a) Valor de la impedancia total del circuito. Cálculo de la impedancia equivalente de las resistencias R 1 R_1 R 1 y R 2 R_2 R 2 en paralelo. Datos
R 1 = 8 Ω R 2 = 8 Ω \begin{gathered}
R_1 = 8 \, \Omega \\ R_2 = 8 \, \Omega
\end{gathered} R 1 = 8 Ω R 2 = 8 Ω Fórmulas
Z R eq = R 1 ⋅ R 2 R 1 + R 2 Z_{R_{\text{eq}}} = \frac{R_1 \cdot R_2}{R_1 + R_2} Z R eq = R 1 + R 2 R 1 ⋅ R 2 Sustitución
Z R eq = 8 Ω ⋅ 8 Ω 8 Ω + 8 Ω = 64 Ω 2 16 Ω Z_{R_{\text{eq}}} = \frac{8 \, \Omega \cdot 8 \, \Omega}{8 \, \Omega + 8 \, \Omega} = \frac{64 \, \Omega^2}{16 \, \Omega} Z R eq = 8 Ω + 8 Ω 8 Ω ⋅ 8 Ω = 16 Ω 64 Ω 2 Resultado
Z R eq = 4 Ω Z_{R_{\text{eq}}} = 4 \, \Omega Z R eq = 4 Ω Cálculo de la impedancia equivalente de las reactancias inductivas X L 1 XL_1 X L 1 y X L 2 XL_2 X L 2 en paralelo. Datos
X L 1 = 10 Ω X L 2 = 10 Ω \begin{gathered}
XL_1 = 10 \, \Omega \\ XL_2 = 10 \, \Omega
\end{gathered} X L 1 = 10 Ω X L 2 = 10 Ω Fórmulas
Z L eq = ( j X L 1 ) ⋅ ( j X L 2 ) j X L 1 + j X L 2 Z_{L_{\text{eq}}} = \frac{(j XL_1) \cdot (j XL_2)}{j XL_1 + j XL_2} Z L eq = j X L 1 + j X L 2 ( j X L 1 ) ⋅ ( j X L 2 ) Sustitución
Z_{L_{\text{eq}}} = \frac{(j 10 \, \Omega) \cdot (j 10 \, \Omega)}{j 10 \, \Omega + j 10 \, \Omega} = \frac{j^2 100 \, \Omega^2}{j 20 \, \Omega} = \frac{-100 \, \Omega^2}{j 20 \, \Omega} = \frac{-100j \, \Omega^2}{j^2 20 \, \Omega} = \frac{-100j \, \Omega^2}{-20 \, \Omega}
Resultado
Z L eq = j 5 Ω Z_{L_{\text{eq}}} = j 5 \, \Omega Z L eq = j 5 Ω Cálculo de la impedancia total del circuito (combinación serie de Z R eq Z_{R_{\text{eq}}} Z R eq , Z L eq Z_{L_{\text{eq}}} Z L eq y Z C 1 Z_{C_1} Z C 1 ). La impedancia del condensador es Z C 1 = − j X C 1 Z_{C_1} = -j XC_1 Z C 1 = − j X C 1 . Datos
Z R eq = 4 Ω Z L eq = j 5 Ω X C 1 = 2 Ω \begin{gathered}
Z_{R_{\text{eq}}} = 4 \, \Omega \\ Z_{L_{\text{eq}}} = j 5 \, \Omega \\ XC_1 = 2 \, \Omega
\end{gathered} Z R eq = 4 Ω Z L eq = j 5 Ω X C 1 = 2 Ω Fórmulas
Z total = Z R eq + Z L eq + ( − j X C 1 ) Z_{\text{total}} = Z_{R_{\text{eq}}} + Z_{L_{\text{eq}}} + (-j XC_1) Z total = Z R eq + Z L eq + ( − j X C 1 ) Sustitución
Z total = 4 Ω + j 5 Ω − j 2 Ω = 4 + j ( 5 − 2 ) Ω Z_{\text{total}} = 4 \, \Omega + j 5 \, \Omega - j 2 \, \Omega = 4 + j(5 - 2) \, \Omega Z total = 4 Ω + j 5 Ω − j 2 Ω = 4 + j ( 5 − 2 ) Ω Resultado
Z total = 4 + j 3 Ω Z_{\text{total}} = 4 + j 3 \, \Omega Z total = 4 + j 3 Ω Valor de la impedancia total en forma polar:
∣ Z total ∣ = 4 2 + 3 2 = 16 + 9 = 25 = 5 Ω ϕ = arctan ( 3 4 ) ≈ 36.87 ∘ \begin{gathered}
|Z_{\text{total}}| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \, \Omega \\ \phi = \arctan\left(\frac{3}{4}\right) \approx 36.87^\circ
\end{gathered} ∣ Z total ∣ = 4 2 + 3 2 = 16 + 9 = 25 = 5 Ω ϕ = arctan ( 4 3 ) ≈ 36.8 7 ∘ Z total = 5 ∠ 36.87 ∘ Ω Z_{\text{total}} = 5 \angle 36.87^\circ \, \Omega Z total = 5∠36.8 7 ∘ Ω b) Valor eficaz de la corriente que circula por el generador. Datos
E = 20 \text{ V} $ (eficaces) $ Z_{\text{total}} = 5 \angle 36.87^\circ \, \Omega
Fórmulas
I = E Z total I = \frac{E}{Z_{\text{total}}} I = Z total E Sustitución
I = \frac{20 \angle 0^\circ \text{ V}}{5 \angle 36.87^\circ \, \Omega} = \frac{20}{5} \angle (0^\circ - 36.87^\circ)
Resultado
I = 4 \angle -36.87^\circ \text{ A} $ El valor eficaz de la corriente es $ |I| = 4 \text{ A}
c) Potencia activa, potencia reactiva y potencia aparente en el generador. Datos
E = 20 V ∣ I ∣ = 4 A Z total = 4 + j 3 Ω ⟹ R total = 4 Ω , X total = 3 Ω ϕ = 36.87 ∘ ⟹ cos ϕ = 0.8 , sin ϕ = 0.6 \begin{gathered}
E = 20 \text{ V} \\ |I| = 4 \text{ A} \\ Z_{\text{total}} = 4 + j 3 \, \Omega \implies R_{\text{total}} = 4 \, \Omega, X_{\text{total}} = 3 \, \Omega \\ \phi = 36.87^\circ \implies \cos \phi = 0.8, \sin \phi = 0.6
\end{gathered} E = 20 V ∣ I ∣ = 4 A Z total = 4 + j 3 Ω ⟹ R total = 4 Ω , X total = 3 Ω ϕ = 36.8 7 ∘ ⟹ cos ϕ = 0.8 , sin ϕ = 0.6 Cálculo de la potencia activa ( P P P ). Fórmulas
P = ∣ I ∣ 2 ⋅ R total P = |I|^2 \cdot R_{\text{total}} P = ∣ I ∣ 2 ⋅ R total Sustitución
P = ( 4 A ) 2 ⋅ 4 Ω = 16 A 2 ⋅ 4 Ω P = (4 \text{ A})^2 \cdot 4 \, \Omega = 16 \text{ A}^2 \cdot 4 \, \Omega P = ( 4 A ) 2 ⋅ 4 Ω = 16 A 2 ⋅ 4 Ω Resultado
P = 64 W P = 64 \text{ W} P = 64 W Cálculo de la potencia reactiva ( Q Q Q ). Fórmulas
Q = ∣ I ∣ 2 ⋅ X total Q = |I|^2 \cdot X_{\text{total}} Q = ∣ I ∣ 2 ⋅ X total Sustitución
Q = ( 4 A ) 2 ⋅ 3 Ω = 16 A 2 ⋅ 3 Ω Q = (4 \text{ A})^2 \cdot 3 \, \Omega = 16 \text{ A}^2 \cdot 3 \, \Omega Q = ( 4 A ) 2 ⋅ 3 Ω = 16 A 2 ⋅ 3 Ω Resultado
Q = 48 VAR Q = 48 \text{ VAR} Q = 48 VAR Cálculo de la potencia aparente ( S S S ). Fórmulas
S = E ⋅ ∣ I ∣ S = E \cdot |I| S = E ⋅ ∣ I ∣ Sustitución
S = 20 V ⋅ 4 A S = 20 \text{ V} \cdot 4 \text{ A} S = 20 V ⋅ 4 A Resultado
S = 80 VA S = 80 \text{ VA} S = 80 VA 
