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ECE 3144 Lecture 7

ECE 3144 Lecture 7. Dr. Rose Q. Hu Electrical and Computer Engineering Department Mississippi State University. Reminder for Lecture 6. Resistors in parallel Current divider Current sources in parallel. or. Problem solving strategy. Find R AB in the circuit given. +. -.

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ECE 3144 Lecture 7

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  1. ECE 3144 Lecture 7 Dr. Rose Q. Hu Electrical and Computer Engineering Department Mississippi State University

  2. Reminder for Lecture 6 • Resistors in parallel • Current divider • Current sources in parallel or

  3. Problem solving strategy Find RAB in the circuit given

  4. + - Wye  Delta Transform I1 R1 R2 V1 R3 • We have learned how to simplify the network with series parallel resistors combination. • In the circuit network given, nowhere is the resistor in series or parallel with another. • Techniques learned so far do not apply here. • We can replace one portion of the network with the equivalent network. The conversion will reduce the network to series parallel combination of resistors, which we are already familiar with. • One conversion technique we are learning here is wye-to-delta or delta-to-wye transformation. R4 R5 R6

  5. a Ra Rc Rb c b Figure (2) Wye  Delta Transform a R1 R2 R3 • Notice that resistors figure (1) forms a  (delta) and resistors in figure (2) forms a Y (wye). • Both of these two configurations are connected to the same terminals a, b,c • It is possible to relate these two networks to each other such that terminal characteristics are the same. The relationship is called wye-delta (Y- ) transformation. • The transformation must keep terminal characteristics the same. At each corresponding pair of terminals, the resistance at the corresponding terminals must be the same. b c Figure (1)

  6. Wye  Delta Transform

  7. Wye  Delta Transform Special case: suppose the delta-connected load is balanced, that is, R1=R2=R3=R. The equivalent wye-connected load is also balanced, so Ra=Rb=Rc=RY. Then we have

  8. Wye  Delta Transform I1 R1 R2 V1 + - R3 R4 R5 R6 Ra Rb Rc V1 + - R4 R5 R6 • Rc and R4 are in series => Rc4 • Rb and R5 are in series => Rb5 • Rc4 and Rb5 are in parallel => Rc4b5 • Rc4b5 and Ra are in series =>Rc4b5a

  9. Is + + - - Rc =2 k Rb=3 k 12 V 4 k 9 k Example Is Ra=6k R1=12 k R2=18 k R3 12 V 6 k 4 k 9 k

  10. Circuits with dependent sources • When writing the KVL and/or KCL equations for the network, treat the dependent source as though it were an independent source. • Write the equations that specifies the relationship of the dependent source to the controlling parameters. • Solve the equations for the unknowns. Be sure that number of linearly independent equations matches the number of the unknowns.

  11. + - Example Determine the voltageVo in the circuit VA = 2000 I1 3 k I1 - + + Vo 12 V 5 k -

  12. + 2 k 3 k 4 I0 10 mA + Vs 4 k V0 I0 - - Example Given the network, find voltage V0

  13. Homework for Lecture 7 • Problems 2.57, 2.64,2.65,2.69, 2.71, 2.72, 2.73, 2.75, 2.77

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