Problem + Hard 32-1    Source:   Adapted from Question 3 - 1st test - Circuit Analysis I - UFRGS - 2016.

Let the quadripole "A" be defined by the following equations:

Determine the values ​​of V₁, V₂, I₁, I₂ and I_F in the circuit shown in the figure below.

Solution of the Problem + Hard 32-1

In order to solve this problem, it is necessary to point out five important points that must never be forgotten.

1) Whenever possible we should simplify the circuit.

2) Resistors in parallel with voltage sources can be eliminated from the circuit.

3) Resistors in series with current sources can be eliminated from the circuit.

4) Current sources in parallel with voltage sources can be eliminated from the circuit.

5) Voltage sources in series with current sources can be eliminated from the circuit.

With that in mind, the circuit was redesigned to draw attention to some details. Appointed some points for the better circuit understanding. See the redrawn circuit in the figure above. notice that the link highlighted by the green line, connecting the points a, c, d, h, is the same, that is, it is a short circuit between these points. In this way, it is concluded that the resistorss of 1, 20 and 30 ohms are in parallel with the sources of voltage. Then, you can exclude them from the circuit (item 2, above). Furthermore, the 10 ohms resistor interconnects the points a-e. But point a and point d are the same. Therefore, this resistor is in parallel with the voltage source of 6 volts and will be removed from the circuit. On the other hand, the 40 ohms resistor is in series with a current source. Therefore, it can also be excluded from the circuit.

Looking more closely at the circuit, the branch formed by the resistors of 50 and 60 ohms and the current source of 1 A, is being short-circuited by the connection between the points c, d, h. In this way, this entire circuit can be eliminated. Also notice that the font I₂ is connected between points b and d, that is, in parallel with the voltage source of 22 volts. Applying the item 4 above, it follows that this current source can be eliminated from the circuit. And finally, after all these simplifications, it is noticed that the two voltage sources, are in series with the 6 A current source. Applying item 5 above, you can eliminate the two voltage sources. Thus, a very simplified circuit was obtained, as can be seen in the figure below.

From the simplified circuit, using nodal analysis, the following relations result:

On the other hand, we have that I_F is given by:

Note that in this problem there are 5 variables. Therefore, 5 equations are needed to solve the system. The last three equations plus the two quadripole equations total five equations. now doing substitution can be reduced to two equations with two unknowns. The unknowns V₁ and V₂. Solving step by step, we replace the equation   ⓶   in   ⓷.   After some algebraic manipulation and isolating I_F we have:

Now, equating the equations  ⓹   and   ⓺,   results in the equation ⓻,  below.

Working with equations  ⓵   and   ⓸,   results in the equation ⓼,  below.

Equating the equations  ⓹   and   ⓼,   results in the equation ⓽,  below.

This procedure is one of the possible ones. Didactically, we opted for this one. Therefore, solving this system formed by the equations  ⓻   and   ⓽,  we get:

With these values ​​in hand, the values ​​of the currents in the circuit are calculated, finding: