A dc generator, battery, and laboratory supply are connected to resistive load RL in Fig. 9.91. a. For each, determine the value of RL for maximum power transfer to RL. b. Under maximum power conditions, what are the current level and the power to the load for each configuration? c. What is the

Question

A dc generator, battery, and laboratory supply are connected to resistive load [latex]R_L[/latex] in Fig. 9.91.
a. For each, determine the value of [latex]R_L[/latex] for maximum power transfer to [latex]R_L[/latex].
b. Under maximum power conditions, what are the current level and the power to the load for each configuration?
c. What is the efficiency of operation for each supply in part (b)?
d. If a load of 1 kΩ were applied to the laboratory supply, what would the power delivered to the load be? Compare your answer to the level of part (b). What is the level of efficiency?
e. For each supply, determine the value of [latex]R_L[/latex] for 75% efficiency.

Accepted Answer

For the 12 V car battery,

[latex]R_L = R_{Th} = R_{int} = 0.05 \Omega [/latex]

For the dc laboratory supply,

[latex]R_L = R_{Th} = R_{int} = 20\Omega [/latex]

b. For the dc generator,

[latex]P_{L_{max}}=\frac{E^{2}_{Th}}{4R_{Th}}=\frac{E^2}{4R_{int}}=\frac{(120V)^2}{4(2.5 Ω)}=1.44 kW [/latex]

For the 12 V car battery,

[latex]P_{L_{max}}=\frac{E^{2}_{Th}}{4R_{Th}}=\frac{E^2}{4R_{int}}=\frac{(12V)^2}{4(0.05 Ω)}=720 W [/latex]

For the dc laboratory supply,

[latex]P_{L_{max}}=\frac{E^{2}_{Th}}{4R_{Th}}=\frac{E^2}{4R_{int}}=\frac{(40V)^2}{4(20 Ω)}=20 W [/latex]

c. They are all operating under a 50% efficiency level because [latex]R_L = R_{Th}[/latex].

d. The power to the load is determined as follows:

[latex]I_L=\frac{E}{R_{int}+R_L}=\frac{40V}{20 Ω + 1000 Ω}=\frac{40V}{1020 Ω}=39.22 mA[/latex]

and [latex]P_L = I^{2}_{L} R_L = (39.22 mA)^2 (1000 Ω) = 1.54 W[/latex]

The power level is significantly less than the 20 W achieved in part (b). The efficiency level is

[latex]\eta \% =\frac{P_L}{P_s}\times 100\%=\frac{1.54 W}{EI_s}\times 100\% =\frac{1.54W}{(40 V)(39.22 mA)}\times 100\%=\frac{1.54W}{1.57W}\times 100\%=98.09 \% [/latex]

which is markedly higher than achieved under maximum power conditions—albeit at the expense of the power level.

e. For the dc generator,

[latex]\eta =\frac{P_o}{P_s}=\frac{R_L}{R_{Th}+R_L} [/latex] (η in decimal form)

and [latex]\eta =\frac{R_L}{R_{Th}+R_L} \[0.5cm] \eta (R_{Th} + R_L) = R_L \[0.5cm] \eta R_{Th}+\eta R_L=R_L\[0.5cm] R_L(1-\eta )=\eta R_{Th} [/latex]

and [latex]R_L=\frac{\eta R_{Th}}{1-\eta } [/latex] (9.7)

[latex]R_L=\frac{0.75(2.5 Ω)}{1 - 0.75 } =7.5 \Omega[/latex]

For the battery,

[latex]R_L=\frac{0.75(0.05 Ω)}{1 - 0.75 } =0.15 \Omega[/latex]

For the laboratory supply,

[latex]R_L=\frac{0.75(20 Ω)}{1 - 0.75 } =60 \Omega[/latex]

Question 9.15: A dc generator, battery, and laboratory supply are connected...

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