Problem + Hard 15-1    Source:    Problem 65 - List of Problems Electrical Circuits I of the School of Engineering - UFRGS - 2011 - Prof. doctor Valner Brusamarello.

In the circuit of the Figure 15-01.1, when the switch "S" is in position 1 the ammeter reads +2A. Determine the current flowing through the 4 ohm resistor when the switch moves to position 2.

Solution of the Problem + Hard 15-1

When the switch "S" is in position 1, points a and b are connected by an ammeter. It's already gone studied that an ideal ammeter has NULL internal resistance. Therefore, it can be stated that the points a and b are in short circuit. This situation can be seen in the Figure 15-01.2. In between the points a and b circulate the current measured by the ammeter, of 2 A. This current, allows you to calculate the voltage V_r. Using Ohm's Law, you find 4 volts. So, knowing V_r, the voltage dependent source, 4 V_r, becomes a voltage source of 12 volts.

Note that the value of e₁ is calculated using Ohm's law, that is, e₁ = 2 x 3 = 6 volts.

And to calculate the current flowing through the 1 ohm resistance, and through the voltage source of 12 volts, which will be called I₁, just do the mesh equation in the sense counterclockwise, as shown by the orange arrow in the figure above. And so:

Solving the equation, calculate I₁, or:

Now, we must find the value of i, making the node equation for point e₂, or:

Solving this equation, the value of i is calculated, or:

As we know e₁ and i, we can calculate e₂, or:

With these values ​​in hand, the value of B can be calculated by writing the equation:

Replacing with numerical values:

We need to calculate the Thevenin equivalent of the circuit seen from points a-b. To do so, remove the ammeter and place a current source of, say 1 ampere, between points a-b. We also remind you that you must eliminate all independent sources, leaving only the dependent sources. notice that by eliminating the current source of 10 A, the three resistors that were in parallel, by a single one of 2 ohms, in which a current of 1.5 i circulates. Furthermore, as the current source has been eliminated, an open circuit appears and therefore no appears in the circuit. The Figure 15-01.3 illustrates this situation.

The basic idea here is to find the value of V_ab. This will allow calculate the value of R_th. Initially, it is necessary to calculate the value of i for this new situation. Write the mesh equation for voltage (Kirchhoff's law) for the indicated circuit by the orange arrow.

Solving the equation, the value of i is:

Knowing the value of i it is easy to calculate V_ab. Enough loop for voltage (left side of loop) clockwise, or:

Solving the equation, the value of V_ab is:

Now it's easy to determine the value of R_th, because:

Once the value of R_th is calculated, the key is returned S to position 1. As stated earlier, the ammeter short circuits the points a-b. This means that the ammeter is reading the short-circuit current between points a-b. Drawing the circuit of the Thevenin equivalent and using this information, you can calculate the value of V_th.

See that with the support of the circuit in the Figure 15-01.4, it is easy to determine the value of V_th, as it is enough to apply Ohm's law to the circuit. Note that the current called I_cc = 2 A is nothing more than the current measured by the ammeter when switch S is in position 1. So:

Now that the Thevenin equivalent of the entire circuit to the right of the points a-b, the key S is returned to position 2 and thus, inserts 4 ohms resistor across terminals a-b. Then, it is easy to calculate the current flowing through this resistor.

The Figure 15-01.5 shows the final circuit with all the values ​​found. A very simple circuit is obtained, consisting of a voltage source and two resistors in series. Now, applying Ohm's law, once again, calculate the current that crosses the 4 ohms resistor, or: