DeCarlo Ch3 solutions

Linear Circuit Analysis: The Time Domain and Phasor Approach

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PROBLEM SOLUTIONS CHAPTER 3. Solution 3.1. Select the bottom node as the reference node, and write a node equation at the positive terminal of the V 1 resistor: V 1 - V 0 3 R + V 1 6 R + V 1 - 4 V 0 6 R = 0 2 V 1 - 2 V 0 + V 1 + V 1 - 4 V 0 = 0 4 V 1 = 6 V 0 V 1 = 1.5 V 0 Solution 3.2 Write a node equation at the top node: 0.6 - V x 100 - 2 V x 100 - V x 50 = 0 ⇒ - V x - 2 V x - 2 V x = - 60 ⇒ - 5 V x = - 60 V x = 12 V Solution 3.3 0.6 - V x 100 + 25 V x 100 - V x 50 - V x - 0.2 V x 40 = 0 ⇒ - 3 V x 100 + 25 V x 100 - 8 V x 400 = - 0.6 V x = - 0.6 × 400 80 V x = - 3 V Solution 3.4 (a) It is evident from the figure that V c = 20. We need to write two equations in V a and V b 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. 15 m - 5 m - 5 m 35 m V a V b = 0.5 0.5
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(b) Solve the matrix equation by inverting the left-most matrix: V a V b = 1 525 - 25 35 m 5 m 5 m 15 m 0.5 0.5 = 1 0.5 35 5 5 15 0.5 0.5 = 40 20 (c) V x = V ab = V a - V b = 20 V , V da = -V a = -40, V db = -V b = -20. (d) P i = 0.5 × 40 = 20W, P v = 20 × (20-20) = 0. P diss = 40 × 40 × 10m + 20 × 20 × 5m + 20 × 20 × 5m = 20 W. Power delivered equals dissipated power. Solution 3.6 Write two nodal equations: V s 1 - V 1 3000 = I s 3 + V 1 - V 2 6000 V 2 30000 = V s 2 - V 2 12000 + V 1 - V 2 6000 Rewrite equations as: 2 V s 1 - 2 V 1 = 6000 I s 3 + V 1 - V 2 2 V 2 = 5 V s 2 - 5 V 2 + 10 V 1 - 10 V 2 Cast into a matrix equation - 3 1 - 10 17 V 1 V 2 = 6000 I s 3 - 2 V s 1 5 V s 2 Solving the matrix equation yields: V 1 V 2 = 181.46 124.39 Power absorbed by the 6k resistor is (V 1 -V 2 ) 2 /R = 0.5429 W . Similarly, P s1 = (V s1 -V 1 )/3000 × V s1 = 4.7W, P s 2 = ( V s 2 - V 2 ) /12000 × V s 2 = - 0.32 W P s 3 = I s 3 × ( V s 2 - V 1 ) = - 1.21 W Solution 3.7 (a) Again, the matrix equation can be written by inspection:
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G 1 + G 2 + G 4 - G 4 - G 4 G 3 + G 4 + G s V B V C = 50 G 1 50 G 3 (b) Substituting the values of conductances and inverting the above matrix equation yields: V B = 34.0132V V C = 33.6842V (c) Power delivered is 80.7566 W . Using the Principle of Conservation of Power: P del = P 1 + P 2 + P 3 + P 4 + P 5 or, P del = 50 × V A - V B 20 + V A - V C 20 = 80.7566 W (d) In this part, we take the above matrix equation and solve it for each value of G s . If we do this, we can get a feel for the behavior of V B and V C w.r.t. changes in G s . The following plot is the voltage difference between the two nodes as a function of G s , 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 R s . Since this resistance itself changes linearly with temperature, this means that V B -V C does not change proportionally to temperature.
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