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Chapter 2. Basic Laws

ENGG 1008Dr. K. K. Y. Wong

HKUEEE

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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 short circuit; R=∞ 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

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• 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

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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

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• 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

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• 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

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• 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

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• 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

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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),

– This is voltage division formula

• In general, for N resistors in series,

1 21 1 2 2

1 2 1 2 ; R Rv iR v v iR v

R R R R= = = =

+ +

1 2 ...eq NR R R R= + + +

1 2 ...n

nN

Rv vR R R

=+ + +

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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 2

1 1 1 1...eq NR R R R

= + + +

• 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

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• From the figures in the previous slide, i1=v/R1 and i2=v/R2• But since v = iReq= iR1R2/(R1+ R2), we have

• 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

• In general, for N parallel conductors (G1, G2,…, GN),

2 11 2

1 2 1 2; R i R ii i

R R R R= =

+ +

1 21 2

1 2 1 2; G i G ii i

G G G G= =

+ +

1 2 ...n

nN

Gi iG G G

=+ + +

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• 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

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• 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

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• 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

– Therefore, Geq=10S

1 1 1 1 1 1 1 1 1 1|| || || ||6 5 8 12 6 5 20 6 4 10eq

R ⎛ ⎞ ⎛ ⎞= + = + = = Ω⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

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• 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:

– Similarly, io can be obtained in two ways• Ohm’s law: vo=3io => io=4/3A• Current division formula:

– Power dissipated in the 3Ω resistor is po=voio=4(4/3)=5.333W

2 (12V) 4V2 4o

v = =+

6 6 42A= A6 3 6 3 3o

i i= =+ +

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• 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,

– vo=9000i1=180V

1

2

18000 (30mA) 20mA9000 18000

9000 (30mA) 10mA9000 18000

i

i

= =+

= =+

– 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

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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

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• 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

• Similarly,

1 3( ) (1)b a cac

a b c

R R RR R RR R R

+= + =

+ +

1 2

2 3

( ) (2)

( ) (3)

c a bab

a b c

a b cbc

a b c

R R RR R RR R RR R RR R RR R R

+= + =

+ ++

= + =+ +

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• Subtracting (3) from (1), we get

• Adding (2) and (4) gives

• Subtracting (4) from (2) yields

• Subtracting (4) from (1), we obtain

• Shortcut for memorizing the conversion rule: – 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

1 2( ) (4)c b a

a b c

R R RR RR R R

−− =

+ +

1 (5)b ca b c

R RRR R R

=+ +

2 (6)c aa b c

R RRR R R

=+ +

3 (7)a ba b c

R RRR R R

=+ +

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• 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

• Dividing (8) by each of (5), (6) and (7) leads to

1 2 2 3 3 1 2( )

( )

= (8)

a b c a b c

a b c

a b c

a b c

R R R R R RR R R R R RR R R

R R RR R R

+ ++ + =

+ +

+ +

1 2 2 3 3 1

1

1 2 2 3 3 1

2

1 2 2 3 3 1

3

(9)

(10)

(11)

a

b

c

R R R R R RRR

R R R R R RRR

R R R R R RRR

+ +=

+ +=

+ +=

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• 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

1

2

3

(10)(25) 515 10 25

(25)(15) 7.515 10 25

(15)(10) 315 10 25

b c

a b c

c a

a b c

a b

a b c

R RRR R R

R RRR R R

R RRR R R

= = = Ω+ + + +

= = = Ω+ + + +

= = = Ω+ + + +

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• 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

– The equivalent circuit is shown in Fig. a)

1 2 2 3 3 1

1

1 2 2 3 3 1

2

1 2 2 3 3 1

3

(10)(20) (20)(5) (5)(10) 3510

350 17.520

350 705

a

b

c

R R R R R RRR

R R R R R RRR

R R R R R RRR

+ + + += = = Ω

+ += = = Ω

+ += = = Ω

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– Combining the three pairs of resistors in parallel, we obtain

– 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.

(70)(30)70 || 30 2170 30(12.5)(17.5)12.5 ||17.5 7.29212.5 17.5

(15)(35)15 || 35 10.515 35

= = Ω+

= = Ω+

= = Ω+