By R. A. DeCarlo and P. Lin
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69. C20 = 50 Ah (a) In eq. 2A (b) Calculate the capacity for n=10 and T=10, this yields 42 Ah. 70. 278W . (b) Following is the graph, and the script used to generate it. *" which means that the division %is performed for each value of RL. *IL); xlabel('Resistance in Ohms'); ylabel('Current in mA'); %The use of subplot lets you subdivide the graphing %window in two halfs. 71. 72. 6e3; © R. A. DeCarlo, P. M. Lin Chap 2 Probs P2 - 17 © R. A. DeCarlo, P. M. 73. 73 R1=20; R2=40; R3=60; R4=30; R5=10; R6=135; R7=150; R8=300; R9=130; R10=200; R11=50; Ga=1/R10+1/R11; Ra=1/Ga; Rb=Ra+R9+(1/(1/R7+1/R8)); c b a R4 R6 R7 R9 Chap 2 Probs P2 - 18 © R.
7566W 20 20 (d) In this part, we take the above matrix equation and solve it for each value of Gs. t. changes in Gs. The following plot is the voltage difference between the two nodes as a function of Gs, and hence as a function of temperature. As can be seen, in this figure, the voltage difference between B and C does not change linearly with Rs. Since this resistance itself changes linearly with temperature, this means that VB-VC does not change proportionally to temperature. 6. 33W. 12 We are required to write the equations in matrix form.
A. DeCarlo, P. M. Lin Chap 2 Probs P2 - 17 © R. A. DeCarlo, P. M. 73. 73 R1=20; R2=40; R3=60; R4=30; R5=10; R6=135; R7=150; R8=300; R9=130; R10=200; R11=50; Ga=1/R10+1/R11; Ra=1/Ga; Rb=Ra+R9+(1/(1/R7+1/R8)); c b a R4 R6 R7 R9 Chap 2 Probs P2 - 18 © R. A. DeCarlo, P. M. 74. 8mA PROBLEM SOLUTIONS CHAPTER 3. 1. 4 (a) It is evident from the figure that Vc = 20. We need to write two equations in Va and Vb and put them in matrix form. In this case, we can write the matrix equation by inspection. Note that the resistors are identified by conductance values.