ch3 - C HAPTER 3 Section 3-1: Node-Voltage Method Problem...

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C HAPTER 3 Section 3-1: Node-Voltage Method Problem 3.1 Apply nodal analysis to find the node voltage V in the circuit of Fig. P3.1. Use the information to determine the current I . 3 Ω 2 Ω 4 Ω 2 Ω 16 V 12 V + _ _ I V Figure P3.1: Circuit for Problem 3.1. Solution: Application of KCL at node V gives: V 16 2 + V 3 + V 12 2 + 4 = 0 V p 1 2 + 1 3 + 1 6 P = 16 2 + 12 6 V = 8 + 2 = 10 V . The current I is related to V by I = ( V 16 ) 2 = ( 10 16 ) 2 = 3 A .
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Problem 3.2 Apply nodal analysis to determine V x in the circuit of Fig. P3.2. 1 Ω 2 Ω 4 Ω 2 Ω 3 A V x V + _ Figure P3.2: Circuit for Problem 3.2. Solution: At node V , application of KCL gives V 2 + 1 3 + V 2 + 4 = 0 , which leads to V = 6 V . By voltage division, V x = V × 4 2 + 4 = 6 × 4 6 = 4 V .
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Problem 3.3 Use nodal analysis to determine the current I x and amount of power supplied by the voltage source in the circuit of Fig. P3.3. 2 Ω 4 Ω 8 Ω 40 V 9 A I x V + _ I Figure P3.3: Circuit for Problem 3.3. Solution: At node V , application of KCL gives 9 + V 2 + V 4 + V 40 8 = 0 V p 1 2 + 1 4 + 1 8 P = 9 + 40 8 7 V 8 = 9 + 5 V = 16 V . The current I x is then given by I x = V 4 = 16 4 = 4 A . To find the power supplied by the 40-V source, we need to first find the current I flowing into its positive terminal, I = V 40 8 = 16 40 8 = 3 A . Hence, P = VI = 40 × ( 3 ) = 120 W (The minus sign confirms that the voltage source is a supplier of power.)
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Problem 3.4 For the circuit in Fig. P3.4: (a) Apply nodal analysis to find node voltages V 1 and V 2 . (b) Determine the voltage V R and current I . 1 Ω 1 Ω 1 Ω 1 Ω 1 Ω V R + _ 16 V V 1 V 2 I + _ Figure P3.4: Circuit for Problem 3.4. Solution: (a) At nodes V 1 and V 2 , Node 1: V 1 16 1 + V 1 1 + V 1 V 2 1 = 0 (1) Node 2: V 2 V 1 1 + V 2 1 + V 2 1 = 0 (2) Simplifying Eqs. (1) and (2) gives: 3 V 1 V 2 = 16 (3) V 1 + 3 V 2 = 0 . (4) Simultaneous solution of Eqs. (3) and (4) leads to: V 1 = 6 V , V 2 = 2 V . (b) V R = V 1 V 2 = 6 2 = 4 V I = V 2 1 = 2 1 = 2 A .
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Problem 3.5 Apply nodal analysis to determine the voltage V R in the circuit of Fig. P3.5. 2 Ω 4 Ω 4 Ω 8 V V R + _ 12 V + _ _ V Figure P3.5: Circuit for Problem 3.5. Solution: At node V : V 12 4 + V 2 + V 8 4 = 0 , which leads to V = 5 V . Hence, V R = 12 V = 12 5 = 7 V .
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Problem 3.6 Use the nodal-analysis method to find V 1 and V 2 in the circuit of Fig. P3.6 and then apply that to determine I x . 6 Ω 12 Ω 6 Ω 3 A 4 A 2 A V 1 V 2 I x Figure P3.6: Circuit for Problem 3.6. Solution: At nodes V 1 and V 2 : Node 1: 2 + V 1 6 + V 1 V 2 12 4 = 0 (1) Node 2: 4 + V 2 6 + V 2 V 1 12 3 = 0 (2) Simplifying Eq. (1): V 1 p 1 6 + 1 12 P V 2 12 = 6 (1) 3 V 1 V 2 = 72 (3) Simplifying Eq. (2): V 1 12 + V 2 p 1 6 + 1 12 P = 1 (2) V 1 + 3 V 2 = 12
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This note was uploaded on 04/19/2010 for the course EE 220 taught by Professor Rawat during the Spring '09 term at Nevada.

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ch3 - C HAPTER 3 Section 3-1: Node-Voltage Method Problem...

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