Problem 7.2   Source:   Problem elaborated by the author of the site.

Knowing that the power dissipated by resistor of 12 ohms is of 12 watts in the circuit shown in Figure 07-02.1, calculate the possible values ​​of the voltage source V and the currents of the circuit.

Solution of the Problem 7.2

Let's recall the equation that allows us to calculate the electric current through a resistance, knowing the value of the power dissipated by it.

However, note the fact that the power dissipated in a resistor does not take into account the direction of the electric current. In our case, if the current flows from the point e₁ to the ground or, if it flows from the ground to e₁, 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 of the point e₁ to ground

Knowing i₃ we can calculate the voltage e₁ because the resistors of 6 and 12 ohms are in series, totaling a equivalent resistance of 18 ohms. Therefore, the value of e₁ is:

Knowing e₁ we calculate i₂. So:

Knowing i₂ and i₃ is possible calculate i₁. Then:

Now that we know i₁ we can compute V making the mesh equation for the first branch, or:

Substituting the known numerical values and solving the equation, we find the value of V, or:

Current of the ground to point e₁

In this case, we must change the signal of the i₃.

Repeating the calculations of the previous item for the new value of i₃ we find the value of e₁.

Knowing e₁ we can to calculate i₂. So:

Knowing i₂ and i₃ we can calculate i₁, or:

Now that we know i₁ we can compute V making the mesh equation for the first branch, or:

Substituting the known numerical values and solving the equation, we find the value of V, or:

Therefore, stay tuned for problems that involve powers dissipated in resistances, because there will be two solutions to it. One for positive values of the electric current and another for negative values.