Prof. Antonio Gabaldón. ETS de Ing. Industrial de Cartagena (Spain)
2007
Chapter 2. Problem 2-7) By Thevenin theorem, and some formal calculation, find the
electrical equivalent at terminals A and B of the network shown in the figure. Obtain the power
demanded by 6Ω resistor. Also, evaluate the power in the 3A current source (on the left). Please,
specify if this source delivers or consumes energy. Notice: all the sources are DC and the
network is in steady state.
Solution:
a) Thevenin/Norton equivalent:
First, and to simplify the calculus of Thevenin/Norton equivalent, it is necessary to
remember some interesting properties of ideal sources (see theory of circuit analysis methods):
- An ideal voltage source in parallel with a generalized branch (real source and or
impedances,...) is equivalent to the same source. Remember that in this kind of branch the
voltage is the ideal source voltage, and the current is undefined (i.e. again an ideal voltage
source).
- An ideal current source in series with a generalized branch (real source and or
impedances,...) is equivalent to the same current source. Remember that in this kind of branch
the current (series) is the ideal source curent, and the voltage is undefined (i.e. again an ideal
current source).
Through the aplication of these ideas,and after we remove the “observer circuit”, we can
reduce the circuit at the terminal E-F (a generalized branch, a dipole. Hint!: apply Thevenin
equivalent again). Notice an ideal source of 6A is in series between E-F dipole and the real
source (7V, 2Ω), see figure 2.7.1
The open circuit voltage Uo (UAB) is done by KCL:
.
U0 = 7 + 2*6 = 19 V
Prof. Antonio Gabaldón. ETS de Ing. Industrial de Cartagena (Spain)
2007
Figure 2.7.1. Equivalent
network (nodes E-F)
The input impedance in the dipole A-B can be found easily. We open current source and
shortcircuit voltage source. So, only 2Ω resistor appear in the network. Our thevenin equivalent
is (19V, 2Ω)
b) Power
Using this equivalent and in steady state (inductor is a short circuit, capacitor is an open
circuit), the absorbed power in the resitor is given by:
P6Ω ' R I 2 ' 6
19
2%6
2
' 33.84W
For the 3A ideal current source, we look into the terminal defined between 5A ideal
current source (C and D) amd 2Ω resistor. The dipole C-D (active circuit) in in series with this
current source, then the Norton equivalent is the 5A ideal current source.
Appliying KVL at node G, and branch equation of (10V, 2Ω) real source:
U3A = 2*8 + 10 = 26 V;
P3A = U3A*3A = -78W (generated)
Notice the power is delivered by source because voltage and current have the same
reference direction (from G to the upper corner) and the result is negative.