This method of analyzing electrical circuits proves to be extremely useful when we have a circuit with some complexity, and through transforming sources we can simplify it to the point of making it easy to understand, and thus find your solution. We present, in the Figure 11-01 the equivalence we can have between the two circuits shown.

What the circuit shows us is that whenever we have a source of voltage in SERIES with a resistor, we can transform it into a current source in PARALLEL with the resistor. We must maintain the resistor value. The value of the current source is given by the division between the value of the voltage source and the value resistor.

The most interesting thing is that this process is REVERSIBLE. In other words: if we have a current source in PARALLEL with a resistor, we can transform it into a voltage source in SERIES with the resistor. We should also keep the resistor value. The value of the voltage source will be given by the product of the current source value and the resistor value. Stay tuned for the fact that in these transformations, the resistor value NEVER changes.

However, DO NOT FORGET if the resistor is in parallel with voltage source we can not make this transformation. In fact, what we must do is eliminate it from the circuit, because its absence does not change at all the solution of the problem. In the same way we must act if the resistor is in series with a current source. We can not do the above transformation, but we can eliminate it from the circuit, as it will not change the solution of the problem.

Let's look at how to solve a circuit like the one shown in the Figure 11-02.

As previously stated, the resistor of 8 ohms, which is in parallel with the voltage source of 12 volts, can be removed from the circuit without prejudice to the solution of the problem. Furthermore, note that in the circuit (figure above) we have two voltage sources in series with resistors. Therefore, it is perfectly possible to perform source transformation.

The voltage source of 12 volts will be transformed into a current source of 12/6 = 2 A in parallel with the resistor of 6 ohms . In turn, the voltage source of 36 volts will be transformed into a current source of 36/12 = 3 A in parallel with the resistor of 12 ohms. Note the direction of the arrow of current sources. The 2 A source points down because the positive pole of the 12 volts voltage source points downwards. Even reasoning applies to the other source of current. See the Figure 11-03 below, as was the new configuration of the circuit.

In the circuit shown in the Figure 11-03 we notice that we have two current sources with opposite directions. Therefore, we can transform them into a single source of current of value equal to the subtraction of their values, that is, 3 - 2 = 1 A. The current source arrow points to the one with the highest numeric value.

On the other hand, to the left of the points a-b we have two resistors in parallel. As a result, we obtain a single resistor with a value equal to 4 ohms. See the Figure 11-04 how the circuit stayed after the transformations.

See how we simplify the circuit making your solution easy. To calculate the current i we can apply a resistive current divider. But because the two resistors have equal values, of course it will circulate half the current of the source of 1 A for each resistor. Soon:

And by applying the Ohm's law we can calculate the voltage V between the points a-b. Like this:

From what has been shown above it is noticed that this method of resolution is very fast to find the solution of the problem.