Problem 13-4 Source:
Adapted from the 1st test - 2016 - Circuit Analysis I - School of Engineering - UFRGS.
For the circuit shown in the Figure 13-04.1, compute by the laws of Kirchhoff the voltage Vo, the current ig and do
a power balance of the circuit.
Figure 13-04.1
Solution of the Problem 13-4 -
Kirchhoff Method
As the problem asks to do a balance of power we must calculate all the currents of the circuit. Let's put all the currents in function of io.
When we calculate the value of io we will be able to calculate the others
currents. Initially we can write to the nodes a , b and c, the following equations:
node a ⇒ ig = i1 - io
node b ⇒ i3 = i1 + i2
node c ⇒ i2 = 0.75 - i4 - io
Starting from the voltage source of 50 volts , passing through
io and i4, in the clockwise, we can write:
- 50 - 800 io + 200 i4 = 0
From the above equation, we can consider the value of i4 as a function of
io, or:
i4 = 4 io + 0.25
Substituting this value into the equation of the node c , we get:
i2 = 0.5 - 5 io
Therefore, we already have two variables in function of io. Now, making the upper loop of the circuit (the one that has no sources), you can write:
- 800 io + 40 i2 - 80 i1 = 0
As we know the relationship between i2 and io, replacing
in the above equation, we find the relation between i1 and io. So:
i1 = 0.25 - 12.5 io
But, notice that we know the equation of the node a , which relates the value of i1 and ig. So by doing the substitution in the above equation, we get:
ig = 0.25 - 13.5 io
However, we also have the equation of the b node, which relates the currents i1 , i2 and i3. Then, making the necessary substitutions, we get:
i3 = 0.75 - 17.5 io
At this moment we know all the currents of the circuit as a function of io. So,
making the mesh close to the voltage source, we obtain:
80 i1 + 50 i3 = 50
Substituting the values found for i1 and i3 in the above equation,
we can calculate the value of io.
80 (0.25 - 12.5 io ) + 50 (0.75 - 17.5 io ) = 50
Therefore, by performing the calculation, we find the value of io, or:
i0 = 0.004 A = 4 mA
With this value, we can calculate the value of Vo and
ig by mere substitution of values. So:
Vo = 800 i0 = 800 x 0.004 = 3.20 volts
ig = 0.25 - 13.5 x (0.004) = 0.196 A
Doing the same for the other currents, we get:
i1 = 0.2 A
i2 = 0.48 A
i3 = 0.68 A
i4 = 0.266 A
This problem could be solved by a system of equations. The results should be the same as those found here. Now let's take stock of power in the circuit.
Power Balance
First, let us compute the dissipated powers on all resistors, or:
P800 = 800 ( io )2 = 800 x 0.0042 = 0.0128 W
P80 = 80 ( i1 )2 = 80 x 0.22 = 3.2 W
P50 = 50 ( i3 )2 = 50 x 0.682 = 23.12 W
P40 = 40 ( i2 )2 = 40 x 0.482 = 9.216 W
P200 = 200 (i4)2 = 200 x 0.2662 = 14.1512 W
Now, summing all the powers dissipated in the resistances is the value of:
P+ = 49.7 W
Note that in the voltage source of 50 volts , the current goes out through the positive pole,
that is, it is supplying power to the circuit. Therefore, its value is negative.
P50 = - 50 ig = - 50 x 0.196 = - 9.80 W
We must calculate what is the potential difference in the current source. It is known that it is the same difference
of potential over 200 ohm resistance , or:
VS = 200 x 0.266 = 53.2 volts
We can see that the voltage on the current source is positive , and
this means that it is supplying power to the circuit. Therefore, its value is negative .
PS = - 0.75 x 53.2 = - 39.9 W
Now, we can add algebraically the powers provided by the sources. Like this:
P - = - 9.8 - 39.9 = - 49.7 W
Finally, we know that the algebraic sum of the powers provided and dissipated
in a circuit must be equal to ZERO , that is:
∑ P = P+ + P - = 49.7 - 49.7 = 0 W
Note that the Power Balance satisfies the
Principle of Energy Conservation .
