Solutions to Chapter 7 Circuits Problems: Applying KVL and Thevenin Equivalent, Assignments of Engineering

Download Solutions to Chapter 7 Circuits Problems: Applying KVL and Thevenin Equivalent and more Assignments Engineering in PDF only on Docsity! Chapter 7, Solution 1. Applying KVL to Fig. 7.1. 0Ridti C 1 t - =+∫ ∞ Taking the derivative of each term, 0 dt di R C i =+ or RC dt i di −= Integrating, RC t- I )t(i ln 0 =      RCt- 0eI)t(i = RCt- 0eRI)t(Ri)t(v == or RCt-0eV)t(v = Chapter 7, Solution 2. CR th=τ where is the Thevenin equivalent at the capacitor terminals. thR Ω=+= 601280||120R th =××=τ -3105.060 ms30 Chapter 7, Solution 3. (a) ms 10102105,510//10 63 ===Ω== −xxxCRkR ThTh τ (b) 6s3.020,208)255//(20 ===Ω=++= xCRR ThTh τ Chapter 7, Solution 4. eqeqCR=τ where 21 21 eq CC CC + =C , 21 21 eq RR RR R + = =τ )CC)(RR( CCRR 2121 2121 ++ Chapter 7, Solution 5. τ= 4)-(t-e)4(v)t(v where 24)4(v = , 2)1.0)(20(RC ===τ 24)-(t-e24)t(v = == 26-e24)10(v V195.1 Chapter 7, Solution 6. Ve4)t(v 25 210x2x10x40RC,ev)t(v V4)24( 210 2)0(vv t5.12 36/t o o − −τ− = ===τ= = + == Chapter 7, Solution 7. τ= t-e)0(v)t(v , CR th=τ where is the Thevenin resistance across the capacitor. To determine , we insert a 1-V voltage source in place of the capacitor as shown below. thR thR 8 Ω i2 i i1 10 Ω0.5 V + v = 1 − + − + − 1 V 1.0 10 1 i1 == , 16 1 8 5.01 i2 = − = 80 13 16 1 1.0iii 21 =+=+= 13 80 i 1 R th == 13 8 1.0 13 80 CR th =×==τ =)t(v V20 813t-e )( 3 2 tui p = )() 3 2( 3 tuAei t += − If i(0) =0, then A + 2/3 = 0, i.e. A=-2/3. Thus )()1( 3 2 3 tuei t−−= Chapter 7, Solution 39. (a) Before t = 0, = + = )20( 14 1 )t(v V4 After t = 0, [ ] τ∞−+∞= t-e)(v)0(v)(v)t(v 8)2)(4(RC ===τ , 4)0(v = , 20)(v =∞ 8t-e)208(20)t(v −+= =)t(v Ve1220 8t-− (b) Before t = 0, 21 vvv += , where is due to the 12-V source and is due to the 2-A source. 1v 2v V12v1 = To get v , transform the current source as shown in Fig. (a). 2 V-8v2 = Thus, =−= 812v V4 After t = 0, the circuit becomes that shown in Fig. (b). 2 F 2 F 4 Ω 12 V + − + − v2 8 V + − 3 Ω 3 Ω (a) (b) [ ] τ∞−+∞= t-e)(v)0(v)(v)t(v 12)(v =∞ , 4)0(v = , 6)3)(2(RC ===τ 6t-e)124(12)t(v −+= =)t(v Ve812 6t-− Chapter 7, Solution 40. (a) Before t = 0, =v V12 . After t = 0, [ ] τ∞−+∞= t-e)(v)0(v)(v)t(v 4)(v =∞ , 12)0(v = , 6)3)(2(RC ===τ 6t-e)412(4)t(v −+= =)t(v Ve84 6t-+ (b) Before t = 0, =v V12 . After t = 0, [ ] τ∞−+∞= t-e)(v)0(v)(v)t(v After transforming the current source, the circuit is shown below. t = 0 4 Ω 2 Ω + − 12 V 5 F 12)0(v = , 12)(v =∞ , 10)5)(2(RC ===τ =v V12 Chapter 7, Solution 41. 0)0(v = , 10)12( 16 30 )(v ==∞ 5 36 )30)(6( )1)(30||6(CR eq === [ ] τ∞−+∞= t-e)(v)0(v)(v)t(v 5t-e)100(10)t(v −+= =)t(v V)e1(10 -0.2t− Chapter 7, Solution 42. (a) [ ] τ∞−+∞= t-oooo e)(v)0(v)(v)t(v 0)0(vo = , 8)12(24 4 )(vo =+ =∞ eqeqCR=τ , 3 4 4||2R eq == 4)3( 3 4 ==τ 4t- o e88)t(v −= =)t(vo V)e1(8 -0.25t− (b) For this case, 0)(vo =∞ so that τ= t-oo e)0(v)t(v 8)12( 24 4 )0(vo =+ = , 12)3)(4(RC ===τ =)t(vo Ve8 12t- Chapter 7, Solution 43. Before t = 0, the circuit has reached steady state so that the capacitor acts like an open circuit. The circuit is equivalent to that shown in Fig. (a) after transforming the voltage source. 40 v 2i5.0 o−= , 80 v i o= Hence, 64 5 320 v 40 v 2 80 v 2 1 o oo ==→−= == 80 v i o A8.0 After t = 0, the circuit is as shown in Fig. (b). τ= t-CC e)0(v)t(v , CR th=τ To find , we replace the capacitor with a 1-V voltage source as shown in Fig. (c). thR 0.5i vC + − 0.5i i 1 V 80 Ω (c) Chapter 7, Solution 55. For t < 0, consider the circuit shown in Fig. (a). + v − 4io io + − io 0.5 H + − 0.5 H + v − + − i 3 Ω 8 Ω 2 Ω 2 Ω 24 V 20 V (a) (b) 24i0i424i3 ooo =→=−+ == oi4)t(v V96 A482 v i == For t > 0, consider the circuit in Fig. (b). [ ] τ∞−+∞= t-e)(i)0(i)(i)t(i 48)0(i = , A2 28 20 )(i = + =∞ Ω=+= 1082R th , 20 1 10 5.0 R L th ===τ -20t-20t e462e)248(2)t(i +=−+= == )t(i2)t(v Ve924 -20t+ Chapter 7, Solution 56. Ω=+= 105||206R eq , 05.0R L ==τ [ ] τ∞−+∞= t-e)(i)0(i)(i)t(i i(0) is found by applying nodal analysis to the following circuit. 0.5 H + − 12 Ω 5 Ω 20 Ω 6 Ω + v − vx i 20 V 2 A 12v 6 v 20 v 12 v 5 v20 2 x xxxx =→++= − + A2 6 v )0(i x == 45||20 = 6.1)4( 64 4 )(i = + =∞ -20t0.05t- e4.06.1e)6.12(6.1)t(i +=−+= Since , 20t-e-20)()4.0( 2 1 dt di L)t(v == =)t(v Ve4- -20t Chapter 7, Solution 57. At , the circuit has reached steady state so that the inductors act like short circuits. −= 0t + − 6 Ω i 5 Ω i1 i2 20 Ω30 V 3 10 30 20||56 30 i == + = , 4.2)3( 25 20 i1 == , 6.0i2 = A4.2)0(i1 = , A6.0)0(i2 =