Equivalent Wye

As always, @Karan Chatrath has given a great solution to the problem. I will comment on the bonus. There are two ways of defining the word "equivalent" in this context:

1) A wye circuit which draws the same currents as the delta for a particular set of boundary voltages
2) A wye circuit which draws the same currents as the delta for all possible sets of boundary voltages

This problem demonstrates that there are infinitely many solutions for definition $1$ . Definition $2$ is much more strict, and there is only one solution in that case. The classic solution for definition $2$ is $Z_A = Z_B = Z_C = 1 + j 0$ , which is what you get from the formulas at the Wikipedia page. In this problem, the classic solution is the only one in which all three impedances are purely resistive.

I posted this problem because people are generally only familiar with the classic solution

I have also posted the solution code below:

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import math

# Constants

VA = complex(10.0,0.0)
VB = complex(5.0,5.0)
VC = complex(5.0,10.0)

ZAB = complex(3.0,0.0)
ZBC = complex(3.0,0.0)
ZCA = complex(3.0,0.0)

###########################################################

# Currents

IA = (VA-VB)/ZAB + (VA-VC)/ZCA
IB = (VB-VA)/ZAB + (VB-VC)/ZBC
IC = (VC-VB)/ZBC + (VC-VA)/ZCA

#print abs(IA + IB + IC)    # check that currents sum to zero
#print ""

###########################################################

# Compute integral

dRA = 10.0**(-6.0)

RA = dRA

Q = 0.0

while RA <= 5.0:      

    flag = 0               # error condition flag

    # VA - ZA*IA = VB - ZB*IB     # equate expressions for the wye neutral voltage
    # VA - ZA*IA = VC - ZC*IC   # this forms the basis of the solution

    ZA = complex(RA,0.0)            # wye impedances
    ZB = (VA - ZA*IA - VB)/(-IB)
    ZC = (VA - ZA*IA - VC)/(-IC)

    RA = ZA.real                # wye resistances
    RB = ZB.real
    RC = ZC.real

    ZT = ZA + ZB + ZC           # total impedance
    ZTmag = abs(ZT)

    if (RA < 0.0) or (RB < 0.0) or (RC < 0.0):   # check for negative resistance
        flag = 1

    if (flag == 0):             # Add to running sum if error flag = 0

        Q = Q + ZTmag*dRA

        #print RA,ZTmag

    RA = RA + dRA

###########################################################

print ""
print ""
print dRA
print Q

#>>> 
#0.0001
#5.60902621372
#>>> ================================ RESTART ================================
#>>> 
#1e-05
#5.60915544503
#>>> ================================ RESTART ================================
#>>> 
#1e-06
#5.60919445718
#>>> 

Very nice problem!

Consider the Delta configuration. The currents through each of the three branches are $I_{AB}$ , $I_{BC}$ , $I_{CA}$ . Based on this, the circuit equations are:

$-V_A + I_{AB}Z_{AB} + V_B = 0$ $-V_C + I_{CA}Z_{CA} + V_A = 0$ $-V_B + I_{BC}Z_{BC} + V_C = 0$

According to Kirchoff's current law:

$I_A + I_{CA} = I_{AB}$ $I_B + I_{AB} = I_{BC}$ $I_C + I_{BC} = I_{CA}$

Using these three equations, one can solve for $I_A$ , $I_B$ and $I_C$ . It is easy to check after solving that:

$I_A+I_B+I_C=0$

Now, consider the Y-configuration. Here, the circuit equations read:

$-V_A + I_AZ_A-I_BZ_B +V_B=0$ $-V_B + I_BZ_B-I_CZ_C +V_C=0$ $-V_C + I_CZ_C-I_AZ_A +V_A=0$

Now, it is easy to see that adding any 2 of the above equations gives the third. This means that the above system of equations has infinite solutions. We use the condition that $Z_A=R_A$ , plug into the result above and solve for $Z_B$ and $Z_C$ . By doing so, one gets:

$Z_B = (3-2R_A) + j(3R_A-3)$ $Z_C = \left(\frac{21-11R_A}{10}\right) + j \left(\frac{3-3R_A}{10}\right)$

Recall that all real parts of $Z_B$ and $Z_C$ cannot be negative. In other words:

$3-2R_A \ge 0$ $\implies R_A \le \frac{3}{2}$ $R_{Amax}=\frac{3}{2}$

Finally:

$Z_T = Z_A+Z_B+Z_C$

Here $Z_T$ is purely a function of $R_A$ . Its absolute value is also only a function of $R_A$ . The simplification to obtain this expression is left out in this solution. And the required integral is, therefore:

$\boxed{\int_{0}^{3/2} \lvert Z_T \rvert \ dR_A \approx 5.6092}$

The integral has a closed-form solution but the final computation is done using Wolfram-Alpha. I have not tackled the bonus question yet.