The question is incomplete! Complete question along with answer and step by step explanation is provided below.
Question:
In the given circuit, V(t)=12cos(2000t+45)V, R1=R2=2Ω, L1=L2=L3=3mH and C1=250μF. You are required to find the Thevenin equivalent of this circuit using phasors.
a. If you write the Thevenin voltage, Vth, in phasor form, what is the magnitude of this phasor? Put your answer in the box below without units.
b. What is the value of the angle associated with the phasor Vth, in degrees?
c. Now, calculate the Thevenin impedance, Zth. What is the magnitude of this phasor?
d. What is the angle associated with the phasor Zth, in degrees?
Answer:
Vth = 6 < 45° V
Zth = 1.414 < 45°
a. The magnitude of the Thevenin voltage is 6 V
b. The phase angle of the Thevenin voltage is 45°
c. The magnitude of the Thevenin impedance is 1.414 V
d. The phase angle of the Thevenin impedance is 45°
Explanation:
The given voltage is
V(t)=12cos(2000t+45)
In phasor form,
V(t) = 12 < 45° V
So the magnitude of voltage is 12 V and the phase angle is 45°
Also the frequency ω = 2000
then the inductance is
L₁ = L₂ = L₃ = jωL = j×2000×0.003 = j6 Ω
and the capacitance is
C₁ = 1/jωC = 1/(j×2000×250x10⁻⁶) = -j2 Ω
and the resistance is
R₁ = R₂ = 2 Ω
Thevenin voltage:
The Thevenin voltage is the voltage that appears across the open-circuited terminals a-b (after removing L₃)
The Thevenin voltage is given by
Vth = V(t) × [ (R₂ + L₂) / (R₁ + L₁) + (R₂ + L₂) ]
Please note that there is no current flow in the capacitor due to open-circuited terminal a-b
Vth = 12 < 45° × [ (2 + j6) / (2 + j6) + (2 + j6) ]
Vth = 12 < 45° × [ (2 + j6) / (4 + j12) ]
Vth = 4.24264 + j4.24264 V
In phasor form,
Vth = 6 < 45° V
a. The magnitude of the Thevenin voltage is 6 V
b. The phase angle of the Thevenin voltage is 45°
Thevenin Impedance:
The Thevenin Impedance is the impedance of the circuit calculated when looking from the terminal a-b
Zth = [ (R₁ + L₁) × (R₂ + L₂) / (R₁ + L₁) + (R₂ + L₂) ] + (-j2)
Zth = [ (2 + j6) × (2 + j6) / (2 + j6) + (2 + j6) ] - j2
Zth = [ (2 + j6) × (2 + j6) / (2 + j6) + (2 + j6) ] - j2
Zth = [ (-32 + j24) / (4 + j12) ] -j2
Zth = [ (1 +j3) ] - j2
Zth = 1 + j Ω
In phasor form,
Zth = 1.414 < 45°
c. The magnitude of the Thevenin impedance is 1.414 V
d. The phase angle of the Thevenin impedance is 45°
