Ω ≈ 11.43 Ω Fórmula:
V R 1 eficaz = I eficaz ⋅ ∣ Z R _ eq ∣ V_{R1_{\text{eficaz}}} = I_{\text{eficaz}} \cdot |Z_{R\_\text{eq}}| V R 1 eficaz = I eficaz ⋅ ∣ Z R _ eq ∣ Sustitución:
V_{R1_{\text{eficaz}}} = (15.85 \text{ A}) \cdot (11.4286 \ \Omega) \approx 181.16 \text{ V}
Resultado:
V R 1 eficaz = 181.16 V V_{R1_{\text{eficaz}}} = 181.16 \text{ V} V R 1 eficaz = 181.16 V 
