13.9D: Bridge Solution by Delta-Star Transform

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Mar 5, 2022
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- 13.9C: The Wien Bridge
- 13.10: The Transformer

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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:

13.9.4 revised.png

That is to say:

13.9.4 revised 2.png
*The impedances are indicated in ohms.

Our question is: What is the impedance of the circuit at a frequency of $\omega = 500 \text{ rad s}^{-1}$ ?

in the star with $Z_1Z_2Z_3Y_1Y_2Y_3$ in the delta and $z_1z_2z_3y_1y_2y_3$ 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 $z_1=\frac{Z_2Z_3}{Z_1+Z_2+Z_3}$ and cyclic variations:

13.11 p21.png
*The impedances are indicated in ohms.

$\begin{align}z_1&=0.642599-3.444043j \\ z_2&=1.931408+0.884477j\\ z_3&=4.693141+1.588448j \\ \end{align}$

$3.931408 - 4.115523j \, \Omega$ .
$4.642599 - 0.444043j \, \Omega$ .
$0.121364 + 0.127048j \, \text{S}$ .
$0.213444 + 0.020415j \, \text{S}$ .
$0.334 808 + 0.147 463j \, \text{S}$ .
$2.501 523 - 1.101 770j \, \Omega$ .
The impedance of AC is $\underline{7.194 664 + 0.486 678j\, \Omega}$ .
$0.138 359 - 0.009 359j \, \text{S}$ .

$\omega = 500 \text{ rad s}^{-1}$ and $\hat{V}= 24 \text{V}$ .

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 $(2.3 + 4.1j)(1.9 - 3.4j)$ $18.31 - 0.03j$ $1/(0.5 + 1.2j)$ $0.2959 - 0.7101j$ . 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 $\text{C}$ to be zero, and I have indicated this by means of a ground (earth) symbol at $\text{C}$ . We are going to try to find the potentials at various other points relative to that at $\text{C}$ .

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 $\text{A}$ is greater than the potential at $\text{C}$ . 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: $I=24\times (0.138359-0.009359j)=(3.320611-0.224620j)\,\text{A}$ .

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

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

$\nonumber\begin{align}&I=I_4+I_5 \\ \nonumber &I_4(z_2+Z_4)=I_5(z_1+Z_5) \\ \end{align}$

From this,

$\nonumber \begin{align}I_4&=\left ( \frac{z_1+Z_5}{z_1+z_2+Z_4+Z_5}\right ) I = \left (\frac{4.642599-0.444043j}{8.574007-4.559567j}\right )\times (3.320611-0.224620j) \\ \nonumber &=\underline{1.514284+0.511681j}\,\text{A}\\ \end{align}$

$I_4$ $\hat{I}_4=1.598397\,\text{A}$ .

$I_5=\left ( \frac{z_2+Z_4}{z_1+z_2+Z_4+Z_5}\right )I=\underline{1.806328-0.736302j \text{ A}}$

$I_5$ lags behind $V$ by 22º.2 $\hat{I}_5=1.950631\, \text{A}$ .

Check: $I_4+I_5=I \checkmark$

$I$ $I_4$ $I_5$ $t = 0$ . The vertical scale is in volts for $V_C$ , and is five times the current in amps for the three currents.

13.11 p23.png

. Let’s first see if we can find some potentials relative to the point $\text{C}$ .

From drawing C we see that

$\underline{V_B=(5.586975-6.548057j)\,\text{V}}$

$V_B$ $0.864 432 \text{ rad }=$ 49º.5 $\hat{V}_B=8.607603 \text{ V}$ .

From drawing C we see that

$\underline{V_D=(9.434216+2.473775j) \text{ V}}$

$\hat{V}_B=9.753153 \text{ V}$ .

$I_3$ (see drawing B) from $I_3=\frac{V_B-V_D}{Z_3}=(V_B-V_D)Y_3$ .

$\underline{I_3=0.981824-1.698178j \text{ A}}$

$I_3$ lags behind $V$ by $1.046 588 \text{ rad }=$ 60º.0. $\hat{I}_3=1.961578 \text{ A}$ .

$V_A$ is 24 V, and (since we have grounded C), its imaginary part is zero. If in doubt about this verify that

$IZ=(3.320611-0.224620 j)(7.194664+0.486678j)=(24+0j)\text{V}.\checkmark$

I find

$\underline{I_1=2.496108-1.186497 j \text{ A}}$

$I_1$ lags behind $V$ by $0.443 725 \text{ rad }=$ 25º.4. $\hat{I}_1=2.763753 \text{ A}$ .

In a similar manner, I find

$\underline{I_2=0.824503+0.961876j \text{ A}}$

$I_2$ leads on $V$ by $0.862 147 \text{ rad }=$ 49º.4. $\hat{I}_2=1.266891 \text{ A}$ .

Summary:

$\underline{I=3.320611-0.224620j \text{ A}}$
$\underline{I_1=2.496108-1.186497j \text{ A}}$
$\underline{I_2=0.824503-0.961876j \text{ A}}$
$\underline{I_3=0.981824-1.698178j \text{ A}}$
$\underline{I_4=1.514284-0.511681j \text{ A}}$
$\underline{I_5=1.806328-0.736302j \text{ A}}$

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

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