Chapter 10 Part 1-Electrical Circuit Analysis-Problem Solutions, Exercises of Electrical Circuit Analysis

Download Chapter 10 Part 1-Electrical Circuit Analysis-Problem Solutions and more Exercises Electrical Circuit Analysis in PDF only on Docsity! February 5, 2006 CHAPTER 10 P.P.10.1 2,010)t2sin(10 =ω°∠⎯→⎯ 4jLjH2 =ω⎯→⎯ 5.2j- Cj 1 F2.0 = ω ⎯→⎯ Hence, the circuit in the frequency domain is as shown below. At node 1, j2.5-2 10 211 VVV − += 21 4j)4j5(100 VV −+= (1) At node 2, 4 3 j2.5-4j 2x212 VVVVV −+ − = where 1x VV = )3(5.2)(4jj2.5- 21212 VVVVV −+−= 21 )5.1j5.2()4j5.7(-0 VV +++= (2) Put (1) and (2) in matrix form. ⎥⎦ ⎤ ⎢⎣ ⎡ =⎥⎦ ⎤ ⎢⎣ ⎡ ⎥⎦ ⎤ ⎢⎣ ⎡ ++ + 0 100 5.1j5.2j4)(7.5- j4-4j5 2 1 V V where °∠=−=+−++=Δ 29.05-74.255.12j5.22j4))-j4)(-(7.5()15.j5.2)(4j5( ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ++ + =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 0 100 5.12j5.22 4j5j47.5 j45.1j5.2 2 1 V V °∠= °∠ °∠ = − + = 01.6032.11)100( 29.05-74.25 96.30915.2 )100( 5.12j5.22 5.1j5.2 1V °∠= °∠ °∠ = − + = 12.5702.33)100( 29.05-74.25 07.285.8 )100( 5.12j5.22 4j5.7 2V -j2.5 Ω + Vx − 4 Ω j4 Ω10∠0° A + − 3Vx2 Ω V1 V2 docsity.com In the time domain, =)t(v1 11.32 sin(2t + 60.01°) V =)t(v2 33.02 sin(2t + 57.12°) V P.P.10.2 The only non-reference node is a supernode. 2j-4j4 15 2211 VVVV ++= − 2211 24jj-15 VVVV ++=− 21 )4j2()j1(15 VV ++−= (1) The supernode gives the constraint of (2) °∠+= 602021 VV Substituting (2) into (1) gives 2)3j3()6020)(j1(15 V++°∠−= °∠= °∠ °∠ = + °∠−− = 7.165376.3 45243.4 72.210327.14 3j3 )6020)(j1(15 2V )32.17j10()8327.0j272.3-(602021 +++=°∠+= VV 154.18j728.61 +=V Therefore, =1V 19.36∠69.67° V, =2V 3.376∠165.7° V P.P.10.3 Consider the circuit below. For mesh 1, 04j)4j2j8( 21 =−+− II (1) 21 4j)2j8( II =+ 6 Ω j4 Ω8 Ω -j2 Ω I1 I2 + − 10∠30° V I3 2∠0° A docsity.com For consider the circuit in Fig. (b). "oI Let , Ω−= 2j81Z Ω+=+ == 769.2j846.1 4j6 24j4j||62Z 53.0j4164.0 77.0j846.9 )769.2j846.1)(2()2( 21 2" o +=+ + = + = ZZ ZI Therefore, 086.1j4961.0"o ' oo +=+= III =oI 1.1939∠65.45° A P.P.10.6 Let , where is due to the voltage source and is due to the current source. For , we remove the current source. " o ' oo vvv += 'ov "ov ' ov 5,030)t5sin(30 =ω°∠⎯→⎯ j- )2.0)(5(j 1 Cj 1 F2.0 == ω ⎯→⎯ 5j)1)(5(jLjH1 ==ω⎯→⎯ The circuit in the frequency domain is shown in Fig. (a). 6 Ω j4 Ω8 Ω -j2 Ω 2∠0° A Io" (b) + Vo' − + − 30∠0° V 8 Ω -j Ω j5 Ω (a) docsity.com Note that -j1.25j5||j- = By voltage division, °∠= − = 81.12-631.4)30( 25.1j8 j1.25-' oV Thus, )12.81t5sin(631.4v'o °−= For , we remove the voltage source. "ov 10,02)t10cos(2 =ω°∠⎯→⎯ 5.0j- )2.0)(10(j 1 Cj 1 F2.0 == ω ⎯→⎯ 10j)1)(10(jLjH1 ==ω⎯→⎯ The corresponding circuit in the frequency domain is shown in Fig (b). Let , 5.0j-1 =Z 9.3j878.410j8 80j 10j||82 +=+ ==Z By current division, )2( 21 2 ZZ Z I + = j3.44.878 j3.9)(4.877j- -j0.5))(2(-j0.5)( 21 2" o + + = + == ZZ Z IV °∠= °∠ °∠ = 86.24-051.1 88.3494.5 51.36-245.6" oV Thus, )24.86t10cos(051.1v"o °−= Therefore, "o'oo vvv += =ov 4.631 sin(5t – 81.12°) + 1.051 cos(10t – 86.24°) V + Vo" − -j0.5 Ωj10 Ω8 Ω 2∠0° I (b) docsity.com P.P.10.7 If we transform the current source to a voltage source, we obtain the circuit shown in Fig. (a). (a) VS + − 4 Ω j Ω-j3 Ω 2 Ω -j2 Ω 1 Ω Io j5 Ω 16j12)3j4)(4j(sss +=−== ZIV We transform the voltage source to a current source as shown in Fig. (b). Let 2j6j23j4 −=++−=Z . Then, 3j5.1 2j6 16j12s s +=− + == Z V I . Io 6 1 Ω Ω IS j5 Ω -j2 -j2 Ω Ω (b) Note that )j1( 3 10 3j6 )5j)(2j6( 5j|| += + − =Z . By current division, )3j5.1( )2j1()j1( 3 10 )j1( 3 10 o + −++ + =I °∠ °∠ = + +− = 1.17602.13 56.11672.44 4j13 40j20 oI =oI 3.288∠99.46° A docsity.com P.P.10.10 To find , consider the circuit in Fig. (a). NZ j13 )3j9)(2j4( )3j9(||)2j4(N − −+ =−+=Z =NZ 3.176 + j0.706 Ω To find , short-circuit terminals a-b as shown in Fig. (b). Notice that meshes 1 and 2 form a supermesh. NI For the supermesh, 0)3j9()3j1(820- 321 =−−−++ III (1) Also, 4j21 += II (2) For mesh 3, 0)3j1(8)j13( 213 =−−−− III (3) Solving for , we obtain 2I °∠ °∠ = − − == 43.18-487.9 11.51-65.79 3j9 62j50 2N II =NI 8.396∠-32.68° A Using the Norton equivalent, we can find as in Fig. (c). oI j2 Ω 4 Ω -j3 Ω 8 Ω 1 Ω ZN a b (a) j2 Ω 4 Ω -j3 Ω8 Ω 1 Ω a b(b) INI2+ − 20∠0° I1 I3 -j4 Ω IN 10 – j5 ΩZN (c) Io docsity.com By current division, )68.32-396.8( 294.4j176.13 706.0j176.3 5j10 NN N o °∠− + = −+ = I Z Z I °∠ °∠°∠ = 18.05-858.13 )32.68-396.8)(53.12254.3( oI =oI 1.971∠-2.10° A P.P.10.11 Ω= ×× = ω ⎯→⎯ k-j20 )1010)(105(j 1 Cj 1 nF10 9-3 1 Ω= ×× = ω ⎯→⎯ k-j10 )1020)(105(j 1 Cj 1 nF20 9-3 2 Consider the circuit in the frequency domain as shown below. As a voltage follower, o2 VV = At node 1, 2020j-10 2 o1o11 VVVVV −+ − = − o1 )j1()j3(4 VV +−+= (1) At node 2, 10j- 0 20 oo1 −= − VVV (2) o1 )2j1( VV += Substituting (2) into (1) gives or o6j4 V= °∠= 90-3 2 oV + − + − 10 kΩ 20 kΩ -j20 kΩ -j10 kΩ Io Vo V2 V1 2∠0° V docsity.com Hence, V)90t5000cos(667.0)t(vo °−= =)t(vo 0.667 sin(5000t) V Now, k20j- 1o o VV I − = But from (2) 3 4- 2j- o1o ==− VVV A66.66j- k20j- 34- o μ==I Hence, A)90t5000cos(67.66)t(io μ°−= =)t(io 66.67 sin(5000t) μA P.P.10.12 Let RCj1 R Cj 1 ||RZ ω+ = ω = ZR R V V o s + = The loop gain is RCj2 RCj1 RCj1 RR R ZR R V V G/1 o s ω+ ω+ = ω+ + = + == where 10)101)(1010)(1000(RC 6-3 =××=ω °∠ °∠ = + + = 69.782.10 29.8405.10 10j2 10j1G/1 G = 1.0147∠–5.6° docsity.com