Tài liệu Bài giải mạch P7 pdf

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- 0 eI)t(i = RCt- 0 eRI)t(Ri)t(v == or RCt- 0 eV)t(v = Chapter 7, Solution 2. CR th =τ where is the Thevenin equivalent at the capacitor terminals. th R Ω = + = 601280||120R th =××=τ -3 105.060 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. eqeq CR=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 2 10x2x10x40RC,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. th R th R 8 Ω i 2 i i 1 10 Ω 0.5 V + v = 1 − + − + − 1 V 1.0 10 1 i 1 == , 16 1 8 5.01 i 2 = − = 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 Chapter 7, Solution 8. (a) 4 1 RC ==τ dt dv Ci- = =→= Ce-4))(10(Ce0.2- -4t-4t mF5 == C4 1 R Ω50 (b) ===τ 4 1 RC s25.0 (c) =×== )100)(105( 2 1 CV 2 1 )0(w 3-2 0C mJ250 (d) () τ −=×= 0 2t- 2 0 2 0R e1CV 2 1 CV 2 1 2 1 w 2 1 ee15.0 00 8t-8t- =→−= or 2e 0 8t = == )2(ln 8 1 t 0 ms6.86 Chapter 7, Solution 9. τ = t- e)0(v)t(v, CR eq = τ Ω = + + =++= 82423||68||82R eq 2)8)(25.0(CR eq ===τ =)t(v Ve20 2t- Chapter 7, Solution 10. 10 Ω 10 mF + v − i 15 Ω i o i T 4 Ω A2 15 )3)(10( ii10i15 oo ==→= i.e. if i , then A3)0( = A2)0(i o = A5)0(i)0(i)0(i oT =+= V502030)0(i4)0(i10)0(v T = + =+= across the capacitor terminals. Ω =+=+= 106415||104R th 1.0)1010)(10(CR -3 th =×==τ -10tt- e50e)0(v)t(v == τ )e500-)(1010( dt dv Ci 10t-3- C ×== = C iAe5- -10t By applying the current division principle, == + = CC i-0.6)i-( 1510 15 )t(i Ae3 -10t Chapter 7, Solution 11. Applying KCL to the RL circuit, 0 R v dtv L 1 =+ ∫ Differentiating both sides, 0v L R dt dv 0 dt dv R 1 L v =+→=+ LRt- eAv = If the initial current is , then 0 I ARI)0(v 0 == τ = t- 0 eRIv, R L =τ ∫ ∞ = t - dt)t(v L 1 i t - t- 0 e L RI- i ∞ τ τ = τ = t- 0 eRI-i τ = t- 0 eI)t(i Chapter 7, Solution 12. When t < 0, the switch is closed and the inductor acts like a short circuit to dc. The 4 Ω resistor is short-circuited so that the resulting circuit is as shown in Fig. (a). 3 Ω i(0 - ) + − 12 V 2 H 4 Ω (a) (b) A4 3 12 )0(i == − Since the current through an inductor cannot change abruptly, A4)0(i)0(i)0(i === +− When t > 0, the voltage source is cut off and we have the RL circuit in Fig. (b). 5.0 4 2 R L ===τ Hence, == τt- e)0(i)t(i Ae4 -2t Chapter 7, Solution 13. th R L =τ where is the Thevenin resistance at the terminals of the inductor. th R Ω = + = + = 37162120||8030||70R th = × =τ 37 102 -3 s08.81 µ Chapter 7, Solution 14 Converting the wye-subnetwork to delta gives 16 Ω R 2 80mH R 1 R 3 30 Ω Ω==Ω==Ω== ++ = 170 10 1700 ,34 50 1700 ,8520/1700 20 105050202010 321 RR xxx R 30//170 = (30x170)/200 = 25.5 Ω , 34//16=(34x16)/50 =10.88 Ω s x R Lx R Th Th m 14.3 476.25 1080 ,476.25 38.121 38.3685 )88.105.25//(85 3 ===Ω==+= − τ Chapter 7, Solution 15 (a) s R L R Th Th 25.020/5,2040//1012 ===Ω=+= τ (b) ms 5.040/)1020(,408160//40 3 ===Ω=+= − x R L R Th Th τ Chapter 7, Solution 16. eq eq R L =τ (a) LL eq = and 31 31312 31 31 2eq RR RR)RR(R RR RR RR + + + = + += =τ 31312 31 RR)RR(R )RR(L ++ + (b) where 21 21 eq LL LL + =L and 21 21213 21 21 3eq RR RR)RR(R RR RR RR + ++ = + += =τ )RR)RR(R()LL( )RR(LL 2121321 2121 +++ + Chapter 7, Solution 17. τ = t- e)0(i)t(i, 16 1 4 41 R L eq ===τ -16t e2)t(i = 16t-16t- o e2)16-)(41(e6 dt di Li3)t(v +=+= =)t(v o Ve2- -16t Chapter 7, Solution 18. If , the circuit can be redrawn as shown below. 0)t(v = + v o (t) − i(t) R e q 0.4 H 5 6 3||2R eq == , 3 1 6 5 5 2 R L =×==τ -3tt- ee)0(i)t(i == τ === 3t- o e-3)( 5 2- dt di -L)t(v Ve2.1 -3t Chapter 7, Solution 19. i 1 i 2 i 2 i 1 i/2 10 Ω 40 Ω − + 1 V i To find we replace the inductor by a 1-V voltage source as shown above. th R 0i401i10 21 =+− But 2iii 2 += and 1 ii = i.e. i2i2i 21 == 30 1 i0i201i10 =→=+− Ω== 30 i 1 R th s2.0 30 6 R L th ===τ =)t(i Ae2 -5t Chapter 7, Solution 20. (a). L50R 50 1 R L =→==τ dt di Lv- = =→= Le-50))(30(Le150- -50t-50t H1.0 == L50R Ω5 (b). === 50 1 R L τ ms20 (c). === 22 )30)(1.0( 2 1 )0(iL 2 1 w J45 (d). Let p be the fraction () τ −=⋅ 0 2t- 00 e1IL 2 1 pIL 2 1 3296.0e1e1p -0.450(2)(10)- =−=−= i.e. = p %33 Chapter 7, Solution 21. The circuit can be replaced by its Thevenin equivalent shown below. R th + − V th 2 H V40)60( 4080 80 V th = + = R 3 80 R80||40R th +=+= R380 40 R V )(i)0(iI th th + ==∞== 1 380R 40 )2( 2 1 IL 2 1 w 2 2 =       + == 3 40 R1 380R 40 =→= + = R Ω33.13 Chapter 7, Solution 22. τ = t- e)0(i)t(i, eq R L =τ Ω =+= 5120||5R eq , 5 2 =τ =)t(i Ae10 -2.5t Using current division, the current through the 20 ohm resistor is 2.5t- o e-2 5 i- -i)( 205 5 i == + = == o i20)t(v Ve04- -2.5t Chapter 7, Solution 23. Since the 2 Ω resistor, 1/3 H inductor, and the (3+1) Ω resistor are in parallel, they always have the same voltage. -1.5)0(i5.1 13 2 2 2 i- =→= + += The Thevenin resistance at the inductor’s terminals is th R 3 4 )13(||2R th =+= , 4 1 34 31 R L th ===τ 0t,e-1.5e)0(i)t(i -4tt- >== τ 4t- oL e/3)-1.5(-4)(1 dt di Lvv === = o v 0t,Ve2 -4t > = + = Lx v 13 1 v 0t,Ve5.0 -4t > Chapter 7, Solution 24. (a) =)t(v u(t)5- (b) [] [ ] )5t(u)3t(u10)3t(u)t(u-10)t(i − − − + −−= = )5t(u10)3t(u20)t(u10- − − − +