- #1

Evari5te

- 8

- 4

- Homework Statement
- Two capacitors each of "10/xF" and a resistor are connected to a 50 Hz three phase power supply, as shown in figure 2.27. The power drawn from the supply is the same whether the switch S is open or closed. Find the resistance of R. - See attached file for diagram and original statement

- Relevant Equations
- Complex Power S = VI*

I assume that the "power drawn" refers to the active power, which I understand as being the real part of S in either scenario.

For power to be the same regardless of switch S being closed (four wire) or open (three wire)

Four wire power = three wire power

=>Re{VcnIc*} = Re{ [Vcn-Vnn’]Ic*}

Where Vcn is the c line to neutral voltage of the supply, Ic is the c line current and Vnn’ is the voltage of the load floating neutral with respect to ground in the three wire scenario.

Vnn’ is determined by Millman’s equation Vnn' = VanYa + VbnYb + VcnYc / (Ya + Yb + Yc)

where voltages are phase to neutral of the supply and Y values the admittances (reciprocal of impedance)

Ya = Yb = jωC = j2πfC

Yc = 1/R

My thoughts were I could use mesh current analysis to derive equations for current hence power in each case in terms of ωC and R, set the equations for real power equal to each other and solve for R. While ωC are known I opted to leave them as letters thinking this might make life easier when reducing the equations.

I assumed "10/xF" is actually 10 micro Farads - I have the Kindle version and am guessing that /x is a printing error because an unusual character, most likely μ, has been kicked out, n and p being the only other common prefixes.

I took Vab as the reference phasor and V(line) is 1 Volt. The actual voltage is irrelevant to the question

Four wire (switch closed):

Three mesh current loops with following voltages:

Vab = 1 + j0

Vbc = -0.5 -j0.866

Vcn = (0 + j)/√3

(2/jωC) x I1 - (1/jωC) x I2 + 0 x I3 = 1 + j0

Can be rearranged

2 x I1 - I2 = jωC (equation 1)

-1/jωC x I1 + (R + 1/jωC) x I2 - R x I3 = -0.5-j0.866 (equation 2)

-R x R2 + R x I3 = j/√3

Ia = I1

Ib = I2 - I1

Ic = I3 - I2

All the above resolved to give me Ic = j/√3R therefore Ic* = -j/√3R

S= VI* = VcnIc*

Vcn = VL/√3 x (0 + j) where VL was set = 1

S =(1/j√3) x -j/√3R

So S is entirely real and equal to 1/3R

Three wire (switch open)

Two mesh current loops with following voltages

Vab = 1 + j0

Vbc = -0.5 -j0.866

2 x I1 - I2 = jωC (eq. 1)

-I1 + I2(1 + jωCR) = ωC(0.866 – j0.5)

Ia = I1

Ib = I2 - I1

Ic = - I2

resolved to give me Ic = -√3ωC / (1 + j2ωCR)

I multiplied top and bottom by (1 - j2ωCR) to give me a complex numerator with real denominator,

Ic = (-√3ωC + j√3(2)ω^2C^2R) / (1 + 4ω^2C^2R^2)

so Ic* = (√3ωC

This was already starting to look pretty ugly.

By the time I had calculated Vnn' using Millman's equation and hence Vcn - Vnn' in order to proceed to complex power S = [Vcn - Vnn']Ic* things really got out hand - equations with R to the fourth power - giving me serious doubt as to whether I was missing a far easier, more obvious solution to this.....

I assumed "10/xF" is actually 10 micro Farads - I have the Kindle version and am guessing that /x is a printing error because an unusual character, most likely μ, has been kicked out, n and p being the only other common prefixes.

I took Vab as the reference phasor and V(line) is 1 Volt. The actual voltage is irrelevant to the question

Four wire (switch closed):

Three mesh current loops with following voltages:

Vab = 1 + j0

Vbc = -0.5 -j0.866

Vcn = (0 + j)/√3

(2/jωC) x I1 - (1/jωC) x I2 + 0 x I3 = 1 + j0

Can be rearranged

2 x I1 - I2 = jωC (equation 1)

-1/jωC x I1 + (R + 1/jωC) x I2 - R x I3 = -0.5-j0.866 (equation 2)

-R x R2 + R x I3 = j/√3

Ia = I1

Ib = I2 - I1

Ic = I3 - I2

All the above resolved to give me Ic = j/√3R therefore Ic* = -j/√3R

S= VI* = VcnIc*

Vcn = VL/√3 x (0 + j) where VL was set = 1

S =(1/j√3) x -j/√3R

So S is entirely real and equal to 1/3R

Three wire (switch open)

Two mesh current loops with following voltages

Vab = 1 + j0

Vbc = -0.5 -j0.866

2 x I1 - I2 = jωC (eq. 1)

-I1 + I2(1 + jωCR) = ωC(0.866 – j0.5)

Ia = I1

Ib = I2 - I1

Ic = - I2

resolved to give me Ic = -√3ωC / (1 + j2ωCR)

I multiplied top and bottom by (1 - j2ωCR) to give me a complex numerator with real denominator,

Ic = (-√3ωC + j√3(2)ω^2C^2R) / (1 + 4ω^2C^2R^2)

so Ic* = (√3ωC

**-**j√3(2)ω^2C^2R) / (1 + 4ω^2C^2R^2)This was already starting to look pretty ugly.

By the time I had calculated Vnn' using Millman's equation and hence Vcn - Vnn' in order to proceed to complex power S = [Vcn - Vnn']Ic* things really got out hand - equations with R to the fourth power - giving me serious doubt as to whether I was missing a far easier, more obvious solution to this.....