Alternating Current Series #3

Using phasor notation for convenience, then $v_1 = V\angle 0^\circ$ , $v_2 = V \angle 120^\circ$ , and $v_3 = V \angle 240^\circ$ . Using the junction $O$ connected to the three voltage sources as reference and considering the sum of current out of node $O$ as zero, and let the magnitude of the impedances be $R$ , then we have:

$\begin{aligned} \frac {v_P-v_1}R + \frac {v_P-v_2}{-jR} + \frac {v_P-v_3}{jR} & = 0 & \small \blue{\text{Let }v_P = V_p \angle \theta} \\ \frac {V_p\angle \theta - V\angle 0^\circ}R + \frac {V_p\angle \theta - V\angle 120^\circ}{-jR} + \frac {V_p\angle \theta - V\angle 240^\circ}{jR} & = 0 \\ \frac {V_p\angle \theta - V\angle 0^\circ}R + \frac {V\angle 120^\circ - V\angle 240^\circ}{R\angle 90^\circ} & = 0 \\ V_p\angle \theta - V\angle 0^\circ + V\angle 30^\circ - V\angle 150^\circ & = 0 \end{aligned}$

$\begin{aligned} \implies V_p \angle \theta & = V\angle 0^\circ - V\angle 30^\circ + V\angle 150^\circ \\ & = V \left(1 - \frac {\sqrt 3}2 - \frac j2 - \frac {\sqrt 3}2 + \frac j2 \right) \\ & = (1-\sqrt 3)V \\ &= (\sqrt 3 - 1) V \angle 180^\circ \end{aligned}$

Therefore $\alpha + \beta = \sqrt 3 - 1 + 180 \approx \boxed{180.732}$ .

The key here is to represent the quantities as complex numbers. The voltages are (we can just consider the magnitude to be one):

$V_1 = e^{j 0^\circ} \\ V_2 = e^{j 120^\circ} \\ V_3 = e^{j 240^\circ}$

The impedances are (take there magnitudes to be unity):

$Z_R = 1 + j 0 \\ Z_C = 0 - j 1 \\ Z_L = 0 + j 1$

The nodal equation at $P$ is:

$\frac{V_P - V_1}{R} + \frac{V_P - V_2}{Z_C} + \frac{V_P - V_3}{Z_L} = 0 \\ V_P \Big( \frac{1}{R} + \frac{1}{Z_C} + \frac{1}{Z_L} \Big) = \frac{V_1}{R} + \frac{V_2}{Z_C} + \frac{V_3}{Z_L}$

Solving for $V_P$ results in:

$V_P = (\sqrt{3} - 1 ) e^{j 180^\circ}$

Solution code is attached:

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import math
import cmath

deg = math.pi/180.0

theta1 = 0.0
theta2 = 120.0*deg
theta3 = 240.0*deg

V1 = complex(math.cos(theta1),math.sin(theta1))
V2 = complex(math.cos(theta2),math.sin(theta2))
V3 = complex(math.cos(theta3),math.sin(theta3))

R = 1.0
ZC = complex(0.0,-1.0)
ZL = complex(0.0,1.0)

right = V1/R + V2/ZC + V3/ZL
left = 1.0/R + 1.0/ZC + 1.0/ZL

VP = right/left

alpha = abs(VP)
beta = cmath.phase(VP)/deg

print alpha
print beta
print (alpha + beta)

#>>> 
#0.732050807569
#180.0
#180.732050808
#>>>