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# 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
 400 mF
 10 mH
 6 mH
 5 W
 2 W
 6 W
 4 W
 3W
 500 mF
 300 mF
 A

That is to say:

 i
 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
 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

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  in the delta and the   in the star with  in the delta and in 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  and cyclic variations:

 i
 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
 C

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 , with w = 500 rad s-1 and = 24 V.   Can we find the currents in each of  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:     .

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.

From this,

I4  leads on V by 18º.7

I5  lags behind V by 22º.2

Check:   ü

I show below graphs of the potential difference between A and C   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.

We don’t really need to know , but we do want to know .   Let’s first see if we can find some potentials relative to the point C.

From drawing C we see that

, which results in

VB  lags behind V by 0.864 432  rad  = 49º.5

From drawing C we see that

, which results in

We can now calculate I3 (see drawing B) from   .  I find

I3 lags behind V by 1.046 588 rad = 60º.0.

I1 can now be found from .  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

ü

I find

I1 lags behind V by 0.443 725 rad = 25º.4.

In a similar manner, I find