Ohm’s Law
Resistor
Materials with a characteristic
behavior of resisting the flow
of electric charge
Resistance: An ability to resist
the flow of electric current,
measured in ohm (Ω)
l
R=ρ
A
where ρ is the resistivity of the material in ohm-meters
l is the length in meters
A is the cross-sectional area in m2
Basic Laws
EEE110 Electric Circuits
Anawach Sangswang
Dept. of Electrical Engineering
KMUTT
2
Ohm’s Law
Ohm’s Law
Georg Simon Ohm (1787-1854: German)
Ohm’s law: the voltage across a resistor (R) is
directly proportional to the current (i) flowing
through the resistor
Mathematical expression: v ∝ i
or R = v
v = iR
Two extreme possible values of R: 0 (zero)
and ∞ (infinite) are related with two basic
circuit concepts: short circuit and open circuit.
Short circuit
Open circuit
i
Note: The direction of current (i) and the polarity
of voltage (v) must conform with the passive
sign convention
v = iR = 0
3
v
=0
R →∞ R
i = lim
4
Fixed resistors
Ohm’s Law
Variable resistors
Linear and nonlinear resistors
Conductance is the ability of an element to
conduct electric current; it is the reciprocal
of resistance R and is measured in mhos or
siemens.
G=
1 i
=
R v
5
Ohm’s Law
Example
The power dissipated by a resistor:
p = vi = i 2 R =
Note
6
v2
R
The power dissipated in a resistor is a nonlinear
function of either current or voltage
The power dissipated in a resistor is always
positive
The resistor always absorbs power and is a passive
element (incapable of generating energy)
7
Calculate the current i, the
conductance G and the
power p
The current
Power
i=
v
30
=
= 6 mA
R 5 × 103
The conductance
G=
1
1
=
= 0.2 mS
R 5 × 103
p = vi = 30(6 × 10−3 ) = 180mW
or
p = i 2 R = (6 × 10−3 ) 2 ⋅ 5 ×103
= 180mW
or
p = v 2G = 302 ⋅ 0.2 ×10−3
= 180mW
8
Branch, Nodes
Loop, Series, Parallel
A loop is a closed path in a circuit
An independent loop contains at least 1 branch
which is not a part of any other independent
loop or path
sets of independent equations
A network with b branches, n nodes, and l
independent loops satisfies b = l + n − 1
Series: 2 or more elements share a single node
and carry the same current
Parallel: 2 or more elements are connected to
the same two nodes and have the same
voltage across them
A branch represents a single element such as
a voltage source or a resistor
A node is the point of connection between
two or more branches
Original circuit
Equivalent circuit
9
Example
10
Kirchhoff’s Laws: KCL
Number of branches, nodes, series and
parallel connection
Kirchhoff’s current law (KCL) states that
“the algebraic sum of currents entering a
node is zero”
N
∑i
n
=0
n =1
Applying KCL: i1 + (−i2 ) + i3 + i4 + (−i5 ) = 0
Rearranging the equation
i1 + i3 + i4 = i2 + i5
“The sum of the currents entering a node is equal to
the sum of the currents leaving the node”
11
12
Kirchhoff’s Laws: KCL
Kirchhoff’s Laws: KCL
Example
KCL also applies to a closed boundary
The total current
entering the closed
surface is equal to
the total current
leaving the surface
13
Kirchhoff’s Laws: KVL
Example: Determine the current I
I + 4-(-3)-2 = 0
⇒I = -5A
We can consider the whole
enclosed area as one “node”.
This indicates that the
actual current for I is
flowing in the
opposite direction.
14
Kirchhoff’s Laws: KVL
Series-connected voltage sources
Kirchhoff’s voltage law (KVL) states that
“the algebraic sum of all voltages around a
closed path (or loop) is zero”
M
∑v
n
−Vab + V1 + V2 − V3 = 0
Vab = V1 + V2 − V3
=0
m =1
−v1 + v2 + v3 − v4 + v5 = 0
or
v2 + v3 + v5 = v1 + v4
Note:
2 different voltages cannot be connected in parallel
2 different currents cannot be connected in series
Sum of voltage drops = Sum of voltage rises
15
16
Kirchhoff’s Laws: KVL
Example
Example: Determine vo and i
Apply KVL around the loop
Determine the current i
−Va + V1 + Vb + V2 + V3 = 0
The Ohm’s law at the each
resistor gives
−12 + 4i + 2vo − 4 − vo = 0
V1 = IR1 , V2 = IR2 , V3 = IR3
The Ohm’s law at the 6-ohm
resistor gives vo = −6i
−Va + IR1 + Vb + IR2 + IR3 = 0
i = −8 A
−12 + 4i + 2(−6i ) − 4 − ( −6i ) = 0
KVL:
I=
vo = 48V
Va − Vb
R1 + R2 + R3
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Series Resistors
18
Series Resistors, Voltage Division
Two or more elements are in series if they are
cascaded or connected sequentially
and consequently carry the same current
The equivalent resistance of any number of
resistors connected in a series is the sum of
the individual resistances
N
Req = R1 + R2 + ⋅ ⋅ ⋅ + R N = ∑ Rn
n =1
The voltage divider can be expressed as
v = v1 + v2 = iR1 + iR2
v1 = iR1 =
i=
v
v
=
R1 + R2 Req
R1
R2
v, v2 = iR2 =
v
R1 + R2
R1 + R2
Req = R1 + R2
vn =
Rn
v
R1 + R2 + ⋅ ⋅ ⋅ + R N
Voltage divider
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20
Parallel Resistors, Current Division
Parallel Resistors, Current Division
Parallel connection: elements
are connected to the same
two nodes and consequently
have the same voltage across
them.
v
v
v = i1 R1 = i2 R2 or i1 = , i2 =
R1
R2
Applying KCL @ node a
i = i1 + i2 =
The equivalent resistance of a circuit with
N resistors in parallel is:
1
1
1
1
=
+
+ ⋅⋅⋅ +
Req R1 R2
RN
Current
v = iReq = i
 1
v
v
1 
+
= v + 
R1 R2
 R1 R2 
1
1
1
= +
Req R1 R2
RR
Req = 1 2
R1 +21R2
Example 2.12
i1 =
R2
v
=
i
R1 R1 + R2
i2 =
R1
v
=
i
R2 R1 + R2
Current
divider
22
Example 2.13
Find io, vo and calculate the power dissipated
in the 3-ohm resistor
Find vo, power supplied/absorbed by each
element
18k
30mA = 20mA
9k + 18k
9k
i2 =
30mA = 10mA
9k + 18k
i1 =
2
(12) = 4V
2+4
4
io = A
vo = 3io = 4V
3
R1 R2
R1 + R2
vo =
4
po = voio = 4   = 5.33W
3
23
vo = 9k ⋅ i1 = 18k ⋅ i2 = 180V
psource = voio = 180 ⋅ 30mA
= 5.4W
p9 k = (20mA) 2 (9k ) = 3.6W
p6 k = (10mA) 2 (6k ) = 0.6W
p12 k = (10mA) 2 (12k ) = 1.2W
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Wye-Delta Transformation
Wye-Delta Transformation
When the resistors are
neither in parallel nor in
series
Delta to wye
Delta-wye conversion
R1 =
Rb Rc
Ra + Rb + Rc
R2 =
Rc Ra
Ra + Rb + Rc
R3 =
Ra Rb
Ra + Rb + Rc
Wye-delta conversion
R12 (Y ) = R1 + R3
R1 + R3 =
Rb ( Ra + Rc )
Ra + Rb + Rc
R12 (∆ ) = Rb //( Ra + Rc )
R1 + R2 =
Rc ( Ra + Rb )
Ra + Rb + Rc
R2 + R3 =
R1 R2 + R2 R3 + R3 R1 =
Ra ( Rb + Rc )
Ra + Rb25+ Rc
Wye-Delta Transformation
R1 R2 + R2 R3 + R3 R1
R1
Rb =
R1 R2 + R2 R3 + R3 R1
R2
Rc =
R1 R2 + R2 R3 + R3 R1
R3
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Example
Wye-delta conversion
Ra =
Ra Rb Rc ( R1 + Rb + Rc )
Ra Rb Rc
=
2
( Ra + Rb + Rc )
Ra + Rb + Rc
Convert the delta network to wye network
27
R1 =
Rb Rc
10 × 25
250
=
=
= 5Ω
Ra + Rb + Rc 15 + 10 + 25 50
R2 =
Rc Ra
25 × 15
=
= 7.5Ω
50
Ra + Rb + Rc
R3 =
Ra Rb
15 × 10
=
= 3Ω
Ra + Rb + Rc
50
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