ProbSolv_Chapter10

ProbSolv_Chapter10 - CHAPTER 10 SINUSOIDAL STEADY-STATE...

Info iconThis preview shows pages 1–21. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 10
Background image of page 11

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 12
Background image of page 13

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 14
Background image of page 15

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 16
Background image of page 17

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 18
Background image of page 19

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 20
Background image of page 21
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CHAPTER 10 - SINUSOIDAL STEADY-STATE ANALYSIS List of topics for this chapter : Nodal Analysis Mesh Analysis Superposition Theorem Source Transformation Thevenin and Norton Equivalent Circuits AC Op Amp Circuits NODAL ANALYSIS Problem 10.1 Given the circuit in Figure 10.1 and i(t) = 5 Sin(1000t) amps, find v0 (t) using nodal analysis. 109 109 Figure 10.1 > Carefully DEFINE the problem. Each component is labeled, indicating the value and polarity. The problem is clear. > PRESENT everything you know about the problem. The goal of the problem is to find V() (t) , which is clearly labeled in Figure 10.1, using nodal analysis. Thus, we need to label the nodes and ground. To find Vo (t) without using derivatives and integrals, we must transform the circuit to the frequency domain. This allows 'us to find the answer using algebra with complex numbers. We can transform the circuit to the frequency domain after setting a reference value. Let us use a reference of A Sin(1000t + 4)) . ' In transforming to the frequency domain, remember that X L = jcoL and X C =1/(jcoC) . Hence, the inductor becomes jcoL = j(103 )(10 x 103) = j10 and the capacitor becomes . l/jcoC = l/[j(103)(50x10‘6 )] = ~j20. 167 Let us draw the circuit after the transformation into the frequency domain and labeling the nodes and ground. ' 10 Q V2 10 Q > Establish a set of ALTERNATIVE solutions and determine the one that promises the greatest likelihood of success. The problem clearly states that the problem be solved using nodal analysis. Thus, the technique to solve the problem is set. There is no reason to look at an alternative at this point. > ATTEMPT a problem solution. Let us begin by writing the node equations. At node 1 : At node 2 : Vl-O Vl—V2 Vz—Vl V2—0 V2—0 + =0 + + = 30 10 10 -j20 10+j10 0 —5+ Simplifying, V2‘V1+flz_+V2(1*J-)=0 v 3v —3V =150 1+ 1 2 10 20 20 ‘4V1’3V2 =150 (2)(V2_V1)+jV2+V2(l"j):O -2V1+3V2 = 0 Thus, the system of simultaneous equations is l4 -3livll_F150l [-2 allvzflol Fvll 1 [3 3i iV2l=(12—6)i2 4i [—_l O l_—_l ll (3va— l—‘—I W A O U] 0 O l_____l ll F—'—l U‘I \] l O L11 ;_1 So, V1 = 7540° or Vl (t) = 75 sin(1000t) volts V2 = 50100 or v2(t) = SOsin(1000t) volts Clearly, and I - ——“— — ——— — —— = (2.5)(1 — = Z - 45°) amps 168 Hence, . V0 : j10 10 = (10290°)(2.5)(J§ 2 —45°) = 2542" 245° = 35.36245o volts Therefore, Vo (t) = 35.36 sin(1000t + 45°) volts > EVALUATE the solution and check for accuracy. Solving the problem with an alternate method, such as mesh analysis in this case, would show that the results of the problem solution are correct. Let us draw the circuit defining the loop currents for mesh analysis. 109 109 Write the loop equations. Loop 1 : I1 = 540° amps Loop2: 30(I2 —Il)+1012—j20(12——I3)=0 . Loop3: —j20(I3—12)+IOI3+j10I3:0 Simplifying the equations for loops 2 and 3, Loop2: —30I1+(40—j20)I2 +j2013 = 0 Loop3: j2012 +(lO—Ij10)I3 = 0 Solve the third loop equation for I3 in terms of 12. __ - j20 ”‘ 10 — j 10 13 I2 = (1—j)12 =1.41422-45°12 Now, substitute I1 and I3 into the second loop equation and simplify to find 12. (—30)(540°) + (40 —- j20) I2 + (j20)(l — j) I2 = 0 (-30)(540°) + (40 —- j20)I2 + (20 + j20) I2 = 0 6012 = (30)(540°) _ (30)(5£0°) I2 60 = 2.510° amps Now, find 13. I3 = (1.41424 —45°) I2 = (1.41424 -45°)(2.540°) = 3.5361 - 45° amps 169 Clearly, I0 :13 = 3.5364—45° amps. Evidently, V0 =j101o = (10490°)(3.536z - 45°) : 35.36445o volts and Von) = 35.36sin(1000t + 45°) volts This answer is the same as the answer obtained using nodal analysis. Our check for accuracy was successful. ' > Has the problem been solved SATISFACTORILY? If so, present the solution; if not, then return to “ALTERNATIVE solutions” and continue through the process again. This problem has been solved satisfactorily. vo (t) = 35.36sin(1000t + 45°) volts Problem 10.2 [10. 5] Given the circuit in Figure 10.2 and vi (t) = 1000800000 volts, use nodal analysis to find vo (t) . 20S) Figure 10.2 ' Let us start by building the ac circuit. The voltage source, vi (t) = 1000500000 volts, becomes 1010°volts, with 0) = 1000. The inductive reactance becomes jooL = j10 . The capacitive reactance becomes 1/ jcoC = 1/[j(1000)(50 X10'6)] = -j20 . Now, draw the ac circuit. 209 V1 —j20§2 V2 j10§2 309 1040° V O 170 At node 1 : At node 2 : . 10-V1 V1+V1—V2 Vl—V2 4V1 V2 =~— . . =—-—-+ . 20 20 120 . 120 20 30+110 10—V1=V1+j(V1—V2) (-4+j)V1=(0.6+j0.8)V2 . . 0.6+‘O.8 10=(2+j)V1—J2+JV2 Vl=—4:j—V2 Note that I0 = V1 / 20 was substituted when writing the equation for node 2. Substituting the equation for node 2 into the equation for node 1, 2+' 0.6+0.8‘ 1OZ< 1x J)V_ 'V _4+j 2 J 2 V _ 170 - or ' 2 ‘ 0.6— 3262 Clearly, 30 3 170 :——"V=_" —=.14,olt V° 30+le 2 [3+jl[0.6—j26.2] 6 5 47026 v0 5 With a reference of A cos(1000t + (1)) , v0 (t) = 6.154 cos(1000t + 70.26°) volts Problem 10.3 Given the circuit in Figure 10.3 and V(t) = 20 cos(1000t) volts, find i0 (t) using nodal analysis. 6 Q 250 HF 10 Q io(t) V(t) a 8 Q E 10 mH Figure 10.3 io (t) = 632.5 cos(1000t - 18.44°) milliamps 171 MESH ANALYSIS Problem 10.4 Given the circuit in Figure 10.4 and i(t) = 2005(1000t) amps, find io (t) using mesh analysis. 10 Q io(t) i(t) a 20 mH Figure 10.4 H" First, we transform the circuit to the frequency domain using a reference of A cos(1000t + and define the mesh currents as seen in the following circuit. Remember that XL = jcoL and XC = — ij ' 10 Q 210° A 0 E j20 Q There is only one unknown loop current, Io. Writing the loop equation, 10(I0 —— 2) +101o +j201o = O (20 + j20)Io = 20 20 1 110° I - -——'— ~ — — m = 0.70711-45° amps ° ‘20+j20‘1+j‘,/§445o Therefore, io (t) = 707.1 cos(1000t —- 45°) milliamps 172 . Problem 10.5 Given the circuit in Figure 10.5 and i(t) = 5 sin(1000t) amps, find io (t) using mesh analysis. 109 109 io(t) 10 mH Figure 10.5 Transform the circuit to the frequency domain using a reference of A sin(1000t + (1)) and define the mesh currents. 109 100 Clearly, IO :12. Use mesh analysis to find I1 and 12. Forloopl: 30(11——5)+1011—j20(11—IZ)=0 Forloop2: -j20(IZ—Il)+1012+j1012=0 Simplifying, (40— j20)I1 + j2OI2 = 150 j2011+(10—j10)12 =0 Simplifying further, (2—j)11+jI2 = 7.5 j211+(l—j)I2 = 0 'Find II in terms of I2 for the second loop equation. -l+j l ,1 1 , .11=( j2 )12 =[§+JE]12 =[3)(1+J)12 . . Substituting this equation into the first loop equation we get, 173 (2 ')[—-—1+j)1+'1 75 ‘J 2 J 2 : ~ - 2 . (2—j)(1+j)12+j212=15 (3+j3)12 =15 15 5 540° Hence, I0 = 3.5364 — 45° amps Therefore, .i0 (t) = 3.536 sin(1000t — 45°) amps Wm. SUPERPOSITION THEOREM The superposition theorem applies to ac circuits the same as it does for do circuits. This theorem is important if the circuit has sources operating at different frequencies. Since the impedances depend on frequency, we must have a different frequency-domain circuit for each. source. The total response is obtained by adding the individual responses in the time domain. Problem 10.6 Given the circuit in Figure 10.6, i(t) = 200$(1000t) amps and v(t) = lOsin(4000/ 3t) volts, find iC (t) . 200 Figure 10.6 Because the two sources have different frequencies, we need to use superposition to solve this problem. Thus, iC (t) = iCl (t) + icz (t). Start with the current source and a reference of A cos(1000t + ¢) . 240° A 20 Q 174 Using current division, 20 2 2 1+' IC1 =( )(2) = 1———,=(—)(—J—)= 1+j = @249: 1.4142245o amps Hence, iCI (t) = 1.4142 COS(1 OOOt + 45°) amps Next, use the voltage source with a reference of A sin(4000 / 3t + (1)) . 20 Q Icz —j15 Q 1040° V Clearly, ' 2 2 4+ '3 2 5436.87° IC2 = = = 956$: = lag—“J = 0.4L36.87° amps Hence, iC2 (t) = O.4sin(4000/3t + 36.87°) amps Recall that iC (t) = iCl (t) + iC2 (t) . Therefore, iC (t) = [1.4142 c0s(1000t) + 0.4 sin(4000/ 3t + 36.87°)] amps Problem 10.7 [10.33] Given the circuit in Figure 10.7 and vi (t) = l2cos(3t) volts, solve for V() (t) using the principle of superposition. 10V Figure 10.7 175 Let V0 : Vl + V2 + V3 where V1, V2 , and V3 are respectively the voltages produced by the 10 volt dc source, the ac current source, and the ac voltage source acting independently. For V1, . consider the circuit shown in Figure (a). 69 2H The capacitor is an open circuit to do while the inductor is a short circuit. Hence, V1 =10 volts and V1 (t) = 10 volts. For V2 , consider the circuit in Figure (b). to = 2 , so the inductor becomes jcoL = j4. - Likewise, the capacitor becomes 1/ ij = - j / (2 / 12) = 1'6. 69 Applying nodal analysis, 4 YA+£+XL£1+A .1}, ‘6 -j6 j4‘6 J6 J4 2 which leads to = 21.45.426.56o volts o (4)02) 2 - J V2: With a reference of A Sin(2t + , V2 (t) = 21.45 sin(2t + 26.56°) volts 176 For V3, consider the circuit in Figure (0). 0) = 3 which leads to ij = j6 for the inductor and 1/ jcoC = - (3/12) = —j4 for the capacitor. At the non-reference node, 12 —- V3 V3 + V3 6 - j4 j6 which leads to (2x12) _ o 1 V3 = . — 107331-2656 volts 2+] With a reference of A cos(3t + (1)) , v3 (t) = 10.733 cos(3t — 26.56°) volts Recall that vo (t) = Vl (t) + v2 (t) + v3 (t). Therefore, V() (t) = [10 + 21.45 sin(2t + 25.56°) + 10.733 cos(3t ~ 25.56°)] volts Problem 10.8 Given the circuit in Figure 10.8 and vs (t) = cos(l 0000 volts, find v(t) . 20 Q 50 uF 20 mH O vs(t) Figure 10.8 "H v(t) = 10 + 20 c0s(1000t — 90°) volts or v(t) = 10 + 20 sin(1000t) volts 177 SOURCE TRANSFORMATION Source transformation in the frequency domain involves transforming a voltage source in series with an impedance to a current source in parallel with an impedance or vice versa. We must keep the following relationship in mind when performing source transformations. Vs vS = 2515 (=9 IS = 2-— S Problem 10.9 Given the circuit in Figure 10.9 and V(t) = 20 cos(1000t) volts, find v0 (t) using source transformations. 6 Q 250 HF 10 Q 8 Q 10 mH v(t) Figure 10.9 Transform the circuit to the frequency domain using a reference of Acos(1000t + . 6Q —j4Q 109 2040° V O 8 Q Reduce the circuit using source transformations. Begin with the 2040°V source in series with 69. which becomes a 10/340°A source in parallel with 69. j10 r2 —j4 Q 10 Q 10/340° A 8 Q j10 (2 NH 178 Now combine 69 || 89 = 24/79. The 10/310°A source in parallel with 24/79 becomes an 80/740°V source in series with 24/79. 24/7 9 —j4 9 10 9 80/740° V 0 To perform the next source transformation, we need to find the parallel equivalent of the resistor and capacitor in series. We know that two series impedances and two parallel impedances are l l 1 = Z + Z d = —— + — 51 52 an Z ZPl ZP2 I Peq j10 o "H Z Seq To find the parallel equivalence of two series impedances, let ZSeq = ZPeq . Then, __1___ ;+_1_ ZSI +ZS2 ZPl ZPZ It can be shown that 1 1 1 24/7— j4 ‘ 8.095 + — j6.939 Thus, the 80/740°V source in series with 24/7 —j49 becomes a 2.169449.4°A source in parallel with 8.095 ~j6.9399. 2.169449.4°A j10 9 Combining 8.0959 || 209 = 5.7639. We now need to find the series equivalent of the resistor in parallel with the capacitor. It can be shown that 1 1 1 5.763 + ~j6.939 z 3.411—j2.833 Thus, the 2.169.449.40A source in parallel with 5.763 — j6.9399 becomes a 9.61849.69°V source in series with 3.411 ~j2.8339. Before redrawing the circuit, let’s combine the series resistances of3.4119 +109 = 13.4119. 179 13.411 Q —j2.833 Q . 9.61849.69°V Now that we have found a less complex circuit via source transformations, it is simple to solve for v0 (t) . 9.61819.69° 9.61849.69° 9.61849.69° I — W w — -— = 0.63254 -18.43° " 13.411e-j2.833+j10 =13.411+j7.167 _15.206428.12° and V0 =j101= (10190°)(0.63251 -18.43°) = 6.325471.57° Recall that we used a reference of A cos(1000t + (1)) . Therefore, v°‘(t) = 6.325 cos(1000t + 71.57 °) volts THEVENIN AND NORTON EQUIVALENT CIRCUITS Problem 10.10 Given the circuit in Figure 10.10 and V(t) = 100 cos(1000t) volts, find ' V111 , IN , and Zeq looking into terminals a and b. Figure 10.10 Because the circuit in Figure 10.10 has a capacitor, we can only determine VTh , IN , and Zeq for an ac circuit in the frequency domain. Hence, the circuit becomes, 180 10040° V 3 Begin by finding the open-circuit voltage, V0c . Note that there is no current flowing through the capacitor. Thus, Voc = V1. Using nodal analysis, V,—100 Vl—O 5 + 5 ~3I=O 100—V1 whereI= . 5 So, Vl—IOO Vl—O [100~V1) 5 + 5 “(3) 5 = . (VI—100)+V1—(3)(100—V1)=0 5V1=400 V1=80volts Hence, VCC = 8OZO° volts Now, find the short—circuit current, Isc . 100£0° V C Use nodal analysis to find V1 and then 181 Writing the nodal equation, V1—100 Vl—O Vl—O + — — 31+ . = 0 5 5 - _] 100 — V1 where I = —--— . 5 So, V1 «100 V1~0 [ICC—VI] . 5 + 5 ~(3) 5 +JV1 — 0 (V1 —— 100) +Vl — (3)(100—V1)+ j5V1 = O (5 + j5) VI = 400 400 ’ 80 (80)(1 — j) V : -———-— = ————. : ———— : —— t 1 5+j5 1+] 2 (40)“ J) Thus, V 40 1— ' ISO = = = (40)(1+ j) = 4045449 amps Finally, °q ’ Isc — 40J§z45° _ 0 ms Therefore, VT11 = V0c = 8010° or VT11 (t) = 80 c0s(1000t) volts IN = Isc = 40fiz45°_ or iN (t) = 56.57 c0s(1000t + 45°) amps zeq = Jig-45° or zeq = (l—j)ohms Problem 10.11 [10. 43] Find the Thevenin and Norton equivalent circuits for the circuit shown in Figure 10.11. 601120° V 5 Q —-j10 Q 2 o j20 Q Figure 10.11 182 To find ZTh , consider the circuit shown in Figure (a). SS) —j10§2 2E2 j20 Q «— (a) zeq = 1'20” (5 —j10)+2 = (16—j12)+2 =18—j12 = 21.634—33.69° ohms To obtain VTh , consider the circuit in Figure (b). 5 Q —j10 s2 2 Q + 604120° V j20 o VDC (1)) 1'20 j4 = —————— 604120° = 420° V°° 5—j10+j20 1+j260 Voc =(l.7889£26.57°)(601120°)=107.334146.57° volts 07.334146.57° = V—‘m— — —1—-————— = 49614180260 amps “I where ISC is the current flowing downward through a short across the terminals. Vm=V oc’ Recall that ZTh = ZN = Z and IN = Isc. eq’ Therefore, zTh = 2N = (18— j12) ohms VTh = 107.334146.57° volts IN = 49614180260 amps 183 Problem 10.12 Given the circuit in Figure 10.12 and V(t) = 100 cos(1000t) volts, find i(t) , the current through R, for R = 09, IQ, 109, 1009, and 10009. 10 mH SQ 109 Figure 10.12 R I i(t) r 0 £2 56574 - 45° A 5.657 c0s(1000t — 45°) A 1 9 5.4391 - 45°A 5.439 cos(1000t - 45°) A 10 (2 4.0411 - 45° A 4.041 cos(1000t - 45°) A 100 Q 1.13144 - 45° A 1.1314 c0s(1000t - 45°) A 1000 Q 0.1381 - 45°A 138 cos(1000t - 45°) mA AC OP AMP CIRCUITS Given the ac circuit in Figure 10.13, find Vout as a function of Vin . Problem 10.13 —j1000 Q Figure 10.13 184 Using nodal analysis at node a, Va - Vin Va — Vout -————+—-————~=0 103 - le3 where Va = Vb = 0. So, - Vin + — Vout __ 103 -fl03 — - jVout : Vin V. Vow = "T = jVin ' J Therefore, Vout : Vin 490° The output is equal to the input except for a phase shift of 90°. Problem 10.14 Given the circuit in Figure 10.14 and Vin (t) = lOsin((ot) volts, find vout (t) for a) = l, 10, 100, 1000 rad/sec. lOuF Figure 10.14 185 Use the following ac circuit, with a reference of A Sin(0)t + , to find V in terms of CD. out —j105/(o Q 1040° Using nodal analysis at node a, Va ~ Va — Vout Va — Vout _ .——‘+ . —0 104 + 1105/0) 3100) where Va = Vb = 0. So, '10 'Vout -Vout —— + =0 104+-j105/03 lem . le4 -100—_]0)V0m+ a) Vow—0 104 [— j0)+--O‘)“]Vout = 100 100 i 100 100 V ——————~—=L= w 490° Now, substitute the values for co into the equation for Vow. 100 At (D :1 rad/sec, ‘70ut E _104 490° :: -90° vent (t) = 10 sin(t ~ 90°) mV 03 103 At a) = 10 rad/sec, Vout 5 W490 : - 9900 v0“t (t) = 101.01 sin(10t — 90°) mV 186 490°: 0.101014 -90° 1 10 At a) = 100 rad/sec, Vout E m190° = ‘0—490° This corresponds. to the case where the LC combination forms a parallel resonant circuit and the output goes to infinity. 05 105 1 At 0) = 1000 rad/sec, Vout E Wig)" = WZ90° = (1101014900 vow (t) = 101.01 sin(1000t + 90°) mV In conclusion, the output, Vout (t) , has a ~90° phase shift for all values of 00 less than 100 and has a 90° phase shift for values of co greater than 100. 187 ...
View Full Document

This note was uploaded on 09/06/2010 for the course EE 50145 taught by Professor Chi during the Spring '10 term at UCSF.

Page1 / 21

ProbSolv_Chapter10 - CHAPTER 10 SINUSOIDAL STEADY-STATE...

This preview shows document pages 1 - 21. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online