Problem 12-5 Source:
Problem 70 - List of Electrical Circuit Exercises I - School of Engineering - UFRGS - 2011 - Prof. Dr. Valner Brusamarello.
In the circuit show in Figure 12-05.1, it is known that if the key S is in the position 1, the ammeter
A indicates a current of 5 A. If the S key is in position 2, then
the ammeter A indicates a current of 14 A.
Find the values of I e K.
Figure 12-05.1
Solution of the Problem 12-5 -
Nodal Voltage Method
First let's look at the circuit when the S key is in the position1.
To do this, we will redesign the circuit to better understand its operation.
Never forget that an ideal ammeter has an internal resistance equal to ZERO,
so it behaves like a short circuit and there is no voltage drop on it.
See in the Figure 12-05.2 the circuit redesigned at position 1.
Figure 12-05.2
In the position 1 the ammeter reads 5 A . This current passing
through the 12 ohms resistor results in a voltage at the point a
of Va = 60 volts . Knowing the voltage in the point a
we can calculate the currents in all the resistors of the circuit. See in the
figure above, in blue, the values of the currents found by dividing the value
of 60 volts by the value of each resistor. Repair
that by the resistor of 4 ohms circulates a current of
5 A and, as a consequence, you will find V = 20 volts .
Now, using Kirchhoff's law
for the a node, we find:
I + 20 K = 20 + 5 + 20 + 5 = 50 A
That is, the two current sources, together, should provide a total of 50 A .
From the first information provided by the problem, the above equation is obtained.
To use the second information, you must
redraw the circuit as shown in the Figure 12-05.3.
Figure 12-05.3
Let's look in more detail at the part of the circuit highlighted in yellow.
Note that we have two current sources. Let's make an explosion of current sources.
See the new circuit configuration in the Figure 12-05.4.
Figure 12-05.4
Note that we can associate the current sources in parallel. In the
Figure 12-05.5 we can see the final circuit with the currents indicated in blue.
Looking at the circuit highlighted in yellow we see that by using Kirchhoff's
law for the
node b , the current passing through the resistor highlighted in green
is equal to ZERO .
Therefore, we can disregard the circuit highlighted in green.
Thus, the point b is on the same level as the earth.
Figure 12-05.5
After these considerations we can redesign the circuit as shown in the figure below.
In the resistor of 2 ohms, where the current i circulates (by definition in the problem statement), and is in parallel with the current source i, causes this current to be confined to these two components. Thus, the current of 14 A that comes out
of the circuit at point c passes through the other resistor of 2 ohms,
causing a voltage drop of 2 x 14 = 28 volts.
Note that this voltage is the same on the 2 ohms resistor where the
current i. Therefore, we can write:
2 i = 28 ⇒ i = 14 A
Figure 12-05.6
Thus, with the determination of the current value i we realize that by the
resistor
of 1 ohm, which is in parallel with the current source i, at the top of the circuit,
does not pass any current, since the 14 A that enters the
pin 2 will all pass through the current source i, and nothing happens through the
1 ohm resistor. This means that the voltage drop on the
the current source i, which appears at the top right of the circuit, is
V1c = ZERO .
Therefore, we can remove from the circuit the resistor of 1 ohm , as well as
the resistor of
2 ohms and the current source i (which find in parallel), as they do
not affect the operation of the circuit.
We could also have eliminated the current source i , because the potential difference over it is
zero. For reasons of dydadic we prefer to show it in the circuit. Then the
circuit is summarized as shown in the Figure 12-05.7.
Figure 12-05.7
Adding all the currents indicated in the circuit by the blue arrows, arriving at the
node b, we find the value of 14 + 14 + 14 + 3.5 = 45.5 A . In addition, the voltage
V on the resistor of 4 ohms is V = 4 x 3.5 = 14 volts . Thus, the current source
K V has the value of 14 K . Looking at the circuit we easily realize
that who provides these
45.5 A are the two current sources, I and KV . Recalling
that V = 14 , we can write the other equation we need to solve the problem, or:
I + 14 K = 45,5
Repeating here the first equation we have a system of two equations with two unknowns.
I + 20 K = 50
Solving this system we find the values of the variables requested in the problem, that is:
K = 9 / 12 = 3 / 4
I = 35 A
Power Balance
Let's take stock of Power.
See in the Figure 12-05.8, as was the circuit with the indication of currents in the
branches.
Figure 12-05.8
Initially we will calculate the powers dissipated in the resistors,
remembering that, by convention, we consider these powers POSITIVAS .
P3 = 3 x 142 = + 588 W
P3 = 3 x 142 = + 588 W
P12 = 12 x 3.52 = + 147 W
P1 = 1 x 142 = + 196 W
P2 = 2 x 142 = + 392 W
Adding all these values, we find:
P+ = 588 + 588 + 147 + 196 + 392 = + 1,911 W
Now, to calculate the power at the sources, let us recall that voltage sources with the current coming out
of the positive pole are considered NEGATIVE powers, because they are supplying power to the circuit. Otherwise
will be POSITIVE powers. The same goes for current sources.
PS35 = - 35 x 42 = - 1,470 W
PS10.5 = - 10.5 x 42 = - 441 W
The power supplied to the circuit is exclusively from the current sources.
P - = - 1,470 - 441 = - 1,911 W
Finally, we know that the algebraic sum of the powers supplied and dissipated in a circuit,
shall be equal to ZERO , since it must satisfy the Law of Conservation of Energy.
∑ P = P+ + P - = + 1,911 - 1,911 = 0 W
Note that we do not take into account the current source power i = 14 A because the potential difference over it is zero . Therefore, the power supplied or absorbed is also
zero. For the current source i in parallel with the resistor of
2 ohms , we also do not calculate, since the resistor
absorbs all power from the source.
