Chapter 2. Basic Laws
ENGG 1008
Dr. K. K. Y. Wong
HKUEEE
1
Ohm’s Law
• Ohm’s law states that the voltage v across a resistor is
directly proportional to the current i flowing through the
resistor
• Mathematically, v=iR, unit of R is Ω
• The proportional constant R, is called the resistance,
denotes the ability to resist the flow of electric current
• R=0 Ù short circuit;
R=∞ Ù open circuit
• Note: not all resistors obey Ohm’s law, but we will only
consider those obey Ohm’s law in this course
• Conductance is defined as G=1/R, unit of G is siemens (S)
• The power dissipated by a resistor p=vi=i2R=v2/R
2
• Example: Calculate the current i, the conductance G and the
power p.
–
–
–
–
–
i = v/R = 30/(5×103) = 6mA
G = 1/R= 1/(5×103) = 0.2mS
p = vi = 30(6 ×10-3) = 180mW
or p = i2R = (6 ×10-3)2 (5×103) = 180mW
or p = v2/R = (30)2/(5×103) = 180mW
• Example: A voltage source of 20sinπt V is connected across
a 5kΩ resistor. Find the current through the resistor and the
power dissipated.
– i = v/R= 20sinπt / (5×103) = 4sinπt mA
– p = vi= 80sin2πt mW
3
Kirchhoff’s Laws
• Basic definitions:
– 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
– A loop is any closed path in a circuit
– Two or more elements are in series if they exclusively share a single
node and consequently carry the same current
– Two or more elements are in parallel if they are connected to the
same two nodes and consequently have the same voltage across them
4
• Kirchhoff’s current law (KCL) states that
the algebraic sum of currents entering a
node (or a closed boundary) is zero
• e.g., in the figure, i1+(-i2)+i3+i4+(-i5)=0
• If we rewrite the above equation, we have
i1+i3+i4= i2+ i5
• Alternative form of KCL: The sum of the
currents entering a node is equal to the sum
of the currents leaving the node
• Note that KCL also applies to a closed
boundary
• A simple application of KCL is combining
current sources in parallel
– Apply KCL to node a => IT+I2=I1+I3
– or equivalently IT=I1 -I2 +I3
5
• Kirchhoff’s voltage law (KVL) states that
the algebraic sum of all voltages around a
closed path (or loop) is zero
• e.g., in the figure, -v1+v2+v3-v4+v5=0
• Rearranging terms gives v2+v3+v5=v1+v4
• Alternative form of KVL: Sum of
voltage drops = sum of voltage rises
• A simple application of KVL is
combining voltage sources
– Applying KVL in a), -Vab+V1+V2-V3=0
– or Vab=V1+V2-V3
• Note that before applying KCL or KVL,
we have to define what is +ve and what is
–ve. Once it is defined, we have to stick
with the convention
6
• Example: Determine vo and i in the circuit in Fig. a)
– First define voltage drop as +ve and voltage rise as -ve
– Apply KVL around the loop as shown in b). The result is
–12 +4i+2vo-4+6i=0
– Applying Ohm’s law to the 6Ω resistor gives vo= -6i
– Substituting the second eq. into the first one yields i=-8A
– vo= -6i = 48V
7
• Example: Find the currents and voltages in the circuit shown
in Fig. a)
–
–
–
–
–
Apply KVL to loop 1, we get –30+8i1 +3i2 = 0
Apply KVL to loop 2, we get -3i2 +6i3 = 0
Apply KCL to node a, i1 = i2 + i3
Solving the equations gives i1 = 3A, i2 = 2A, i3 = 1A
This also gives v1=24V, v2=6V, v3=6V
• Practice problem: Find the currents and voltages in the
circuit shown in Fig. below.
– Ans: v1=3V, v2=2V, v3=5V
– i1=1.5A, i2=0.25A, i3=1.25A
8
Series resistors and voltage division
• Consider a simple series circuit
• Apply KVL to the loop, we have
v = iR1+iR2
• Define Req= R1+R2 , then v = iReq
• Therefore, two resistor can be
replaced by an equivalent resistor
• Since i=v/(R1+R2),
v1 = iR1 =
R1
R2
v ; v2 = iR2 =
v
R1 + R2
R1 + R2
– This is voltage division formula
• In general, for N resistors in series, Req = R1 + R2 + ... + RN
Rn
vn =
v
R1 + R2 + ... + RN
9
Parallel resistors and current division
•
•
•
•
Consider a simple parallel circuit as shown
Apply KCL to node a, we have i=v/R1+v/R2
Therefore, i = v(1/R1+1/R2) = v/Req
In general, for N resistors in parallel
1
1
1
1
=
+
+ ... +
Req R1 R2
RN
• Special case 1: Two resistors in parallel, Req = R1R2/(R1+ R2)
• Special case 2: N identical resistors in parallel, Req=R/N
• It is often more convenient to use conductance rather than
resistance when dealing with resistors in parallel
• Since G=1/R, Geq= G1 + G2 + … + GN
10
• From the figures in the previous slide, i1=v/R1 and i2=v/R2
• But since v = iReq= iR1R2/(R1+ R2), we have
i1 =
•
•
•
•
R2i
R1i
; i2 =
R1 + R2
R1 + R2
This is the current division rule
Extreme case 1: R2=0 => Req=0
Extreme case 2: R2=∞ => Req=R1
If we divide both numerator and
denominator by R1R2, we have
G1i
G2i
i1 =
; i2 =
G1 + G2
G1 + G2
• In general, for N parallel conductors (G1, G2,…, GN),
Gn
in =
i
G1 + G 2 + ... + G N
11
• Example: Find Req for the circuit shown
in the right upper corner.
– The 6Ω and 3Ω resistors are in parallel, their
equivalent resistance is 6Ω || 3Ω =
(6)(3)/(6+3)=2Ω
– The 1Ω and 5Ω resistors are in series, their
equivalent resistance is 1+5=6Ω
– The circuit reduces to that in Fig. a)
– The 2Ω and 2Ω resistors are in series, their
equivalent resistance is 2+2=4Ω
– The 4Ω and 6Ω resistors are now in parallel,
their equivalent resistance is 4Ω || 6Ω =
(4)(6)/(4+6)=2.4Ω
– The circuit in Fig. a) can be replaced with
that in Fig. b)
– The three resistors are in series, and the
equivalent resistance is 4+2.4+8=14.4 Ω
12
• Example: Find the equivalent resistance
Rab in the circuit shown at the right upper
corner.
– Note that the 6Ω and 3Ω resistors are in
parallel, their equivalent resistance is 2Ω
– The 12Ω and 4Ω resistors are also in parallel,
their equivalent resistance is 12Ω || 4Ω =
(12)(4)/(12+4)=3Ω
– The 1Ω and 5Ω resistors are in series, their
equivalent resistance is 6Ω
– The original circuit can be replaced by that of
Fig. a)
– The 3Ω and 6Ω resistors are now in parallel,
giving an equivalent resistance 2Ω
– The 1Ω and the equivalent 2Ω are in series,
giving an equivalent resistance 3Ω
– The circuit reduces to that of Fig. b)
– Finally, Rab= 2Ω || 3Ω + 10Ω =11.2Ω
13
• Example: Find the equivalent
conductance Geq for the circuit in Fig a).
– 8S and 12S are in parallel, their equivalent
conductance is 8+12=20S
– This 20S is now in series with 5S as shown in
Fig. b), the combined conductance is
(20)(5)/(20+5)=4S
– This 4S is in parallel with 6S, hence
Geq=6+4=10S
– Note that Fig. a) is the same as Fig. c), where
the conductances are expressed in resistances
Req =
1 ⎛1 1 1 ⎞ 1 ⎛1 1 ⎞ 1 1 1
|| ⎜ + || ⎟ = || ⎜ + ⎟ = || = Ω
6 ⎝ 5 8 12 ⎠ 6 ⎝ 5 20 ⎠ 6 4 10
– Therefore, Geq=10S
14
• Example: Find io and vo in Fig. a) and the
power dissipated in the 3Ω resistor.
– The 6Ω and 3Ω resistors are in parallel, so their
combined resistance is 6||3=(6)(3)/(6+3)=2Ω
– The circuit reduces to Fig. b)
– vo can be obtained in two ways
• Ohm’s law: i=12/(4+2)=2A, and vo=2i=4V
• voltage division formula: v = 2 (12V) = 4V
o
2+4
– Similarly, io can be obtained in two ways
• Ohm’s law: vo=3io => io=4/3A
6
6
4
• Current division formula: io =
i=
2A= A
6+3
6+3
3
– Power dissipated in the 3Ω resistor is
po=voio=4(4/3)=5.333W
15
• Example: For Fig. a), determine vo, the
power supplied by the current source and
the power absorbed by each resistors
– The 6kΩ and 12kΩ resistors are inseries so
that their combined value is 6+12=18kΩ
– Fig. a) reduces to Fig. b)
– Apply the current division,
18000
(30mA) = 20mA
9000 + 18000
9000
i2 =
(30mA) = 10mA
9000 + 18000
i1 =
– vo=9000i1=180V
–
–
–
–
–
Power supplied by the source is po=voio=180(30)mW=5.4W
Power absorbed by the 12kΩ resistor is p=i22R=(10×10-3)2(12000)=1.2W
Power absorbed by the 6kΩ resistor is p=i22R=(10×10-3)2(6000)=0.6W
Power absorbed by the 9kΩ resistor is p=vo2/R=(180)2/9000=3.6W
Note that power supplied = power absorbed
16
Wye-Delta Transformations
• Situations often arise in circuits analysis
when the resistors are neither in parallel
nor in series
• e.g., the bridge circuit
• How do we combine R1 to R6?
• Many circuits of this type can be
simplified using three-terminal
networks:
– Wye (Y) or tee (T) network
– Delta (∆) or pi (Π) network
17
• Question: Suppose we identified a ∆ network, but it is more
convenient to work with a Y network, can we transform a ∆
network into a Y network and how?
• First superimpose a ∆ network with a Y network
• Rac(Y)=R1+R3, Rac(∆)= Rb||(Ra+Rc)
• Setting Rac(Y) = Rac(∆) gives
Rb ( Ra + Rc )
Rac = R1 + R3 =
(1)
Ra + Rb + Rc
• Similarly,
Rc ( Ra + Rb )
Rab = R1 + R2 =
Ra + Rb + Rc
(2)
Ra ( Rb + Rc )
Rbc = R2 + R3 =
Ra + Rb + Rc
(3)
18
• Subtracting (3) from (1), we get
Rc ( Rb − Ra )
R1 − R2 =
Ra + Rb + Rc
(4)
Rb Rc
• Adding (2) and (4) gives R1 =
Ra + Rb + Rc
(5)
Rc Ra
• Subtracting (4) from (2) yields R2 =
Ra + Rb + Rc
Ra Rb
• Subtracting (4) from (1), we obtain R3 =
Ra + Rb + Rc
• Shortcut for memorizing the conversion rule:
(6)
(7)
– Each resistor in the Y network is the product of the resistors in the two
adjacent ∆ branches, divided by the sum of the three ∆ resistors
19
• If we identified a Y network and want to transform it into a ∆
network, how can we do that?
• From (5), (6) and (7), we have
R1R2 + R2 R3 + R3 R1 =
=
Ra Rb Rc ( Ra + Rb + Rc )
( Ra + Rb + Rc )2
Ra Rb Rc
Ra + Rb + Rc
(8)
• Dividing (8) by each of (5), (6) and (7) leads to
R1R2 + R2 R3 + R3 R1
Ra =
R1
(9)
R1R2 + R2 R3 + R3 R1
Rb =
R2
(10)
R1R2 + R2 R3 + R3 R1
Rc =
R3
(11)
20
• Shortcut for memorizing the Y to ∆ conversion rule:
– Each resistor in the ∆ network is the sum of all possible products of Y
resistors taken two at a time, divided by the opposite Y resistor
• Special case:
– when R1 = R2 = R3 = RY or Ra = Rb = Rc = R∆
– Under these conditions, conversion formulas becomes R∆=3 RY or RY=
R∆/3
– The networks are said to be balanced
• Example: Convert the ∆ network in Fig. a) into a Y network
R1 =
Rb Rc
(10)(25)
=
= 5Ω
Ra + Rb + Rc 15 + 10 + 25
R2 =
Rc Ra
(25)(15)
=
= 7.5Ω
Ra + Rb + Rc 15 + 10 + 25
Ra Rb
(15)(10)
R3 =
=
= 3Ω
Ra + Rb + Rc 15 + 10 + 25
21
• Example: Obtain the equivalent
resistance Rab for the circuit and use it to
find i
– Convert the Y network comprising the 5Ω,
10Ω and 20Ω resistors by using R1=10Ω,
R2=20Ω and R3=5Ω
– Using (9), (10) and (11), we have
R R + R2 R3 + R3 R1 (10)(20) + (20)(5) + (5)(10)
Ra = 1 2
=
= 35Ω
R1
10
R R + R2 R3 + R3 R1 350
=
= 17.5Ω
Rb = 1 2
20
R2
R R + R2 R3 + R3 R1 350
=
= 70Ω
Rc = 1 2
5
R3
– The equivalent circuit is shown in Fig. a)
22
– Combining the three pairs of resistors in parallel, we obtain
(70)(30)
= 21Ω
70 + 30
(12.5)(17.5)
12.5 ||17.5 =
= 7.292Ω
12.5 + 17.5
(15)(35)
15 || 35 =
= 10.5Ω
15 + 35
70 || 30 =
– The equivalent circuit reduces to Fig. b)
– Rab=(7.292+10.5)||21=(17.792)(21)/(17.792+21)=9.632Ω
– i=vs/ Rab = 120/9.632 =12.458A
• Follow up challenge: Try starting with converting ∆ a-c-n to
an equivalent Y network and then obtain Rab. You should get
the same answer.
23