<jupyter_start><jupyter_text>Ejercicios de clase - Thevenin__UNLZ - Facultad de Ingeniería__       __Electrotecnia__      __Alumno:__ <NAME>   Problema N°11.1 Hallar el circuito equivalente de Thevenin entre los puntos A y B.1.2 Hallar el circuito equivalente de Norton entre los puntos A y B.   Solución * __Cálculo de $Z_{th}$__        1)  $Z_1 || Z_2$     2)  $(Z_1 || Z_2) + Z_3$      3) $[(Z_1 || Z_2) + Z_3] || Z_4$ 1) Hacemos $Z_1 || Z_2$  $$ \frac{Z_1 \cdot Z_2}{Z_1 + Z_2} = \frac{5e^{j0^\circ} \cdot 5e^{j90^\circ}}{(5 + j0) + (0 + j5)} = \frac{25^{j90^\circ}}{5 + j5} = \frac{25e^{j90^\circ}}{\sqrt{50} e^{j45^\circ}} = 3,54e^{j45^\circ} \, \Omega \quad \rightarrow 2,5 + j2,5 \, \Omega $$ 2)  $(Z_1 || Z_2) + Z_3$  $$ (2,5 + j2,5) + (2 + j2) = 4,5 + j4,5 \, \Omega \quad \rightarrow  6,36e^{j45} \, \Omega $$ 3) $\left[(Z_1 || Z_2) + Z_3 \right]  \, || \, Z_4$  $$ = \frac{6,36e^{j45^\circ} \cdot 6e^{j0^\circ}}{(4,5 + j4,5) + (6 + j0)} = \frac{38,18e^{j45^\circ}}{10,45 + j4,5} = \frac{38,18e^{j45^\circ}}{11,42e^{j23,2^\circ}} = 3,34e^{j21,8^\circ} $$ Por lo tanto$$ Z_{th} =3,34e^{j21,8} \, \Omega = (3,1 + j1,2) \, \Omega $$     * __Cálculo de $V_{th}$__  $V_{th} = V_{AB} = V_{Z_4}$ 1) $Z_4 + Z_3$  $$ = (6 + j0) + (2 + j2) = 8 + j2 \quad \rightarrow \sqrt{68} e^{j14^\circ} $$ 2) $(Z_4 + Z_3) || Z_2$  $$ = \frac{\sqrt{68} e^{j14^\circ} \cdot 5e^{j90^\circ}}{(8 + j2) + (0 + j5)} = \frac{5 \sqrt{68} e^{j104^\circ} }{8 + j7} = \frac{ 5 \sqrt{68} e^{j104^\circ} }{\sqrt{113} e^{j41,18^\circ} } = 3,88e^{j62,85^\circ} $$ $$ \rightarrow 3,88e^{j62,85^\circ} = (1,77 + j3,45) \, \Omega $$ 3) $[(Z_4 + Z_3) || Z_2] + Z_1 $  $$ Z_{eq} = (1,77 + j3,45) + (5 + j0) = (6,77 + j3,45) \, \Omega \quad \rightarrow 7,60e^{j27^\circ} \, \Omega $$  __Cálculo  de $I_t$__$$ I_t = \frac{U}{Z_{eq}} = \frac{55,8e^{-j17,4^\circ}}{7,60e^{j27}} = 7,34e^{-j44,4^\circ} \, \mathrm{A} $$ __Cálculo de $V_{Z1}$__ $$ V_{Z_1} = I_t Z_1 = 7,34e^{-j44,4^\circ} 5e^{j0^\circ} = 36,7e^{-j44,4^\circ} \, \mathrm{V} $$ __Cálculo de $V_{Z_2}$__ $$ V_{Z_2} = V - V_{Z_1} = (27,02 + j9,01) \, \mathrm{V} \quad \rightarrow 28,48e^{18,44^\circ} \, \mathrm{V} $$ __Cálculo de $I_3$__ $$ I_3 = \frac{V_{Z_2}}{Z_3 + Z_4} = \frac{28,48e^{j18,44^\circ}}{\sqrt{68}e^{j14}} = 3,45e^{j4,40^\circ} \, \mathrm{V} $$ Cálculo de $V_{th}$$$ V_{th} = V_{AB} = V_{Z_4} = I_3 Z_4 = 20,72 e^{j4,40^\circ} \, \mathrm{V}$$      Usando python para calcular (opcional)<jupyter_code>import math, cmath
# Datos:
U = cmath.rect(55.8, math.radians(-17.4) )
Z1 = 5 + 0j
Z2 = 0 + 5j
Z3 = 2 + 2j
Z4 = 6 + 0j
# Cálculo de Z1 || Z2
ans1 = (Z1*Z2)/(Z1+Z2)

# Cálculo de (Z1 || Z2) || Z3
ans2 = ans1 + Z3

# Cálculo de Zth
Zth = (ans2*Z4)/(ans2 + Z4)
Zthpol = cmath.polar(Zth)

# Cálculo de Z3 + Z4
ans3 = Z4 + Z3

# Cálculo (Z4 + Z3) || Z2 
ans4 = (ans3*Z2)/(ans3 + Z2)

# Cálculo de Zeq
Zeq = ans4 + Z1

# Cálculo de It
It = U/Zeq

# Cálculo de V_z1
V_z1 = It * Z1

# Cálculo de V_z2
V_z2 = U - V_z1

# Cálculo de I3
I3 = V_z2/(Z3 + Z4)
I3pol = cmath.polar(I3)

# Cálculo de Vth
Vth = I3*Z4
Vth_pol = cmath.polar(Vth)
print('Resultados:')
print('-----------------------------------------')
print('Zth = (%.2f < %.2f°) Ohm'%(Zthpol[0], math.degrees(Zthpol[1]) ) )
print('Zth = {:.2f} Ohm'.format(Zth))
print('-----------------------------------------')
print('Vth = (%.2f < %.2f°) V'%(Vth_pol[0], math.degrees(Vth_pol[1]) ))
print('Vth = {:.2f} V'.format(Vth))
print('-----------------------------------------')
%reset -s -f<jupyter_output><empty_output><jupyter_text>Problema N°2Obtener el equivalente de Thevenin (entre los puntos a y b) y calcular la corriente que circularía si, en estos terminales, se coloca una resistencia de $30 \, \Omega$ $(R_c)$.   Solución __Cálculo de $Z_{th}$__1) $Z_0 || Z_1$; siendo $Z_0 = 0 + j0 = 0e^{-j0^\circ}$ $$ \frac{Z_0 Z_1}{Z_0 + Z_1} = \frac{0e^{j0} \cdot 111,8e^{j63,43^\circ}}{(0 + j0) + (50 + j100)} = 0 $$  2) $Z_2 || Z_3$ $$ \frac{Z_2 Z_3}{Z_2 + Z_3} = \frac{111,8e^{j63,43} \cdot 50e^{-j90^\circ}}{(50 + j100) + (0 - j50)} = 79,06e^{-j71,57^\circ} \, \Omega \quad \rightarrow (25 - j75) \, \Omega$$  3) $(Z_2 || Z_3) + Z_4$  $$ (25 - j75) + (25 + j0) = (50 - j75) \, \Omega $$ Por lo tanto$$ Z_{th} = (50 - j75) \, \Omega \quad \rightarrow \quad 90,14e^{-j56,31^\circ} \, \Omega $$     __Cálculo de $V_{th}$__ $$ V_{th} = V_{AB} = V_{Z_3} $$  1) $Z_3 + Z_2$$$ (0 - j50) + (50 + j100) = 50 + j50 \, \Omega \quad \rightarrow 70,71e^{j45^\circ} \, \Omega $$  2) $(Z_4 + Z_3) \, || Z_1$$$ \frac{70,71e^{j45^\circ} \cdot 111,8e^{j63,43^\circ}}{(50 + j50) + (50 + j100)} = 43,85e^{j52,13^\circ} \, \Omega $$ Por lo tanto$$ Z_{eq} = 43,85e^{j52,13^\circ} \, \Omega $$ Cálculo $I_t$  $$ I_t = \frac{V}{Z_{eq}} = \frac{220e^{j0^\circ}}{43,85e^{51,13^\circ}} = 5,02e^{-j52,13^\circ} \, \mathrm{A} \quad \rightarrow (3,08 - j3,96) \, \mathrm{A}  $$ Además$$ I_2 = I_t - I_1 $$ Entonces$$ I_1 = \frac{U}{Z_1} = \frac{220e^{j0^\circ}}{111,8e^{j63,43^\circ}} = 1,97e^{-j63,43^\circ} \, \mathrm{A} \quad \rightarrow (0,88 - j1,76) \, \mathrm{A} $$ Cálculo de $I_2$$$ I_2 = I_t - I_2 = (2,20 - j2,20) \, \mathrm{A} \quad \rightarrow 3,11e^{-j45^\circ} \, \mathrm{A}  $$ Por lo tanto$$ V_{th} = V_{AB} = V_{Z_3} = I_2 Z_3 = 155,56e^{-j135^\circ} \, \mathrm{V} $$      Cálculo de $I_{RC}$$$ I_{RC} = \frac{V_{th}}{Z_{th} + R_c} = 1,42e^{-j91,85^\circ} \, \mathrm{A} $$      Usando python para calcular (opcional)<jupyter_code>import math, cmath
# Datos
Z1 = 50+100j
Z2 = Z1
Z3 = 0-50j
Z4 = 25+0j
U = 220+0j
Rc = 30
# Cálculo de Zth
#1)   Z0 || Z1 ; siendo Z0 = 0 + 0j
Z0 = 0
ans1 = (Z0*Z1)/(Z0 + Z3)

#2)   Z2 || Z3
ans2 = (Z2*Z3)/(Z2+Z3)

#3)   (Z2 || Z3) + Z4
Zth = ans2 + Z4
Zthpol = cmath.polar(Zth)

# Cálculo de Zeq = (Z2 + Z3) || Z1
Zeq = ((Z2 + Z3)*Z1)/(Z1 + Z2 + Z3)

# Cálculo de It
It = U/Zeq
I1 = U/Z1
I2 = It - I1

# Cálculo de Vth
Vth = I2*Z3
Vth_pol = cmath.polar(Vth)

# Cálculo de I_rc
I_rc = Vth/(Zth + Rc)
I_rcpol = cmath.polar(I_rc)
print('Resultados:')
print('-----------------------------------------')
print('Zth = (%.2f < %.2f°) Ohm'%(Zthpol[0], math.degrees(Zthpol[1]) ) )
print('Zth = {:.2f} Ohm'.format(Zth))
print('-----------------------------------------')
print('Vth = (%.2f < %.2f°) V'%(Vth_pol[0], math.degrees(Vth_pol[1]) ))
print('Vth = {:.2f} V'.format(Vth))
print('-----------------------------------------')
print('I_rc = (%.2f < %.2f°) A'%(I_rcpol[0], math.degrees(I_rcpol[1]) ))
print('I_rc = {:.4f} A'.format(I_rc))
print('-----------------------------------------')
%reset -s -f<jupyter_output><empty_output><jupyter_text>------------------<jupyter_code># Esta celda da el estilo al notebook
#from IPython.core.display import HTML
#css_file = 'styles/aeropython.css'
#HTML(open(css_file, "r").read())<jupyter_output><empty_output>
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