Resolved Problems about Electrical Power

Problem 7.4 Source: Problem elaborated by the author of the site.

Repeat the previous problem by calculating the eletric power dissipated in the potentiometer P₁.

Solution of the Problem 7.4

In this case, the same simplifications made for the previous problem are worth (problem 7-3). Therefore, let's start by analyzing the circuit as shown in the Figure 07-04.2.

Since we want to calculate the power dissipated in P₁, we can assume that it is a load for the voltage source in series with R₁. As the statement of the problem asks to draw a graph of power in P₁, then we will calculate several power values. Therefore, we should give several values to P₁. Notice that the value of R₁ is constant and assumes the role of internal resistance of the voltage source.

For each value of P₁, we calculate the electric current I, indicated by the blue arrow in the circuit and thus we will have to calculate the power dissipated in P₁. Initially we will do P₁ = 0. Then the current will be the voltage of 50 volts divided by R₁, or:

I = V / R₁= 50 / 20 = 2.5 A

With value of I we can calculate the power dissipated in P₁, or:

P_P1 = P₁ I² = 0 2.5² = 0 watt

Doing similarly for other values of P₁ we can assemble the Table 07-04.

Table 07-04

Valor de P₁	Corrente Elétrica (I)	Potência Elétrica (P_P1)
0 ohm	2.50 A	0 W
5 ohms	2.00 A	20 W
10 ohms	1.67 A	27.89 W
20 ohms	1.25 A	31.25 W
25 ohms	1.11 A	30.80 W
30 ohms	1.00 A	30.00 W
40 ohms	0.83 A	27.56 W
50 ohms	0.71 A	25.20 W

Notice the highlighting of the table row where P₁ assumes the value 20 ohms. This condition allows P₁ to dissipate the maximum power. This is in accordance with the maximum power transfer theorem , which states that for maximum power dissipated in the load this should have the same value of the internal resistance of the source. In this case represented by R₁. To emphasize this statement more strongly, we resolve make four graphs. In addition to varying the value of P₁, we will provide four different values for R₁. The values chosen for R₁ were: 10 , 15 , 20 and 30 ohms. In the Figure 07-04.3 we see the graph, and we easily realize that the maximum power dissipated in P₁ in any case, occurs exactly when P₁ has the same value as R₁.

To draw this graph we use the relation I = V / (R₁ + P₁). On the other hand, we know that P_P1 = P₁. I ². So, doing the substitution:

In this equation we replace P₁ with x. Thus, we can enter this equation in any graphical program such as gnuplot, geogebra, etc ... and we will get the graph above. Of course we should replace the values of V and R₁ by the problem. Note that for values of P₁ much larger than the R₁ all the curves converge to a final value. This suggests that there is a convergence zone when the above condition is satisfied.

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