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Electric circuits-chapter-2 Basic Laws
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Transcript of Electric circuits-chapter-2 Basic Laws
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Chapter 2Basic Laws
04/09/23
DKS1113 Electric Circuits
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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.
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Ohm’s Law
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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)
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Ohm’s Law
v=iR
*pay careful attention to current direction
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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
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Ohm’s Law
Conductance (G) Unit mho or Siemens (S).
Reciprocal of resistance R
G = 1 / R
Has the ability to conduct electric current
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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.
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Ohm’s Law
Example 1: Determine voltage (v), conductance (G) and power
(p) from the figure below.
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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.
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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
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Nodes, Branches & Loops
Example 3: Determine how many branches and nodes for the
following circuit.
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Nodes, Branches & Loops
5 Branches 1 Voltage Source 1 Current Source 3 Resistors
3 Nodes a b c
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Nodes, Branches & Loops
Example 4: Determine how many branches and nodes for the
following circuit.
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Kirchoff’s Laws
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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
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Kirchoff’s Laws
Example 5: Given the following circuit, write the equation for
currents.
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Kirchoff’s Laws
Example 6: Current in a closed boundary
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Kirchoff’s Laws
Example 9: Use KCL to obtain currents i1, i2, and i3 in the circuit.
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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
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Kirchoff’s Laws
Example 10: Use KVL to obtain v1, v2 and v3.
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Kirchoff’s Laws
Example 11: Use KVL to obtain v1, and v2.
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Kirchoff’s Laws
Example 12: Calculate power dissipated in 5Ω resistor.
10
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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
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Series Resistors & Voltage Division
Voltage Division:
Previously: v1 = iR1 & v2 = iR2 i = v/(R1+R2 )
Thus: v1=vR1/(R1+R2) v2=vR2/(R1+R2)
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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 )
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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 )
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Conductance (G)
Series conductance: 1/Geq = 1/G1 +1/G2+…
Parallel conductance: Geq = G1 +G2+…
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Voltage and Current Division
Example 13: Calculate v1, i1, v2 and i2.
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Voltage and Current Division
Example 14: Determine i1 through i4.
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Voltage and Current Division
Example 15: Determine v and i.
Answer v = 3v, I = 6 A.
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Voltage and Current Division
Example 16: Determine I1 and Vs if the current through 3Ω
resistor = 2A.
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Voltage and Current Division
Example 17: Determine Rab.
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Voltage and Current Division
Example 18: Determine vx and power absorbed by the 12Ω
resistor. Answer v = 2v, p = 1.92w.
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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
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Wye-Delta Transformations
Y network T network
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Wye-Delta Transformations
Δ network π network
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Wye-Delta Transformations
Delta (Δ) to wye (y) conversion.
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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
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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
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Wye-Delta Transformations
Example 19: Transform the circuit from Δ to y. Answer R1=18, R2=6, R3=3.
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Wye-Delta Transformations
Example 20: Determine Rab. Answer Rab=142.32.
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Wye-Delta Transformations
Example 21: Determine Io.
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