Problem + Hard 7-1.    Source:  Prob. 1 - 1st Test - 1975 - School of Engineering - UFRGS - Electric Circuits

What is the value of I so that R₁ = 2 Ω dissipates a power 12.5 watts?

Solution of the Problem + Hard 7.1

Note that this problem features a 4 ohm resistor in parallel with the voltage source of 10 volts. Then it can be removed from the circuit without any prejudice to the solution of the problem. In the figure below, you can see how the circuit was without it.

It is also possible to simplify the circuit, transforming the voltage source of 10 volts, in series with the 2 ohms resistor, in a source of current in parallel with that resistor.

The current source will have the value of 10 / 2 = 5 A. And the 2 ohm resistor will be in parallel with the other 2 ohm resistor, resulting in a resistor equivalent of 1 ohm.

See in the figure to the side how the circuit was with all the transformations carried out. By the statement of the problem the information given is that the resistor R₁ dissipates a power of 12.5 watts.

Notice that the branch containing R₁ was named i₁ the current going from point a to point b and the current that passes through the 1 ohm resistor, i₂.

As we have the power dissipated in R₁, it is easily calculated i₁, using the equation below:

In this problem there is a very important point to emphasize, because the power dissipated in a resistor does not taking into account the direction of electric current. In this case, if the current flows from point a to point b or, from b for a, the dissipated power will be the same, provided that the electric current is also the same. In this way, we will calculate the values ​​of the currents and voltages in the circuit analyzing the two situations.

Current flows from point a to b

Based on the figure above, the positive value of the current i₁ going from point a to point b. Calculating the value of this current and substituting the numerical values ​​in the equation above:

Now that you know the value of i₁, you can calculate the value of V_ab and after, the value of i₂.

So the value of i₂ is:

Now, applying the law of knots to the knot a, we find the value of I, or:

Substituting for numerical values ​​and solving the equation:

Current flows from point b to a

Let's analyze the circuit assuming that the value of i₁ is negative. In this case, the meaning of i₁ is shown in the figure on the side, that is, going from point b to point a.

To solve the problem, apply the same steps as in the previous item, that is, the value of V_ab is calculated. Like this:

So the value of i₂ will be:

Now, applying the law of nodes to the node a, we find the value of I, or:

Substituting for numerical values ​​and solving the equation:

Conclusion

Notice that when we changed the sign of the value of i₁, it changed completely the value of the current source I, including its direction. It should be noted that in the first case, the two current sources provided power to the circuit. In the second case, only the 15 A source supplies power to the circuit, while the 5 A source absorbs potency.