(b) (a) (d) (c) (e) Figure 10-N1. (f) Solution:

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1 Example: The inpu o each of he circuis shown in Figure 10-N1 is he volage source volage. The oupu of each circui is he curren i( ). Deermine he oupu of each of he circuis. (a) (b) (c) (d) (e) Figure 10-N1 (f) Soluion: Case (a) This circui will be a seady sae unil ime = 0. Because he inpu is consan before ime = 0, all of he elemen volages and currens will be consan. A ime = 0, he inpu changes abruply, disurbing he seady sae. Evenually he disurbance dies ou and he circui is again a seady sae. All of he elemen volages and currens will again be consan, bu hey will have differen consan values, because he inpu has changed. The hree sages can be illusraed as shown in Figure 10-N2. Figure 10-N2a represens he circui for < 0. The source volage is consan and he circui is a seady sae so he inducor acs like a shor circui. The inducor curren is

2 4 2 = = A 3 In paricular, immediaely before = 0, i ( 0 ) = 0.7 A. The curren in an inducor is coninuous, so ( ) i( ) i 0 + = 0 = 0.7 A (a) (b) (c) Figure 10-N2 Figure 10-N2b represens he circui immediaely afer = 0. The inpu is consan bu he circui is no a seady sae, so he inducor does no ac like a shor circui. The par of he circui ha is conneced o he inducor has he form of a Thevenin equivalen circui, so we recognize ha Consequenly R = Ω and v oc = V i = = 2 A. The ime consan of he circui is Finally, L 2 1 τ = = = R 3 () ( ( ) ) τ ( ) 3 3 i = i + i 0 + i e = e = e A As increases, he exponenial par of ( ) The exponenial par of () i i ges smaller. When = 5τ = 1.7 seconds, 3( 1.7) e = = A seady sae and ha ha new seady sae curren is has become negligible so we recognize ha he circui is again a i( ) = 2 A.

3 Figure 10-N2c represens he circui afer he disurbance has died ou and he circui has reached seady sae, ha is, when > 5τ. The source volage is consan and he circui is a seady sae so he inducor acs like a shor circui. As expeced, he inducor curren is 2 A. Case (b) This circui does no conain a swich and he inpu does no change abruply, so we expec he circui o be a seady sae. The inpu is sinusoidal a a frequency of 5 rad/s so all of he elemen currens and volages will be sinusoidal a a frequency of 5 rad/s. We can find he seady response by represening he circui in he frequency domain using impedances and phasors. Figure 10-N3 Figure 10-N3 shows he frequency domain represenaion of he circui. Ohm s law gives 0 0 I ( ω) = = = A + j The corresponding curren in he ime domain, is () = ( ) i 1.03cos 5 59 A Case (c) The volage source, resisor and inducor in his circui are conneced in parallel. The elemen volage of he resisor and inducor are each equal o he volage source volage. The curren in he resisor is given by Ohm s law o be The curren in he inducor is 5 e 5 ir () = = 2 e A 1 1 5τ il() = v( ) d il( 0) e d il( 0) L τ τ + = τ = + = Finally, using KCL gives ( ) ( 5 ) ( ) 5 e 1 i ( ) L 0 1.2e 1.2 il 0

4 5 5 5 = i () + i () = 2e 1.2e i ( 0) = 0.8e i ( 0) R L L L Case (d) This circui will be a seady sae unil he swich opens a ime = 0. Because he source volage is consan, all of he elemen volages and currens will be consan. A ime = 0, he swich opens, disurbing he seady sae. Evenually he disurbance dies ou and he circui is again a seady sae. All of he elemen volages and currens will be consan, bu hey will have differen consan values, because he circui has changed. The hree sages can be illusraed as shown in Figure 10-N4. Figure 10-N4a represens he circui for < 0. The closed swich is represened as a shor circui. The source volage is consan and he circui is a seady sae so he inducor acs like a shor circui. The inducor curren is i( ) = 0 A In paricular, immediaely before = 0, i ( 0 ) = 0 A. The curren in an inducor is coninuous, so ( ) i( ) i 0+ = 0 = 0 A (a) (b) (c) Figure 10-N4 Figure 10-N4b represens he circui immediaely afer = 0. The inpu is consan bu he circui is no a seady sae, so he inducor does no ac like a shor circui. The par of he circui ha is conneced o he inducor has he form of a Thevenin equivalen circui, so we recognize ha Consequenly R = Ω and v oc = V i = = 2 A. The ime consan of he circui is

5 Finally, L 2 1 τ = = = R 3 () ( ( ) ) τ ( ) 3 3 i = i + i 0+ i e = e = 2 2 e A As increases, he exponenial par of ( ) The exponenial par of () i i ges smaller. When = 5τ = 1.7 seconds, 3( 1.7) e seady sae and ha he seady sae curren is = 2 2 = A has become negligible so we recognize ha he circui is again a i( ) = 2 A. Figure 10-N4c represens he circui afer he disurbance has died ou and he circui has reached seady sae, ha is, when > 5τ. The source volage is consan and he circui is a seady sae so he inducor acs like a shor circui. As expeced, he inducor curren is 2 A. Case (e) This circui does no conain a swich and he inpu does no change abruply, so we expec he circui o be a seady sae. Because he source volage is consan, all of he elemen volages and currens will be consan. Because he source volage is consan and he circui is a seady sae, he inducor acs like a shor circui. (We ve encounered his circui wice before in his example, afer he disurbance died ou in cases b and d.) The curren is given by = = 2 A Case (f) We expec ha his circui will be a seady sae before he swich opens. As before, opening he swich will change he circui and disurb he seady sae. Evenually he disurbance will die ou and he circui will again be a seady sae. We will see ha he seady sae curren is consan before he swich opens and sinusoidal afer he swich opens. Figure 10N5a shows he circui before he swich opens. Applying KVL gives d = 0 d Consequenly, he inducor curren is i( ) = 0 before he swich opens. The curren in an inducor is coninuous, so ( ) i( ) i 0+ = 0 = 0 A

6 (a) (b) (c) Figure 10-N5 Figure 10-N5b represens he circui afer he swich opens. We can deermine he inducor curren by adding he naural response o forced response and hen using he iniial condiion o evaluae he consan in he naural response. Firs, we find he naural response. The par of he circui ha is conneced o he inducor has he form of he Thevenin equivalen circui, so we recognize ha The ime consan of he circui is The naural response is R = Ω L 2 1 τ = = = R 3 n ( ) i = Ke 3 A We can use he seady sae response as he forced response. As in case b, we obain he seady sae response by represening he circui in he frequency as shown in Figure 10N-5c. As before, we find I ω =. The forced response is ( ) A Then A = 0, ( ) = ( ) if 1.03cos 5 59 A 3 () () ( ) ( ) i = i + i = Ke cos 5 59 A. n f

7 so 0 ( ) ( ) i 0 = Ke cos 59 = K () = + ( ) i 0.53e 1.03cos 5 59 A