This document covers fundamental circuit analysis concepts including:
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2) Series and parallel resistor combinations are examined along with voltage and current division techniques.
3) Wye-delta transformations allow the analysis of resistor networks that are neither purely series nor parallel.
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Electric circuits-chapter-2 Basic Laws
1. 08/01/12
Chapter 2
Basic Laws
DKS1113 Electric Circuits
2. Introduction
Fundament laws that govern electric circuits:
Ohm’s Law.
Kirchoff’s Law.
These laws form the foundation upon which electric
circuit analysis is built.
Common techniques in circuit analysis and design:
Combining resistors in series and parallel.
Voltage and current divisions.
Wye to delta and delta to wye transformations.
These techniques are restricted to resistive circuits.
08/01/12 DKS1113 Electric Circuits 2/43
4. Ohm’s Law
Relationship between current and voltage
within a circuit element.
The voltage across an element is directly
proportional to the current flowing through it
v α i
Thus::v=iR and R=v/i
Where:
R is called resistor.
Has the ability to resist the flow of electric current.
Measured in Ohms (Ω)
08/01/12 DKS1113 Electric Circuits 4/43
5. Ohm’s Law
*pay careful attention to current direction
v=iR
08/01/12 DKS1113 Electric Circuits 5/43
6. Ohm’s Law
Value of R :: varies from 0 to infinity
Extreme values == 0 & infinity.
Only linear resistors obey Ohm’s Law.
Short circuit Open Circuit
08/01/12 DKS1113 Electric Circuits 6/43
7. Ohm’s Law
Conductance (G)
Unit mho or Siemens (S).
Reciprocal of resistance R
G=1/R
Has the ability to conduct electric current
08/01/12 DKS1113 Electric Circuits 7/43
8. Ohm’s Law
Power:
P = iv i ( i R ) = i2R watts
(v/R) v = v2/R watts
R and G are positive quantities, thus power
is always positive.
R absorbs power from the circuit Passive
element.
08/01/12 DKS1113 Electric Circuits 8/43
9. Ohm’s Law
Example 1:
Determine voltage (v), conductance (G) and power
(p) from the figure below.
08/01/12 DKS1113 Electric Circuits 9/43
10. Ohm’s Law
Example 2:
Calculate current i in figure below when the switch
is in position 1.
Find the current when the switch is in position 2.
08/01/12 DKS1113 Electric Circuits 10/43
11. Nodes, Branches & Loops
Elements of electric circuits can be
interconnected in several way.
Need to understand some basic concepts of
network topology.
Branch: Represents a single element
(i.e. voltage, resistor & etc)
Node: The meeting point between two
or more branches.
Loop: Any closed path in a circuit.
08/01/12 DKS1113 Electric Circuits 11/43
12. Nodes, Branches & Loops
Example 3:
Determine how many branches and nodes for the
following circuit.
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13. Nodes, Branches & Loops
5 Branches 3 Nodes
1 Voltage Source a
1 Current Source b
3 Resistors c
08/01/12 DKS1113 Electric Circuits 13/43
14. Nodes, Branches & Loops
Example 4:
Determine how many branches and nodes for the
following circuit.
08/01/12 DKS1113 Electric Circuits 14/43
16. Kirchoff’s Laws
Kirchoff’s Current Law (KCL)
The algebraic sum of current entering /
leaving a node (or closed boundary) is zero.
Current enters = +ve
Current leaves = -ve
∑ current entering = ∑ current leaving
08/01/12 DKS1113 Electric Circuits 16/43
17. Kirchoff’s Laws
Example 5:
Given the following circuit, write the equation for
currents.
08/01/12 DKS1113 Electric Circuits 17/43
18. Kirchoff’s Laws
Example 6:
Current in a closed boundary
08/01/12 DKS1113 Electric Circuits 18/43
19. Kirchoff’s Laws
Example 9:
Use KCL to obtain currents i1, i2, and i3 in the circuit.
08/01/12 DKS1113 Electric Circuits 19/43
20. Kirchoff’s Laws
Kirchoff’s Voltage Law (KVL)
Applied to a loop in a circuit.
According to KVL The algebraic sum of voltage (rises
and drops) in a loop is zero.
+ v1 - +
+
vs V2
-
- v3 + -
08/01/12 DKS1113 Electric Circuits 20/43
21. Kirchoff’s Laws
Example 10:
Use KVL to obtain v1, v2 and v3.
08/01/12 DKS1113 Electric Circuits 21/43
22. Kirchoff’s Laws
Example 11:
Use KVL to obtain v1, and v2.
08/01/12 DKS1113 Electric Circuits 22/43
23. Kirchoff’s Laws
Example 12:
Calculate power dissipated in 5Ω resistor.
10
08/01/12 DKS1113 Electric Circuits 23/43
24. Series Resistors & Voltage Division
Series resistors same current flowing
through them.
v1= iR1 & v2 = iR2
KVL:
v-v1-v2=0
v= i(R1+R2)
i = v/(R1+R2 ) =v/Req
or v= i(R1+R2 ) =iReq
iReq = R1+R2
08/01/12 DKS1113 Electric Circuits 24/43
25. Series Resistors & Voltage Division
Voltage Division:
Previously:
v1 = iR1 & v2 = iR2
i = v/(R1+R2 )
Thus:
v1=vR1/(R1+R2)
v2=vR2/(R1+R2)
08/01/12 DKS1113 Electric Circuits 25/43
26. Parallel Resistors & Current Division
Parallel resistors Common voltage across it.
v = i1R1 = i2R2
i = i1+ i2
= v/R1+ v/R2
= v(1/R1+1/R2)
=v/Req
v =iReq
1/Req = 1/R1+1/R2
Req = R1R2 / (R1+R2 )
08/01/12 DKS1113 Electric Circuits 26/43
27. Parallel Resistors & Current Division
Current Division:
Previously:
v = i1R1 = i2R2
v=iReq = iR1R2 / (R1+R2 )
and i1 = v /R1 & i2 =v/ R2
Thus:
i1= iR2/(R1+R2)
i2= iR1/(R1+R2 )
08/01/12 DKS1113 Electric Circuits 27/43
29. Voltage and Current Division
Example 13:
Calculate v1, i1, v2 and i2.
08/01/12 DKS1113 Electric Circuits 29/43
30. Voltage and Current Division
Example 14:
Determine i1 through i4.
08/01/12 DKS1113 Electric Circuits 30/43
31. Voltage and Current Division
Example 15:
Determine v and i.
Answer v = 3v, I = 6 A.
08/01/12 DKS1113 Electric Circuits 31/43
32. Voltage and Current Division
Example 16:
Determine I1 and Vs if the current through 3Ω
resistor = 2A.
08/01/12 DKS1113 Electric Circuits 32/43
33. Voltage and Current Division
Example 17:
Determine Rab.
08/01/12 DKS1113 Electric Circuits 33/43
34. Voltage and Current Division
Example 18:
Determine vx and power absorbed by the 12Ω
resistor.
Answer v = 2v, p = 1.92w.
08/01/12 DKS1113 Electric Circuits 34/43
35. Wye-Delta Transformations
Given the circuit, how to combine R1 through R6?
Resistors are neither in series nor parallel…
Use wye-delta transformations
08/01/12 DKS1113 Electric Circuits 35/43
39. Wye-Delta Transformations
Thus Δ to y conversion ::
R1 = RbRc/(Ra+Rb+Rc)
R2 = RaRc/(Ra+Rb+Rc)
R3 = RaRb/(Ra+Rb+Rc)
# Each resistors in y network is the
product of two adjacent
branches divide by the 3 Δ
resistors
08/01/12 DKS1113 Electric Circuits 39/43
40. Wye-Delta Transformations
Y to Δ conversions:
Ra = (R1R2 +R2 R3 +R1R3)/R1
Rb = (R1R2 +R2 R3 +R1R3)/R2
Rc= (R1R2 +R2 R3 +R1R3)/R3
08/01/12 DKS1113 Electric Circuits 40/43
41. Wye-Delta Transformations
Example 19:
Transform the circuit from Δ to y.
Answer R1=18, R2=6, R3=3.
08/01/12 DKS1113 Electric Circuits 41/43
42. Wye-Delta Transformations
Example 20:
Determine Rab.
Answer Rab=142.32.
08/01/12 DKS1113 Electric Circuits 42/43