DeCarlo Ch5 Solutions

Linear Circuit Analysis: The Time Domain and Phasor Approach

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PROBLEM SOLUTIONS CHAPTER 5. Solution 5.1. (a) V s = 10 V, P = 20 W and P = V s × I s implies I s = 2 A. (b) R in = V s /I s = 10/2 = 5 (c) By the linearity/proportionality property: V s new V s old = I s new I s old which implies 2 10 = I s new 2 implies I s new = 0.4 A. (d) P new = V s new × I s new = 2 × 0.4 = 0.8 watts. Observe that P new P old = 0.8 20 V s new V s old = 2 10 It follows that the proportionality property does not hold for power calculations. Solution 5.2 First note that the ratio I R /V S is constant. With the given values of voltage and current, this ratio is: I R /V S = 0.25/25 = 0.01 Power dissipated in the resistor is P = I R 2 R = 2.5 L I R 2 = 2.5/R = 0.25 L I R = 0.5 Since I R is always 0.01 × V S , it follows that V S = 50V. Solution 5.3 Label the resistances R 1 , R 2 , and so on in the manner shown in Example 5.11. In this problem, we have R 1 to R 10 (the last being the 2 Ohm resistance at the voltage source). First, assume that V 1 (the voltage across R 1 ) is 1V. Then evaluate the rest of the currents and voltages until you deduce the
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resulting V S . It should be noted that the equivalent resistance looking into R 3 , R 5 , R 7 , and R 9 is always 2 . V 1 = 1 I 1 = V 1 4 = 0.25 I 2 = 0.25 V 2 = I 2 × 2 = 0.5 V V 3 = V 1 + V 2 = 1.5 L I 3 = V 3 3 = 0.5 L I 4 = I 3 + I 2 = 0.75 L V 4 = I 4 × 4 = 3 V 5 = V 3 + V 4 = 4.5 L I 5 = V 5 3 = 1.5 L I 6 = I 5 + I 4 = 2.25 L V 6 = I 5 × 4 = 9 V 7 = V 6 + V 5 = 13.5 L I 7 = V 7 3 = 4.5 L I 8 = I 7 + I 6 = 6.75 L V 8 = I 8 × 4 = 27 V 9 = V 8 + V 7 = 40.5 L I 9 = V 9 3 = 13.5 L I 10 = I 9 + I 8 = 20.25 L V 10 = I 10 × 2 = 40.5 V S = V 9 + V 10 = 40.5 + 40.5 = 81 V Thus, an 81 V input produces a 1 V output L V out = (1/81) × V S = 2 V. Solution 5.4 Label the resistances R 1 to R 10 progressively from right to left just like in the previous problem. Then, assume I out = 1 and proceed as follows: I out = 1 L V 1 = I out × 4 = 4 L I 2 = V 1 4 = 1 I 3 = I 1 + I 2 = 2 L V 3 = I 3 × 4 = 8 L V 4 = V 3 + V 1 = 12 L I 4 = V 4 3 = 4 I 5 = I 4 + I 3 = 6 L V 5 = I 5 × 4 = 24 L V 6 = V 5 + V 4 = 36 L I 6 = V 6 3 = 12 I 7 = I 6 + I 5 = 18 L V 7 = I 7 × 4 = 72 L V 8 = V 7 + V 6 = 108 L I 8 = V 8 3 = 36 I 9 = I 8 + I 7 = 54 L V 9 = I 9 × 4 = 216 L V 10 = V 9 + V 8 = 324 L I 10 = V 10 3 = 108
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I S = I 10 + I 9 = 162 L I out /I S = 1/162 L I out = (1/162) × 40.5 = 0.25A. Solution 5.5 (a) MATLAB code given in problem. (b) Subsitute to obtain V 1 = 10V. (c) R eq = V S /I S = 11.6667 . (d) First, define r1 = 1:0.25:10; then create an outermost loop around the code of part (a) as: for j=1:length(r1) then, in the statement defining R, do R = [R1(j), R2, R3, R4, R5, R6, R7, R8]’; Finally, replace the last statement with Vs(j) = V(n) + V(n-1); end; The following is the resulting plot: Solution 5.6 (a) The following code can be used: n = 9;
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v = zeros(n,1); i = zeros(n,1); r = [r1 r2 r3 r4 r5 r6 r7 r8 r9]’; i(1) = 1; v(1) = i(1)*r(1); i(2) = i(1); for k=2:2:n-2 v(k) = r(k)*i(k); v(k+1) = v(k)+v(k-1); i(k+1) = v(k+1)/r(k+1); i(k+2) = i(k+1) + i(k); end; v(8) = i(8)*r(8); v(9) = v(8) + v(7); i(9) = v(9)/r(9); Is = i(9) + i(8); It follows that I s = 16.9877A. (b) By the proportionality property: I 1 new I 1 old = I s new I s old I 1 new = 1 16.9877 × 200mA = 11.77mA (c) R eq = v(9)/Is = 38.15 . Solution 5.7 V a = 12V, i B = 60m By inspection: V out_a = 300/900 × 12 = 4V V out_b = (300||600) × 60m = 12
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_ V out = 4 + 12 = 16V.
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