Problem 15-18    Source:    Adapted from Problem 4.87 - page 153 -    NILSSON - RIEDEL -    Book: Electric Circuits - 10th edition - Ed. Pearson Education of Brasil - 2016.

In the circuit shown in Figure 15-18.1, the resistor R_L is adjusted until it absorbs the maximum power of the circuit. Determine:

a ) The value of R_L:

b ) The maximum power dissipated by R_L:

c ) Make a power balance on the circuit.

Solution of the Problem 15-18

Item a

To find the value of R_L that dissipates the maximum power, we must calculate the equivalent resistance of the entire circuit that R_L "see". In other words, we must calculate the value of the Thévenin resistance. To do this, we will remove R_L from the circuit and in its place we introduce a voltage source. We choose a convenient voltage, such as 11 V. Furthermore, to facilitate the calculation, we performed a source transformation with the controlled voltage source 3 i and the resistor R₃. See the modifications to the circuit as shown in Figure 15-18.2.

We can find a relationship between V (voltage over R₁) and the current i, because over R₁ circulates the current i + 0.1 V. Like this:

Making the external loop passing through the voltage source of 11 V, R₁ and R₂, we obtain the value of the current i, or:

Now we can find the value of I by applying the Kirchhoff current law to the node a.

So, dividing the voltage of 11 V by the current I we find the value of R_th.

As the maximum power transfer theorem states that for there to be maximum power transfer of power to the load, this must have the same value as the Thévenin resistance. Therefore, we conclude that R_L  =  R_th. So, answering item a), we have:

Item b

To calculate the power over R_L we must find the Thévenin voltage. To do this, we must remove R_L of the circuit. The Thévenin voltage will be the tension that appears between points a - b. Thus, we obtain Figure 15-18.3.

In this case, the equation eq. 15-18-1 still applies, repeated below for greater clarity.

We can easily calculate the value of I in the circuit in Figure 15-18.3, as we see that R₃ circulates through the resistor the current:

And since V_th = - 8 I, then we have:

With the value of V_th = - 11 i and making the outer mesh in a counterclockwise direction, starting with V_th, it will be possible to calculate the value of i, that is:

Therefore, the value of V_th = - 11 i  is:

Therefore, the power dissipated by the load R_L is:

Carrying out the calculation, we find:

Item c

To answer item c we will base ourselves on the circuit shown in Figure 15-18.4.

Note that the voltage at point a we already know, as it is the Thévenin voltage calculated in item b. We also know that V = - 6 i. Soon, making the outer mesh in a counterclockwise direction and starting from point b, we will find the value of i when we include R_L to the circuit.

With this value we can calculate the voltage V = - 6 i = 45 V. The voltage across the resistor R₂ is V_R2 = 5 x 7.5 = 37.5 V. Soon, V₁ = V_a + V_R2 = 27.5 + 37.5 = 65 V. All these calculations allow us to make a power balance in the circuit.

As these powers are dissipated by resistors, these powers are considered positive. Now let's calculate the powers absorbed or supplied by the sources that are part of the circuit.

All calculated powers are negative meaning that all sources are supplying power to the circuit. If all the calculations are correct, then the algebraic sum of the powers involved in the circuit must be equal to zero. Let's calculate the sum of positive and negative powers.

Adding the values found above, we have:

Therefore, we have proven that the calculations are correct.