DeCarlo Ch10b Solutions

# Linear Circuit Analysis: The Time Domain and Phasor Approach

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1 PROBLEM SOLUTIONS CHAPTER 10 S OLUTION 10.50. The input voltage phasor is = 2000 rad/s and V S = 20 0 ° V. Now, do a source transformation on the phasor circuit: where I S = V S Z L = 20 0 ° j 2000 × 10 × 10 - 3 = 1 ∠- 90 ° = - j A Now, Y eq = 1 20 + 1 j 20 + j 2000 × 50 × 10 - 6 = 0.0707 45 ° S and V C = I S Y eq = 1 ∠- 90 ° 0.0707 45 ° = 14.14 ∠ - 135 ° V S OLUTION 10.51. Use superposition. First, find response to current source using circuit below: V x_1 = I s1 Z RC = I s 1 3 × ( - j 3) 3 - j 3 = 2 0 °× 2.121 ∠- 45 ° = 4.242 ∠- 45 ° Now, find the response due to the voltage source using the following circuit:

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2 The voltage across the inductor is the same as the input source, and this voltage divides between the series combination of capacitor and resistor: V x_2 = j 3 3 - j 3 V s2 = j 3 3 - j 3 3 90 o = 2.121 ∠ - 135 o V Combining the two contributions implies that: V x = 4.242 ∠ - 45 o + 2.121 ∠- 135 o = 4.74 ∠ - 71.6 o V S OLUTION 10.52. (a) As stated, V L = a V s 1 + b I s 2 . To find a, set I s 2 = 0 and use voltage division: V L =- j 30 j 30 + j 30 V s1 = - 0.5 V s1 = - a V s1 To find b, set V s 1 = 0 and use parallel impedance and Ohm's law: V L = j 30/ / j 30 ( 29 I s2 = j 15 I s2 = b I s2 Hence V L = –0.5 V s 1 +j15 I s 2 (b) For this part, V s1 = 10 and I s2 = 0.5 ∠ - 90 o . Hence from the formula, V L =- 0.5 V s 1 + j 15 I s2 = - 0.5 × 10 + j 15 × 0.5( - j ) =- 5 + 7.5 = 2.5 V
3 Therefore v L (t) = 2.5 cos(100 π t) V. S OLUTION 10.53. In this problem, we can make use of the linearity property for phasors. Specifically, from the given information, we can write V 1 I in = 20 45 ° 10 0 ° = 2 45 ° = a and V 1 V in = 5 90 ° 10 45 ° = 0.5 45 ° = b Hence, V 1 = a I in + b V in Substituting the new values of input current and voltage, we obtain: V 1 = 10 0 ° + 10 45 ° = 18.48 22.5 ° V S OLUTION 10.54. (a) For V out to be zero, we want R C R C - j C = j L R L + j L R C R L = L C (b) Substituting R C C = 2 s and R L = 3 into the above expression gives: L = R C CR L = 6 H. (c) The bridge circuit can be represented by an impedance Z bridge (j ϖ ). The voltage that appears across the bridge, say V bridge , is obtained by voltage division. Hence, by the voltage substitution theorem, the problem may be solved as in part (a) with this new source voltage V bridge appearing across Z bridge (j ϖ ). S OLUTION 10.55 The input phasor is: = 1000 rad/s and I in = 2 45 ° A, assuming peak value. First compute Y th ( j 10 3 ) = 0.25 - j 1000 × 4 × 10 - 3 + j 1000 × 0.25 × 10 - 3 = 0.25S Z th ( j 10 3 ) = 4 Then, V oc = I in × 4 = 8 45 ° V

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4 in which case v oc ( t ) = 8cos(10 3 t + 45 o ) V. The final equivalent is a voltage source (having value V oc ) in series with a resistance of 4 . S OLUTION 10.56. (a) First note that the frequency is given in Hz, so, = 1281.77rad/s and I in = 10 0 ° A. Then, in MATLAB, »R = 0.25; L = 1.17e-3; C = 520e-6; »w = 2*pi*204; »Yin = j*w*C +1/(R + j*w*L) Yin = 1.0815e-01 + 1.7737e-02i »Zin = 1/Yin Zin = 9.0039e+00 - 1.4766e+00i »abs(Zin) ans = 9.1242e+00 »angle(Zin)*180/pi ans = -9.3134e+00 Therefore Z th = 9.1242 ∠- 9.313 ° . Finally, V oc = Z th I in = 91.2 ∠- 9.313 ° V. (b) Now, the circuit looks like the following: Simple voltage division can yield: V L = Z L Z L + Z TH V oc = 46.219 0 o v L t ( 29 = 46.2cos 1281.77 t ( 29 V
5 S OLUTION 10.57 . For this problem we short the V-source and compute Z th and use voltage division to find V oc . Specifically Z th = 1 j ϖ C + 1 0.1 + j ϖ L = 10 3 - j 10 and V oc = 1 j

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