Three-Phase with Floating Neutrals

Electricity and Magnetism Level 5

A three-phase AC circuit contains three ideal sinusoidal voltage sources and a load consisting of three resistors. The voltage sources share a common neutral, and the resistors share a common neutral. The two neutrals are floating, and are not directly connected to each other.

The voltage sources have values ( V A ⃗ , V B ⃗ , V C ⃗ ) = ( 1 ∠ 0 ∘ , 1 ∠ − 120 ∘ , 1 ∠ 12 0 ∘ ) (\vec{V_A},\vec{V_B},\vec{V_C}) = (1\angle0 ^\circ, \, 1\angle{-120} ^\circ, 1\angle120 ^\circ) . The resistor values are shown in the diagram.

Consider the source neutral to be the voltage reference for the circuit ( V = 0 ) (V = 0) . Let V N ⃗ \vec{V_N} be the voltage at the load neutral with respect to the source neutral.

What is the magnitude of V N ⃗ \vec{V_N} , to 3 decimal places?

Note: All voltage magnitudes are given in RMS, and the answer is expected in the same format. The symbol " ∠ \angle " indicates the complex phase angle of a sinusoid.

Hint: The answer can be found by writing and solving one complex equation, equivalent to two real-valued equations.


The answer is 0.328.

Steven Chase by Steven Chase , 37, USA

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2 solutions

Chew-Seong Cheong
Mar 16, 2017

Similar solution with Steven Chase 's

Consider the current sum of the common neutral of the load and using Ohm's law for the currents.

V A ⃗ − V N ⃗ 1 + V B ⃗ − V N ⃗ 2 + V C ⃗ − V N ⃗ 3 = 0 V A ⃗ + 1 2 V B ⃗ + 1 3 V C ⃗ = ( 1 + 1 2 + 1 3 ) V N ⃗ 1 ∠ 0 ∘ + 1 2 ∠ − 12 0 ∘ + 1 3 ∠ 12 0 ∘ = 11 6 V N ∠ θ 1 + 1 2 ( − 1 2 − 3 2 i ) + 1 3 ( − 1 2 + 3 2 i ) = 11 6 V N ∠ θ 7 12 − 3 12 i = 11 6 V N ∠ θ 7 − 3 i = 22 V N ∠ θ ⟹ V N ∠ θ = 7 − 3 i 22 = 13 11 ∠ − tan ⁡ − 1 3 7 ⟹ V N ≈ 0.328 \begin{aligned} \frac {\vec{V_A}-\vec{V_N}}1 + \frac {\vec{V_B}-\vec{V_N}}2 + \frac {\vec{V_C}-\vec{V_N}}3 & = 0 \\ \vec{V_A} + \frac 12 \vec{V_B} + \frac 13 \vec{V_C} & = \left(1+\frac 12+\frac 13\right) \vec{V_N} \\ 1\angle 0^\circ + \frac 12 \angle -120^\circ + \frac 13 \angle 120^\circ & = \frac {11}6 V_N \angle \theta \\ 1 + \frac 12 \left(-\frac 12 - \frac {\sqrt 3}2i \right) + \frac 13 \left(-\frac 12 + \frac {\sqrt 3}2i \right) & = \frac {11}6 V_N \angle \theta \\ \frac 7{12} - \frac {\sqrt 3}{12}i & = \frac {11}6 V_N \angle \theta \\ 7 - \sqrt 3 i & = 22 V_N \angle \theta \\ \implies V_N \angle \theta & = \frac {7 - \sqrt 3i}{22} \\ & = \frac {\sqrt{13}}{11} \angle - \tan^{-1} \frac {\sqrt 3}7 \\ \implies V_N & \approx \boxed{0.328} \end{aligned}

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Steven Chase
Mar 16, 2017

The currents through the three resistors must sum to zero at the neutral. These currents can be calculated using Ohm's Law as follows:

V N ⃗ − V A ⃗ 1 + V N ⃗ − V B ⃗ 2 + V N ⃗ − V C ⃗ 3 = 0 V N ⃗ ( 1 + 1 2 + 1 3 ) = V A ⃗ + V B ⃗ 2 + V C ⃗ 3 \large{\frac{\vec{V_N} - \vec{V_A}}{1} + \frac{\vec{V_N} - \vec{V_B}}{2} + \frac{\vec{V_N} - \vec{V_C}}{3} = 0 \\ \vec{V_N}(1 + \frac{1}{2} + \frac{1}{3}) = \vec{V_A} + \frac{\vec{V_B}}{2} + \frac{\vec{V_C}}{3}}

Plugging in numbers gives V N ⃗ ≈ 0.328 ∠ − 13. 9 ∘ \vec{V_N} \approx 0.328\angle -13.9^\circ .

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