Problem 11-4    Source:   Problem elaborated by the author of the site.

In the circuit show in Figure 11-04.1, calculate:

a) The current i_x and i_y.

b) The power dissipated on 6 ohms resistors.

Solution of the problem using Superposition Method   click here!

Solution of the Problem 11-4 -  Method of Transforming Sources

Item a

Recalling that by the method of transforming sources, any voltage source that is in series with one or more resistors, may be replaced by a current source in parallel with a resistor whose value is the sum of the value (s) of the resistor (s) which is found in series with the voltage source. What's more, the value of the current source is the quotient between the value of the voltage source and the value of the voltage source. resistance in series.

Thus, applying this principle to the yellow-highlighted circuit results in a current source of 12 A in parallel with the 3 and 6 ohms.

For the circuit highlighted in green we have two current sources in parallel with resistors. In this case, the same principle applies. We can turn them into a voltage source in series with a resistance. If there is more than one resistor in parallel with the current source, the value of the series resistors is the equivalent resistance value of this parallel. In our case, since we only have one resistor, then its value is unchanged. Thus, after these transformations we can see the circuit in the Figure 11-04.2.

Attention should be paid to two points in the circuit above. On the left a 2 ohms resistor appears in parallel with the current source of 12 A. The value of this resistor is the result of the parallel between the resistors of 3 and 6 ohms, this result, due to the transformation performed in the circuit highlighted in yellow of the previous figure. And the value of the current source of 12 A is the result of the division of the voltage source of 36 volts by the resistor value of 3 ohms. On the right side of the circuit there is a resistor of value 10 ohms result of the series of two resistors of 5 ohms each. It should point out that the two voltage sources are in series with opposite polarities. Therefore we can replace them with a single voltage source with the value equal to the difference between them, that is, 20 volts. The positive pole of the source points to the side of the positive pole of the highest value source, in this case, that of 60 volts. See the Figure 11-04.3 for the circuit.

Note on the left side of the circuit the transformation performed with the current source and the two resistors that were connected to the point a. Multiplying the 12 A current source with the 2 ohms resistor was performed by finding a voltage source of 24 volts. The 2 ohms resistor is in series with the 8 ohms resistor. Thus this series association results in a resistor of value equal to 10 ohms, as is explained in the circuit of the Figure 11-04.4. Notice that the current i_x circulates by this resistance, as this current goes towards the point b. Then again we have voltage sources in series with resistors. Again, using the principle reported above, these two voltage sources can be transformed in series with the resistors into two current sources and two resistors in parallel.

See the figure above for the modified circuit. From the transformation resulted in two sources of current: one of 2 A and another of 2.4 A, which together resulted in a single source of 4.4 A. And the two 10 ohms resistors that stayed in parallel, result one of 5 ohms. Notice that no change in the 6 ohms resistor was made since we want to calculate the current i_y that flows through it. So we just use a current divider and we can calculate i_y, or:

Now knowing the value of i_y we can calculate the value of V_b, ou:

How we know V_b then you can calculate i_x subtracting this voltage from the 24 volt source. The result is divided by value of the 10 ohms resistor, or:

Item b

Now knowing i_y we can calculate the power at 6 ohms resistor by which this current circulates, that is:

To calculate the power at the other resistor of 6 ohms notice that we know the value of i_x = 1.2 A and V_b = 12 volts. Thus, one can calculate the value of V_a:

Then the electric current flowing through the resistor of 6 ohms connected to the point a, is:

Therefore the power dissipated at 6 ohms connected to the point a, is: