Chapt07PP - 2009 CHAPTER 7 The circuit in Fig. (a) is...

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© 2009 CHAPTER 7 P.P.7.1 The circuit in Fig. (a) is equivalent to the one shown in Fig. (b). Ω = + = 12 6 || 12 8 R eq s 4 ) 3 / 1 )( 12 ( C R eq = = = τ 45e –0.25t V = = = τ 4 t - t - c c e 45 e ) 0 ( v v = + = c x v 8 4 4 v 15e –0.25t V V e 30 - v v v v v v -0.25t c x o c o x = = ⎯→ + = = = 8 v i o o –3.75 e –0.25t A P.P.7.2 When t < 0, the switch is closed as shown in Fig. (a). V 8 ) 24 ( 6 3 3 ) 0 ( v c = + = Ω = = 3 12 || 4 R eq + v c (0) 12 Ω 4 Ω + 24 V 6 Ω (a) 12 Ω + v c + v x + v o 8 Ω i o 1/3 F 6 Ω (a) 1/3 F R eq + v (b)
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When t > 0, the switch is open as shown in Fig. (b). s 2 / 1 ) 6 / 1 )( 3 ( C R eq = = = τ V e 8 t 2 - = = τ t - c e ) 0 ( v ) t ( v = × × = = 64 6 1 2 1 ) 0 ( Cv 2 1 ) 0 ( w 2 c c 5.333J P.P.7.3 This can be solved in two ways. Method 1 : Find th R at the inductor terminals by inserting a voltage source. Applying mesh analysis gives 0 v 2 i 2 i 3 1 x 2 1 = + + , where 1 x i 1 v = Loop 1: 1 i 2 i 5 2 1 = ( 1 ) 0 v 2 i 2 i 8 x 1 2 = Loop 2: = 8i 2 – 2i 1 – 2i 1 1 2 i 2 1 i = ( 2 ) From (1) and (2), 5i 1 – 1i 1 = 1 or 1 o i i = = (1/4) A + + v o = 1 V + v x i o i 1 1 Ω 2 Ω 6 Ω 2v x i 2 3 Ω + 24 V 6 Ω 1/6 F (b) t = 0
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4 i v R o o th = = 2 1 4 2 R L = = = τ , s A e 5 ) t ( i -2t = Method 2 : We can obtain i using mesh analysis. Applying KVL to the loops, we obtain 1 Ω + v x i 1 i 2 Ω 6 Ω 2v x i 2 2 H + Loop 1: 0 v 2 i 2 i 3 dt di 2 1 = + + x 2 1 where 1 x i 1 v = 0 i 2 i 5 dt di 2 2 1 1 = + (3) oop 2: 0 v 2 i 2 i 8 x 1 2 = L 1 2 i 2 1 i = ( 4 ) ubstituting (4) into (3) yields S 0 i 1 i 5 dt di 2 1 = + 1 1 or 0 i 2 dt di 1 1 = + t 2 - 1 Ae i = = = t 2 - 1 Be i - i B 5 ) 0 ( i = = -2t = herefore, 5e –2t A ( i A e 5 ) t T ) t ( i = and ) t ( v x –5e –2t V = = i(t) -1
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P.P.7.4 For t < 0, the equivalent circuit is shown in Fig. (a). 8 Ω (b) 12 Ω 5 Ω 2 H 8 Ω A 2 12 / ) 6 / 24 x 6 ( 12 / )}] 8 / 1 ( ) 12 / 1 ( ) 24 / 1 /{( 1 [ 6 ) 0 ( i = = + + = or t > 0, the current source and 24-ohm is cut off and the RL circuit is shown in Fig. (b). F Ω = = + = 4 5 || 20 5 || ) 8 12 ( R , eq 5 . 0 2 L = = = τ 4 R eq = = τ t - e ) 0 ( i ) t ( i 0 t , A e 2 t -2 > .P.7.5 For t < 0, the switch is closed. The inductor acts like a short so the P equivalent circuit is shown in Fig. (a). A 12 ) 18 ( 2 4 4 i = + = , A 6 i o = , V 24 i 2 v o = = or t > 0, the current source is cut off so that the circuit becomes that shown in Fig. (b). F The Thevenin equivalent resistance at the inductor terminals is Ω = + = 2 3 || ) 2 4 ( R , th 2 1 L = = τ R th 2t - o e 4 i 3 1 - 3 6 ) i - ( 3 i = = + = and -2t o o e 8 -2i v = = 2 Ω 4 Ω (a) 18 A i i o (b) 4 Ω 3 Ω 2 Ω 1 H i i o 24 Ω 12 Ω (a) i(t) 6 A
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Thus, > < 0 t A e 12 0 t A 12 2t - > < 0 t A e 4 0 t A 6 2t - > < 0 t V e 8 0 t
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Chapt07PP - 2009 CHAPTER 7 The circuit in Fig. (a) is...

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