Transmission Line with Load Fault

Electricity and Magnetism Level pending

A balanced three-phase voltage source feeds a set of load resistors through a transmission line. The three phases of the transmission line are magnetically coupled, which can be modeled using self and mutual impedances as follows:

V A S − Z S I A − Z M I B − Z M I C = V A R V B S − Z M I A − Z S I B − Z M I C = V B R V C S − Z M I A − Z M I B − Z S I C = V C R V_{AS} - Z_S I_A - Z_M I_B - Z_M I_C = V_{AR} \\ V_{BS} - Z_M I_A - Z_S I_B - Z_M I_C = V_{BR} \\ V_{CS} - Z_M I_A - Z_M I_B - Z_S I_C = V_{CR}

In the circuit, the star (neutral) point of the voltage sources is taken as the voltage reference. By convention, the currents flow across the transmission line from left to right. The system parameters are:

V A S = 100 e j 0 V B S = 100 e − j 2 π / 3 V C S = 100 e j 2 π / 3 Z S = 1 + j 10 Z M = 1 + j 5 R = 10 R N = 2 V_{AS} = 100 \, e^{j 0} \\ V_{BS} = 100 \, e^{-j 2 \pi/3} \\ V_{CS} = 100 \, e^{j 2 \pi/3} \\ Z_S = 1 + j 10 \\ Z_M = 1 + j 5 \\ R = 10 \\ R_N = 2

There is a fault which short-circuits the phase A A load resistor (as shown in red in the diagram).

What is the magnitude of voltage V C R V_{CR} ?

Bonus: Is this result intuitive?


The answer is 111.22.

Steven Chase by Steven Chase , 37, USA

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1 solution

Karan Chatrath
Oct 31, 2019

The first step is writing down the circuit equations by applying Kirchoff's laws.

V A S = ( V A S − V A R ) + ( I A + I B + I C ) R N V_{AS} = (V_{AS} - V_{AR}) + (I_A + I_B + I_C)R_N V B S = ( V B S − V B R ) + I B R + ( I A + I B + I C ) R N V_{BS} = (V_{BS} - V_{BR}) + I_BR + (I_A + I_B + I_C)R_N V C S = ( V C S − V C R ) + I C R + ( I A + I B + I C ) R N V_{CS} = (V_{CS} - V_{CR}) + I_CR + (I_A + I_B + I_C)R_N

Replacing the equation V A S − V A R V_{AS} - V_{AR} by the given transmission line model, in each equation, re-writing and simplifying gives:

[ V A S V B S V C S ] = [ D − R N N N D N N N D ] [ I A I B I C ] ⟹ V = Z I \left[\begin{matrix} V_{AS} \\V_{BS} \\V_{CS} \end{matrix}\right] = \left[\begin{matrix} D-R&N&N\\N&D&N\\N&N&D \end{matrix}\right] \left[\begin{matrix} I_{A} \\I_{B} \\I_{C} \end{matrix}\right] \implies V = ZI

Where:

N = Z M + R N N = Z_M + R_N D = Z S + R N + R D = Z_S + R_N+R

Let, Z Z be referred to as an 'impedance matrix'. From here, the currents can be computed by:

I = Z − 1 V \boxed{I = Z^{-1}V}

Having computed the currents I A I_A , I B I_B and I C I_C , the final step is the computation of V C R V_{CR} which is as follows:

V C R = V C S − Z M I A − Z M I B − Z S I C V_{CR} = V_{CS} - Z_MI_A - Z_MI_B -Z_SI_C

∣ V C R ∣ ≈ 111.2201 \boxed{\mid V_{CR} \mid \approx 111.2201}

The above computations are done using a computer to save time. Also, voltage magnitude is computed relative to a point. I assumed that the circuit after R N R_N and before the sources is grounded ( 0 V 0 V ).

1 Helpful 1 Interesting 1 Brilliant 0 Confused

As for the bonus question, I would say it is counter-intuitive. I expected the voltage magnitude V C R V_{CR} to be lower than V C S V_{CS} .

Karan Chatrath - 1 year, 7 months ago

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Indeed, the mutual coupling can yield some surprising results. Thanks for the solution

Steven Chase - 1 year, 7 months ago

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