Problem 11-13    Source:    Problem created by the site's authors.

In the circuit shown in Figure 11-13.1 and using source transformation, determine the value of the current I.

Solution of the Problem 11-13 -  Method of Transforming Sources

Before solving the problem, it is possible to see that we can make some simplifications in the circuit. Since the resistor R₅ is in series with the current source, it can be eliminated from the circuit. Furthermore, R₄ and R₆ are in series. Therefore, they can be replaced by a single resistor with a value equal to 12 Ω. Then, the circuit can be redesigned as shown in Figure 11-13.2.

From the circuit above we see that point c is 12 V above point b. So, let's separate point c into two: point c1 and point c2. Now we introduce the voltage sources in such a way that the voltages at the points considered do not change. See Figure 11-13.3 how the circuit looked after the modifications.

Note the two voltage sources in series with opposite polarities. In this case, there is zero tension between the two points b, as shown in the figure above. So, is it possible replace the two voltage sources with a short circuit, as we can see in Figure 11-13.4

As we have a voltage source in series with the resistor R₃, it is clear that we can do one more source transformation and obtain the circuit shown in Figure 11-13.5. So we are left with a current source of 12/6 = 2 A and two resistors in parallel that can be replaced by a single resistor with a value equal to 4 Ω.

This way, we can do another source transformation and obtain the circuit shown in Figure 11-13.6.

Now, we must pay attention to the fact that we obtain a current source in series with a voltage source and a resistor. As we know, in this case, the current source prevails and we can eliminate the voltage source and the resistor R_P and obtain the circuit shown in Figure 11-13.7.

In the circuit above, a current source of 6 A appears, which is the result of the following sequence: we transform the voltage source of 12 V in series with the resistor R₂ in a current source of 12/6 = 2 A. This source is in parallel with the 4 A source. So, adding the two current sources we obtain a single source current equal to 6 A. And, naturally, the two resistors are in parallel. Now applying a current divider to the circuit we easily find the value of I, or:

Carrying out the calculation, we obtain:

So, after some source transformations we found the value of I with just one "operation". Otherwise we would have to calculate a system of three equations in three unknowns much more difficult.