# 1 ECE 221 Electric Circuit Analysis I Chapter 8 Example 4.3, Problem 4.11 Node-Voltage Method Herbert G. Mayer, PSU Status 1/19/2015.

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1 ECE 221 Electric Circuit Analysis I Chapter 8 Example 4.3, Problem 4.11 Node-Voltage Method Herbert G. Mayer, PSU Status 1/19/2015

2 Syllabus Example 4.3 Example 4.3 Problem 4.11 Problem 4.11

3 Example 4.3 See [1] Example 4.3 on p. 95 in 10 th edition Analyze a circuit with 2 voltage sources, one constant, one dependent, several resistors, and compute the currents in all branches First compute the voltages v1 in the 20 Ω and v2 in the 10 Ω resistors, using the Node- Voltage Method

4 Circuit for Example 4.3

5 Example 4.3: 2 Node-Voltages KCL in node n1 above 20 Ω: (1) (v1 - 20)/2 + v1/20 + (v1 - v2)/5=0 KCL in node n2 above 10 Ω: (2) v2/10 + (v2 - v1)/5 + (v2 - 8*iα)/2=0 Third Equation needed due to Dependent Voltage Source: (3) iα=(v1 - v2)/5

6 Example 4.3: Students Solve in Class (3) Substituted in (2) v2/10 + v2/5 - v1/5 + v2/2 – 4*(v1 - v2)/5=0... v2=v1 * 5 / 8... V1=16 V V2=10 V

7 Problem 4.11 See [1] Problem 4.11 on p. 131 in 10 th edition Analyze a circuit with 2 constant voltage sources, several resistors, and compute the currents in all branches First compute the voltages v1 in the 60 Ω, and v2 in the 80 Ω resistors, using the Node- Voltage Method

8 Circuit for Problem 4.11

9 Problem 4.11: 2 Node-Voltages KCL in node n1 above 60 Ω: (1)(v1 - 128)/5 + v1/60 + (v1 - v2)/4=0// *60 12*v1 - 12*128 + v1 + 15*v1 – 15*v2=0 v1*( 12 + 1 + 15 ) – 12*128=15*v2 v2=v1*28/15 - 12*128/15= v1*28/15 - 4*128/5 KCL in node n2 above 80 Ω: (2)(v2 - v1)/4 + (v2 - 320)/10 + v2/80=0 v2*( 1/4 + 1/10 + 1/80 ) – v1/4 – 32=0

10 Problem 4.11: 2 Node-Voltages Substitute (1) in (2): (2) (v1*28/15 – 4*128/5)*(1/4 + 1/10 + 1/80 ) – v1/4 = 32 v1 *0.676667 – 37.12 – v1/4 = 32 v1=162 Volt v2 = ( v1 / 4 + 32 ) / 0.3625 v2=200 Volt ia=( 128 - 162 ) / 5= -6.8 A ib=v1 / 60=2.7 A id=v2 / 80=2.5 A ic=ia – ib=-9.5 A

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