CHAPTER 3 PROBLEM SOLVING ADDEMDUM

1
CHAPTER 3PROBLEM SOLVING ADDEMDUM

• NETWORKS 1
• Professor Linda Head
• ROWAN UNIVERSITY, COLLEGE OF ENGINEERING
• DEPARTMENT OF ELECTRICAL COMPUTER ENGINEERING
• FALL SEMESTER, 2001

2
PROBLEM SET-UP
va
vb
_
_
node3
node1
node2


Rb
Ra
ib
ia

ivs

vc

ic
vis
vs
Rc
is
_
_
_
node4
3
STEPS TAKEN

• Apply P.S.C. to passive elements.
• Show current direction at voltages sources.
• Show voltage direction at current sources.
• Name nodes and loops.
• Name elements and sources.
• Name currents and voltages.

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WRITE THE KCL EQUATIONS
node1
node3
node2
node4
5
WRITE THE KVL EQUATIONS
loop1
loop2
6
WRITE SUPPLEMENTARYEQUATIONS
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CIRCUIT REDUCTION
8

• Begin with loop on far right.
• Combine the three resistors that are in series.
• Req 4550100 195?

9

• Again using the loop on the far right.
• The 90 ? and 195 ? resistors are in parallel.
• Req (90)(195)/(90195) 61.58 ?

10

• Still working with the loop on the far right.
• The 30 ? and the 61.58 ? resistors are in series.
• Req 30 61.58 91.58 ?

11

• Again, the far right loop.
• The 15 ? and 91.58 ? resistors are in parallel.
• Req (15)(91.58)/(1591.58) 12.9 ?

12

• Now there is only one loop.
• All the resistors are in series.
• Req 1012.95 27.9 ?

13
a
b

• Use Ohms Law to determine iT.
• iT 5/27.9 0.179A
• iT flows in all three resistors, the 12.9 ?
  resistor is the equivalent resistance of the
  entire circuit beyond points a and b.

14
a
ix

• iT divides at a to flow through the 15 ? and the
  91.58 ? resistors (the 91.58 ? is an equivalent
  resistance for the rest of the circuit).
• Use current divider ix (0.179)(15)/(1591.58)
  0.0252A.

15
a
0.0252A
b

• No calculations are required at this step because
  the 0.0252A is flowing through both resistors in
  the right loop.
• This circuit must be drawn however, because the
  61.58 ? resistor is an equivalent for the circuit
  to the right of a and b.

16
a
0.0252A
b

• Use the current divider equation again to
  determine i1.
• i1 (0.0252)(195)/(90195) 0.01724A 17.24mA.
• The current through the 195 ? resistor is 0.0252
  - 0.01724 7.96mA