Skip to main content
\(\require{cancel}\)
Physics LibreTexts

13.9.4: Bridge Solution by Delta-Star Transform

   In the above examples, we have calculated the condition that there is no current in the detector -. i.e. that the bridge is balanced.  Such calculations are relatively easy.  But what if the bridge is not balanced?  Can we calculate the impedance of the circuit?  Can we calculate the currents in each branch, or the potentials at any points?  This is evidently a little harder.  We should be able to do it.  Kirchhoff’s rules and the delta-star transform still apply for alternating currents, the complication being that all impedances, currents and potentials are complex numbers. 

 

 

  Let us start by trying the following problem:

 

i

w  =   500  rad s-1

400 mF

10 mH

6 mH

5 W

2 W

6 W

4 W

3W

500 mF

300 mF

A

File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image001.gif  

 

 

 

  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

That is to say:

 

 

 

i

w  =   500  rad s-1

400 mF

10 mH

6 mH

5 W

2 W

6 W

4 W

3W

500 mF

300 mF

The impedances are indicated in

ohms.

i

      File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image002.wmz

w  =   500  rad s-1

Z1=5+5j

Z4=2-5j

Z5=4+3j

Z2=6-10j

Z3=3-4j

A

B

C

D

I

I1

I2

I4

I5

I3

 

 

B

 
  File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image003.gif

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

      Our question is:  What is the impedance of the circuit at a frequency of w  =   500  rad s-1?

 

     We refer now to Chapter 4, Section 12.  We are going to replace the left hand “delta” with its equivalent “star”.  Recall equations 4.12.2 - 4.12.13.  With alternating currents, We can use the same equations with alternating currents, provided that we replace the File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image005.gif in the delta and the  File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image007.gif in the star with File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image009.gif in the delta and File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image011.gifin the star.  That is, we replace the resistances with impedances, and conductances with admittances.   This is going to need a little bit of calculation, and familiarity with complex numbers. I used a computer - hand calculation was too tedious and prone to mistakes.  This is what I got, using File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image013.gif and cyclic variations:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

i

w  =   500  rad s-1

Z4=2-5j

Z5=4+3j

The impedances are indicated in

ohms.

z1 = 0.642 599 - 3.444 043j

 

z2 = 1.931 408 + 0.884 477j

 

z3 = 4.693 141 + 1.588 448j

z2

z3

z1

B

D

C  Potential here taken to be zero.

P

A

i2  =  I4

I4

I5

i1  =  I5

File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image014.wmz

C

 

 

File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image015.gif  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

The impedance of PBC is  3.931 408  -  4.115 523j     W.

The impedance of PDC is  4.642 599  -  0.444 043j     W.

The admittance of PBC is  0.121 364  +  0.127 048j     S.

The admittance of PDC is  0.213 444  +  0.020 415j     S.

 

The admittance of PC   is  0.334 808  +  0.147 463.j      S.

The impedance of PC   is  2.501 523  -  1.101 770j      W.

 

The impedance of AC  is  7.194 664  +   0.486 678j     W.

The admittance of  AC  is 0.138 359  -   0.009 359j      S.

 

   Suppose the applied voltage is File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image017.gif, with w = 500 rad s-1 and File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image019.gif= 24 V.   Can we find the currents in each of File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image021.gif and the potential differences between the several points? This should be fun.  I’m sitting in front of my computer.  Among other things, I have trained it instantly to multiply two complex numbers and also to calculate the reciprocal of a complex number.  If I ask it for (2.3 + 4.1j)(1.9 - 3.4j) it will instantly tell me 18.31 - 0.03j.  If I ask it for 1/(0.5 + 1.2j) it will instantly tell me 0.2959 -  0.7101j.  I can instantaneously convert between impedance and admittance.

   

   The current through any element depends on the potential difference across it.  We can take any point to have zero potential, and determine the potentials at other points relative to that point. I choose to take the potential at C to be zero, and I have indicated this by means of a ground (earth) symbol at C.  We are going to try to find the potentials at various other points relative to that at C.

 

   In drawings B and C I have indicated the currents with arrows.  Since the currents are alternating, they should, perhaps, be drawn as double-headed arrows. However, I have drawn them in the direction that I think they should be at some instant when the potential at A is greater than the potential at C.  If any of my guesses are wrong, I’ll get a negative answer in the usual way.

 

The total current I is V times the admittance of the circuit. 

That is:     File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image023.gif.

 

 The peak current will be 3.328200 A  (because the modulus of the admittance is 0.138675 S), and the current lags behind the voltage by 3º.9.  

 

I hope the following two equations are obvious from drawing  C.

 

                                                File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image025.gif

From this,

 

File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image027.gif

 

I4  leads on V by 18º.7    File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image029.gif

 

File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image031.gif

 

I5  lags behind V by 22º.2     File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image033.gif

 

 

Check:   File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image035.gifü

 

  I show below graphs of the potential difference between A and C  File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image037.gif and the currents I, I4 and I5.  The origin for the horizontal scale is such that the potential at A is zero at t = 0.  The vertical scale is in volts for VC, and is five times the current in amps for the three currents.

 

File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image039.gif

 

We don’t really need to know File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image041.gif, but we do want to know File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image043.gif.   Let’s first see if we can find some potentials relative to the point C. 

 

From drawing C we see that

 

File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image045.gif, which results in  File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image047.gif 

 

VB  lags behind V by 0.864 432  rad  = 49º.5    File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image049.gif

 

 

From drawing C we see that

 

File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image051.gif, which results in  File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image053.gif 

 

VD  leads on V by  0.256 440 rad = 14º.7    File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image055.gif

 

 

We can now calculate I3 (see drawing B) from  File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image057.gif .  I find

File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image059.gif

 

I3 lags behind V by 1.046 588 rad = 60º.0.    File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image061.gif

 

 

I1 can now be found from File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image063.gif.  The real part of VA is 24 V, and (since we have grounded C), its imaginary part is zero.  If in doubt about this verify that

File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image065.gifü

I find

 

File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image067.gif

 

I1 lags behind V by 0.443 725 rad = 25º.4.    File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image069.gif

 

In a similar manner, I find

 

File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image071.gif

 

I2 leads on V by 0.862 147 rad = 49º.4.    File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image073.gif

 

Summary:

File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image075.gif

File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image067.gif

File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image071.gif

File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image059.gif

File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image077.gif                               

File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image079.gif

 

These may be checked by verification of Kirchhoff’s first rule at each of the points A, B, C, D.