Post on 30-Nov-2014
description
transcript
Chapter 2Basic Laws
04/09/23
DKS1113 Electric Circuits
04/09/23 2/43
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.
DKS1113 Electric Circuits
Ohm’s Law
04/09/23 3/43
DKS1113 Electric Circuits
04/09/23 4/43
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)
DKS1113 Electric Circuits
04/09/23 5/43
Ohm’s Law
v=iR
*pay careful attention to current direction
DKS1113 Electric Circuits
04/09/23 6/43
Ohm’s Law
Value of R :: varies from 0 to infinity
Extreme values == 0 & infinity.
Only linear resistors obey Ohm’s Law.
Short circuitShort circuit Open Circuit Open Circuit
DKS1113 Electric Circuits
04/09/23 7/43
Ohm’s Law
Conductance (G) Unit mho or Siemens (S).
Reciprocal of resistance R
G = 1 / R
Has the ability to conduct electric current
DKS1113 Electric Circuits
04/09/23 8/43
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.
DKS1113 Electric Circuits
04/09/23 9/43
Ohm’s Law
Example 1: Determine voltage (v), conductance (G) and power
(p) from the figure below.
DKS1113 Electric Circuits
04/09/23 10/43
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.
DKS1113 Electric Circuits
04/09/23 11/43
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.DKS1113 Electric Circuits
04/09/23 12/43
Nodes, Branches & Loops
Example 3: Determine how many branches and nodes for the
following circuit.
DKS1113 Electric Circuits
Nodes, Branches & Loops
5 Branches 1 Voltage Source 1 Current Source 3 Resistors
3 Nodes a b c
04/09/23 13/43
DKS1113 Electric Circuits
04/09/23 14/43
Nodes, Branches & Loops
Example 4: Determine how many branches and nodes for the
following circuit.
DKS1113 Electric Circuits
04/09/23 15/43
Kirchoff’s Laws
DKS1113 Electric Circuits
04/09/23 16/43
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
DKS1113 Electric Circuits
04/09/23 17/43
Kirchoff’s Laws
Example 5: Given the following circuit, write the equation for
currents.
DKS1113 Electric Circuits
04/09/23 18/43
Kirchoff’s Laws
Example 6: Current in a closed boundary
DKS1113 Electric Circuits
04/09/23 19/43
Kirchoff’s Laws
Example 9: Use KCL to obtain currents i1, i2, and i3 in the circuit.
DKS1113 Electric Circuits
04/09/23 20/43
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 -
- v3 +
+
V2
-
vs
DKS1113 Electric Circuits
04/09/23 21/43
Kirchoff’s Laws
Example 10: Use KVL to obtain v1, v2 and v3.
DKS1113 Electric Circuits
04/09/23 22/43
Kirchoff’s Laws
Example 11: Use KVL to obtain v1, and v2.
DKS1113 Electric Circuits
04/09/23 23/43
Kirchoff’s Laws
Example 12: Calculate power dissipated in 5Ω resistor.
10
DKS1113 Electric Circuits
04/09/23 24/43
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
DKS1113 Electric Circuits
04/09/23 25/43
Series Resistors & Voltage Division
Voltage Division:
Previously: v1 = iR1 & v2 = iR2 i = v/(R1+R2 )
Thus: v1=vR1/(R1+R2) v2=vR2/(R1+R2)
DKS1113 Electric Circuits
04/09/23 26/43
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 )
DKS1113 Electric Circuits
04/09/23 27/43
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 )
DKS1113 Electric Circuits
04/09/23 28/43
Conductance (G)
Series conductance: 1/Geq = 1/G1 +1/G2+…
Parallel conductance: Geq = G1 +G2+…
DKS1113 Electric Circuits
04/09/23 29/43
Voltage and Current Division
Example 13: Calculate v1, i1, v2 and i2.
DKS1113 Electric Circuits
04/09/23 30/43
Voltage and Current Division
Example 14: Determine i1 through i4.
DKS1113 Electric Circuits
04/09/23 31/43
Voltage and Current Division
Example 15: Determine v and i.
Answer v = 3v, I = 6 A.
DKS1113 Electric Circuits
04/09/23 32/43
Voltage and Current Division
Example 16: Determine I1 and Vs if the current through 3Ω
resistor = 2A.
DKS1113 Electric Circuits
04/09/23 33/43
Voltage and Current Division
Example 17: Determine Rab.
DKS1113 Electric Circuits
04/09/23 34/43
Voltage and Current Division
Example 18: Determine vx and power absorbed by the 12Ω
resistor. Answer v = 2v, p = 1.92w.
DKS1113 Electric Circuits
04/09/23 35/43
Wye-Delta Transformations
Given the circuit, how to combine R1 through R6? Resistors are neither in series nor parallel…
Use wye-delta transformationsDKS1113 Electric Circuits
04/09/23 36/43
Wye-Delta Transformations
Y network T network
DKS1113 Electric Circuits
04/09/23 37/43
Wye-Delta Transformations
Δ network π network
DKS1113 Electric Circuits
04/09/23 38/43
Wye-Delta Transformations
Delta (Δ) to wye (y) conversion.
DKS1113 Electric Circuits
04/09/23 39/43
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
DKS1113 Electric Circuits
04/09/23 40/43
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
DKS1113 Electric Circuits
04/09/23 41/43
Wye-Delta Transformations
Example 19: Transform the circuit from Δ to y. Answer R1=18, R2=6, R3=3.
DKS1113 Electric Circuits
04/09/23 42/43
Wye-Delta Transformations
Example 20: Determine Rab. Answer Rab=142.32.
DKS1113 Electric Circuits
04/09/23 43/43
Wye-Delta Transformations
Example 21: Determine Io.
DKS1113 Electric Circuits