Problem 55-21
Source: Adapted from Problem 8.63 - page 497 - Ulaby, Maharbiz & Furse - Book: Circuit Analysis and Design - 2018.
Determine the equivalent impedance between points a - b in the circuit shown in Figure 55-21.1.
Solution of the Problem 55-21
To solve this problem we will calculate the impedances between the points a-m, b-m and d-m. Note that the impedances between points b-m and d-m
will be equal, as the values of the components are the same. Therefore, we have:
See Figure 55-21.2 for the new circuit with the impedances calculated above.
In the figure above, it is noticeable that the reactive elements of the circuit are connected in a star configuration.
Now we can transform the star circuit into a delta circuit with what was learned
in chapter 5 see here!
Then, using the equations mentioned in chapter 5, and naming the impedance between the points a-d of the star configuration as Z'ad, we obtain:
Substituting these values into the circuit in Figure 55-21.2, we obtain the circuit shown in Figure 55-21.3.
Now it is necessary to calculate the parallel between the components that appear in the circuit shown in Figure 55-21.3.
Let us call the parallel of the components between
the points a-b of Zab. This value will be equal to Zad, as the components have the same
values. Calculating, we obtain:
Zab = Zad = 20 ( j 20) / ( 20 + j 20) = 10 + j 10
Zbd = 20 ( -j 20) / ( 20 - j 20) = 10 - j 10
We must pay attention to the fact that the impedances Zad and Zdb are connected in series configuration. And we will call this association
ZS. Thus, adding the two impedances, we obtain:
ZS = 10 + j 10 + 10 - j 10 = 20 Ω
And obviously, ZS is connected in parallel with Zab. Calculating this parallel we will find the value of the equivalent impedance between the points
a-b, which we will call Zeq. Then: