In Fig. 9.93, a fixed load of 16 Ω is applied to a 48 V supply with an internal resistance of 36 Ω. a. For the conditions in Fig. 9.93, what is the power delivered to the load and lost to the internal resistance of the supply? b. If the designer has some control over the internal resistance level

Question

In Fig. 9.93, a fixed load of 16 Ω is applied to a 48 V supply with an internal resistance of 36 Ω.

a. For the conditions in Fig. 9.93, what is the power delivered to the load and lost to the internal resistance of the supply?
b. If the designer has some control over the internal resistance level of the supply, what value should he or she make it for maximum power to the load? What is the maximum power to the load? How does it compare to the level obtained in part (a)?
c. Without making a single calculation, find the value that would result in more power to the load if the designer could change the internal resistance to 22 Ω or 8.2 Ω. Verify your conclusion by calculating the power to the load for each value.

Accepted Answer

a. [latex]I_L=\frac{E}{R_s+R_L}=\frac{47V}{36 Ω + 16 Ω}=\frac{48V}{52 Ω}=923.1 mA\[0.5cm] P_{R_s}=I^{2}_{L}R_s=(923.1 mA)^2(36 Ω) = 30.68 W\[0.5cm] P_L= I^{2}_{L}R_L=(923.1 mA)^2(16 Ω) = 13.63 W [/latex]

b. Be careful here. The quick response is to make the source resistance [latex]R_s[/latex] equal to the load resistance to satisfy the criteria of the maximum power transfer theorem. However, this is a totally different type of problem from what was examined earlier in this section. If the load is fixed, the smaller the source resistance [latex]R_s[/latex], the more applied voltage will reach the load and the less will be lost in the internal series resistor. In fact, the source resistance should be made as small as possible. If zero ohms were possible for [latex]R_s[/latex], the voltage across the load would be the full supply voltage, and the power delivered to the load would equal

[latex]P_L=\frac{V^{2}_{L}}{R_L} =\frac{(48V)^2}{16 Ω}=144 W [/latex]

which is more than 10 times the value with a source resistance of 36 Ω.

c. Again, forget the impact in Fig. 9.86: The smaller the source resistance, the greater is the power to the fixed 16 Ω load. Therefore, the 8.2 Ω resistance level results in a higher power transfer to the load than the 22 Ω resistor.

For [latex] R_s = 8.2 Ω[/latex]

[latex]I_L=\frac{E}{R_s+R_L}=\frac{48V}{8.2 Ω + 16 Ω} =\frac{48V}{24.2 Ω}=1.983 A [/latex]

and [latex]P_L = I^{2}_{L}R_L = (1.983 A)^2 (16 Ω) ≅ 62.92 W[/latex]

For [latex]R_s = 22 Ω[/latex]

[latex]I_L=\frac{E}{R_s+R_L}=\frac{48V}{22 Ω + 16 Ω} =\frac{48V}{38 Ω}=1.263 A [/latex]

and [latex]P_L = I^{2}_{L} R_L = (1.263 A)^2 (16 Ω) ≅ 25.52 W[/latex]

Question 9.17: In Fig. 9.93, a fixed load of 16 Ω is applied to a 48 V supp...

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