Unity-Gain Phase Shifting Transformer

Electricity and Magnetism Level pending

A three-phase AC transformer is connected in a delta configuration on the left side. The winding voltages on the left side are:

V A B = V A − V B = e j 0 V B C = V B − V C = e − j 2 π / 3 V C A = V C − V A = e j 2 π / 3 V_{AB} = V_A - V_B = e^{j 0} \\ V_{BC} = V_B - V_C = e^{-j 2 \pi/3} \\ V_{CA} = V_C - V_A = e^{j 2 \pi/3}

The transformer has a turns ratio of N N , resulting in voltages N V A B N V_{AB} , N V B C N V_{BC} , and N V C A N V_{CA} across the windings on the right side. Each coil on the right side is divided into two sub-windings as shown in the diagram, with R R being the ratio of the number of turns in the lower sub-winding to the total number of winding turns. The sub-winding voltages are proportional in the same way, and these are explicitly given in the diagram. The lower winding portions on the right are connected in a delta configuration.

Determine the values of N N and R R such that the phase-to-phase voltage on the right side has the same magnitude as that on the left side, and leads by a phase angle of π / 12 \pi / 12 .

V a − V b V A − V B = V a b V A B = e j π / 12 \frac{V_a - V_b}{V_A - V_B} = \frac{V_{ab}}{V_{AB}} = e^{j \pi / 12}

Enter your answer as N + R N + R .

Details and Assumptions:
1) The winding fraction R R must be a real number between 0 0 and 1 1 .
2) Transformer turns ratio N N must be a positive real number
3) e e is Euler's number, and j j is the imaginary unit


The answer is 1.4505.

Steven Chase by Steven Chase , 37, USA

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1 solution

Karan Chatrath
May 22, 2021

The line indicated in the diagram is grounded. This implies that:

V a − 0 = N ( 1 − R ) V A B V_a - 0 = N(1-R)V_{AB} 0 − V 2 = N R V A B 0 - V_2 = NRV_{AB} V b − V 2 = N V B C ( 1 − R ) V_b - V_2 = NV_{BC}(1-R)

It is given that V A B = 1 V_{AB} = 1 . Therefore:

V a − 0 = N ( 1 − R ) V_a - 0 = N(1-R) V b = N V B C ( 1 − R ) + V 2 V_b = NV_{BC}(1-R) + V_2 V b = N V B C ( 1 − R ) − N R V_b = NV_{BC}(1-R) -NR

Therefore:

V a − V b V A B = V a − V b = N ( 1 − R ) − N V B C ( 1 − R ) + N R \frac{V_a - V_b}{V_{AB}} = V_a - V_b = N(1-R)-NV_{BC}(1-R) +NR V a − V b = cos ⁡ ( π 12 ) + j sin ⁡ ( π 12 ) = N − N ( 1 − R ) ( − 1 2 − j 3 2 ) V_a - V_b = \cos\left(\frac{\pi}{12}\right) + j \sin\left(\frac{\pi}{12}\right) = N - N(1-R)\left(-\frac{1}{2} - j \frac{\sqrt{3}}{2}\right) ⟹ cos ⁡ ( π 12 ) + j sin ⁡ ( π 12 ) = ( 3 N 2 − N R 2 ) + j ( 3 N 2 − 3 N R 2 ) \implies \cos\left(\frac{\pi}{12}\right) + j \sin\left(\frac{\pi}{12}\right) = \left(\frac{3N}{2} - \frac{NR}{2}\right) + j \left(\frac{\sqrt{3}N}{2} - \frac{\sqrt{3}NR}{2}\right)

Equating the real and imaginary parts allows us to solve for N N and R R .

N = cos ⁡ ( π 12 ) 3 − sin ⁡ ( π 12 ) 3 N = \frac{\cos\left(\frac{\pi}{12}\right)\sqrt{3} - \sin\left(\frac{\pi}{12}\right)}{\sqrt{3}} R = 3 N 2 − sin ⁡ ( π 12 ) 3 N 2 R =\frac{ \frac{\sqrt{3}N}{2} - \sin\left(\frac{\pi}{12}\right)}{ \frac{\sqrt{3}N}{2}}

N + R = 1.4505 N + R = 1.4505

3 Helpful 3 Interesting 3 Brilliant 0 Confused

Nice solution

Talulah Riley - 3 weeks ago

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