ch04_ism - 4 Techniques of Circuit Analysis Assessment...

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4 Techniques of Circuit Analysis Assessment Problems AP 4.1 [a] Redraw the circuit, labeling the reference node and the two node voltages: The two node voltage equations are 15 + v 1 60 + v 1 15 + v 1 v 2 5 = 0 5+ v 2 2 + v 2 v 1 5 = 0 Place these equations in standard form: v 1 ± 1 60 + 1 15 + 1 5 ² + v 2 ± 1 5 ² =1 5 v 1 ± 1 5 ² + v 2 ± 1 2 + 1 5 ² = 5 Solving, v 1 =60 V and v 2 =10 V; Therefore, i 1 =( v 1 v 2 ) / 5=10 A [b] p 15A = (15 A ) v 1 = (15 A )(60 V )= 900 W = 900 W(delivered) [c] p 5A =(5 A ) v 2 A )(10 V )=50 W = 50 W(delivered) 4–1
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4–2 CHAPTER 4. Techniques of Circuit Analysis AP 4.2 Redraw the circuit, choosing the node voltages and reference node as shown: The two node voltage equations are: 4 . 5+ v 1 1 + v 1 v 2 6+2 = 0 v 2 12 + v 2 v 1 + v 2 30 4 = 0 Place these equations in standard form: v 1 ± 1+ 1 8 ² + v 2 ± 1 8 ² = 4 . 5 v 1 ± 1 8 ² + v 2 ± 1 12 + 1 8 + 1 4 ² = 7 . 5 Solving, v 1 =6 V v 2 =18 V To Fnd the voltage v , Frst Fnd the current i through the series-connected 6Ω and 2Ω resistors: i = v 1 v 2 = 6 18 8 = 1 . 5 A Using a KVL equation, calculate v : v =2 i + v 2 =2( 1 . 5)+18=15 V AP 4.3 [a] Redraw the circuit, choosing the node voltages and reference node as shown: The node voltage equations are: v 1 50 6 + v 1 8 + v 1 v 2 2 3 i 1 = 0 v 2 4 + v 2 v 1 2 +3 i 1 = 0
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Problems 4–3 The dependent source requires the following constraint equation: i 1 = 50 v 1 6 Place these equations in standard form: v 1 ± 1 6 + 1 8 + 1 2 ² + v 2 ± 1 2 ² + i 1 ( 3) = 50 6 v 1 ± 1 2 ² + v 2 ± 1 4 + 1 2 ² + i 1 (3) = 5 v 1 ± 1 6 ² + v 2 (0) + i 1 (1) = 50 6 Solving, v 1 =32 V; v 2 =16 i 1 =3 A Using these values to calculate the power associated with each source: p 50V = 50 i 1 = 150 W p 5A = 5( v 2 ) = 80 W p 3 i 1 i 1 ( v 2 v 1 ) = 144 W [b] All three sources are delivering power to the circuit because the power computed in (a) for each of the sources is negative. AP 4.4 Redraw the circuit and label the reference node and the node at which the node voltage equation will be written: The node voltage equation is v o 40 + v o 10 10 + v o +20 i 20 =0 The constraint equation required by the dependent source is i = i 10 Ω + i 30 Ω = 10 v o 10 + 10+20 i 30 Place these equations in standard form:
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4–4 CHAPTER 4. Techniques of Circuit Analysis v o ± 1 40 + 1 10 + 1 20 ² + i (1) = 1 v o ± 1 10 ² + i ± 1 20 30 ² = 1+ 10 30 Solving, v o =24 V i = 3 . 2 A AP 4.5 Redraw the circuit identifying the three node voltages and the reference node: Note that the dependent voltage source and the node voltages v and v 2 form a supernode. The v 1 node voltage equation is v 1 7 . 5 + v 1 v 2 . 5 4 . 8=0 The supernode equation is v v 1 2 . 5 + v 10 + v 2 2 . 5 + v 2 12 1 =0 The constraint equation due to the dependent source is i x = v 1 7 . 5 The constraint equation due to the supernode is v + i x = v 2 Place this set of equations in standard form: v 1 ± 1 7 . 5 + 1 2 . 5 ² + v ± 1 2 . 5 ² + v 2 (0) + i x (0) = 4 . 8 v 1 ± 1 2 . 5 ² + v ± 1 2 . 5 + 1 10 ² + v 2 ± 1 2 . 5 +1 ² + i x (0) = 12 v 1 ± 1 7 . 5 ² + v (0) + v 2 (0) + i x (1) = 0 v 1 (0) + v (1) + v 2 ( 1) + i x (1) = 0 Solving this set of equations for v gives v =8 V v 1 =15
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This note was uploaded on 03/26/2008 for the course ELEN 214 taught by Professor Zou during the Spring '08 term at Texas A&M.

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ch04_ism - 4 Techniques of Circuit Analysis Assessment...

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