Problem 11-6 Source:
Problem 13 - List of Electrical Circuit Exercises I - School of Engineering - UFRGS - 2011 - Prof. Dr. Valner Brusamarello.
In the circuit show in Figure 11-06.1, by transformation and
displacement of sources, reduce to a single source associated with a single
resistor.
Figure 11-06.1
Solution of the Problem 11-6 -
Method of Transforming Sources
Initially we will transform the voltage source of 60 volts in series with the
resistor
of 10 ohms in a current source of 60/10 = 6 A in parallel with
the 10 ohm resistor. See the Figure 11-06.2 for the transformation.
Figure 11-06.2
We can see in the figure above that the sources of 2 A and 6 A are in parallel,
(notice the red line, interconnecting the two current sources to the terminal a), as well as the two resistors of 10 ohms.
Therefore, since the two current sources point in the same direction, we can sum
their values by finding an equivalent source of 8 A. And, of course, the two
resistors in parallel result in a single resistor
of 5 ohms. With this information we can redesign the circuit
as shown in the Figure 11-06.3.
Figure 11-06.3
In the figure above it is possible to perform a further transformation with the current source of
8 A and the 5 ohms resistor. Let's turn them into a source of
8 x 5 = 40 volts and the resistor is in series with it. Note that in this way the two 5 ohms resistors will be in series. Therefore, we can sum them, resulting in an equivalent resistance of
10 ohms. See the Figure 11-06.4 how the circuit was transformed.
Figure 11-06.4
Let's continue with the source transformations. This time we will transform the
voltage source of
40 volts and the 10 ohm resistor into a current source
of 40/10 = 4 A and the resistor being in parallel with the current source
and the 15 ohms resistor.
Figure 11-06.5
After this transformation, in the Figure 11-06.5, there are two resistors in parallel: one of
10 ohms and one of 15 ohms. As a result we get an equivalent resistor
of
10 x 15 / (10 + 15) = 6 ohms. Therefore we can transform the current source
of 4 A and this resistor into a voltage source of 4 x 6 = 24 volts
and the resistor is in series. It is obvious that the resistor of 6 ohms
will be in series with the 2 ohms resistor. Therefore, we can add their values
and obtain a single resistor with a value equivalent to 8 ohms. In the
Figure 11-06.6 we see how these transformations were.
Figure 11-06.6
Let's make some considerations about the circuit shown in the figure above. Note that we have two voltage sources in series. In this case, we can replace them with
a single voltage source and its value will be the algebraic sum of the values of each source. Note that the polarities of the sources are opposite. This means that we must subtract its absolute values and the polarity of the resulting source will be the polarity of the source of the highest absolute value. Then, the equivalent source value will be 28 - 24 = 4 volts.
See the Figure 11-06.7 for the circuit.
Figure 11-06.7
We have already been able to reduce to a single source. Finally, we will transform the voltage source of
4 volts in series with the 8 ohms resistor into a current source of 4/8 = 0.5 A and how the two resistors will be in parallel,
we can associate them in a single resistor with a value equal to 4 ohms.
See the Figure 11-06.8 for the final result.
Figure 11-06.8
This final circuit is what we know as Norton Equivalent. If in the problem
statement there is a need to present a Thevenin Equivalent, just make one
more source transformation and arrive at the
circuit shown in Figure 11-06.9.
