Professor Wan Wang g Youyi, PhD. Office: S2-B2b-47 Email: eyywang@ntu.edu.sg emes er
Pr escr ibe bed d Te T ext and Re Referenc nce es • – "Electric Circuits", by James W Nilsson & Susan A , th , ,
•
REFERENCES – "Engineering Circuit Analysis", by William H Hayt, Jack E Kemmerly & Steven M Durbin, McGraw Hill, 7th Rev. Edition, 2007. – "Fundamentals of Electric Circuits" by Charles K Alexander & Matthew Sadiku, 3rd Edition, McGraw-Hill, 2006.
Pr escr ibe bed d Te T ext and Re Referenc nce es • – "Electric Circuits", by James W Nilsson & Susan A , th , ,
•
REFERENCES – "Engineering Circuit Analysis", by William H Hayt, Jack E Kemmerly & Steven M Durbin, McGraw Hill, 7th Rev. Edition, 2007. – "Fundamentals of Electric Circuits" by Charles K Alexander & Matthew Sadiku, 3rd Edition, McGraw-Hill, 2006.
Course Syllabus • – 6 lectures, 3 tutorials – Chapters 1-5 – 4 lectures, 2 tutorials – Chapters 6-8
•
nuso a
c rcu ts ts
– 4 lectures, 1 ½ tutorials – Chapters 9-10
•
Lapl La plac acee tr tran ansf sfor orm m te tech chni niqu ques es (L (LKG KG)) – 6 lectures, 3 tutorials
•
Netw Ne twor ork k fun funct ctio ions ns an and d tw twoo-po port rt ne netw twor orks ks (W (WYY YY)) – 6 lectures, 2 tutorials
Course Assessments •
– – – Closed book – 2 hours – – Conducted in exam hall at end of semester
•
uz– – Closed book – 30 minutes –
o
mu p e-c o ce ques ons
– Conducted in your registered tutorial group – Week 8 - the week following semester break –
a e up qu z • Only for those with valid reason/s and supporting document/s • To be completed within 1 week (i.e. by end of Week 9)
Basic Concepts •
, transmit and measure electric signals. ’ the mathematician’s tools for manipulating those models to produce the systems that meet practical needs.
•
Electrical systems are part of our lives.
•
rcu ana ys s ecomes paramoun mpor an s nce w e p n un ers an ng the behavior of circuit model and its ideal circuit component.
•
e e mos rus ra ng par s o c rcu ana ys s are o now ow o s ar problem and obtain a complete set of equations and organizing them in such a way as to appear manageable.
Basic Strategy
Read the problem statement slowly and carefully.
•
.
overcoming the common obstacles of starting a problem and managing the solution process •
Collect the known information.
Devise a plan.
Several of the steps may seem obvious, but it is the chronological order as well as the
.
success. •
The real key to success in circuit analysis is , , environment.
Determine if additional information s requ re . No Attempt a solution.
, learning from mistakes will always be a part of the process of becoming a .
No
Verify the solution. Is it reasonable or ex ected? Yes End.
Yes
EE2001 – L ECTURE 1 Voltage and Current (1.4) Ideal Basic Circuit Element (1.5) . Passive Sign Convention (1.5, 1.6) Circuit Elements (2.1, 2.2)
Voltage and Current • – Electric charge is the basis for describing all electrical phenomena – Charge is bipolar (positive and negative charges) – – Charge exists in discrete quantities which are integral multiples of the electronic charge 1.6022 x 10-19 Coulombs. –
•
In circuit theory, the separation of charge creates an electric force (voltage), and – The concepts of voltage and current are useful because they can be expressed quantitatively. –
.
•
Voltage is the energy per unit charge created by the separation of the positive and negative charges. It is expressed as follows:
dw v= dq •
where
v – voltage in volts w – the energy in joules q – the charge in coulombs
The electrical effects caused by charges in motion depend on the rate of charge flow and this is known as electric current. It can be ex ressed as follows:
dq i= dt
where
i – the current in amperes q – the charge in coulombs – time in seconds
•
These two equations are definitions for the magnitude of voltage and current .
•
The bipolar nature of electric charge requires that we assign polarity references to these variables.
Ideal basic circuit element • – it has only two terminals, which are points of connection to other circuit components; – current and/or voltage; and – it cannot be subdivided into other elements.
•
is volta e across terminals of the element – Polarity reference for voltage is indicated by plus and minus signs
i is current in the element
•
–
i
Positive
Negative
voltage drop from terminals 1 to 2 or voltage rise from terminals 2 to 1
voltage rise from terminals 1 to 2 or voltage drop from terminals 2 to 1
positive charge flowing from terminals 1 to 2 or nega ve c arge ow ng rom erm na s o
positive charge flowing from terminals 2 to 1 or nega ve c arge ow ng rom erm na s o
•
The flow of current is conventionally regarded as a flow of positive charges, although charge flow in metal conductors results from electrons with a negative . – Direction of current is opposite to the direction of electron flow
•
The assignment of the reference polarity of voltage and the reference direction for curren are en re y ar rary.
•
Example 1.1: – i1 is a current flowing from terminal ‘a’ to ‘b’ – i1 = - i2, i.e., the variable i2 represents the same current i1.
•
i1 b
a i2
Example 1.2: -
– Numeral values show the actual direction
•
Example 1.3: – For the element in the figure, v1 = 17V. Determine v2. – (Ans: v2 = -v1 = - 17 V)
+
-
v1
v2
-
+
Power and energy •
.
dw p= dt •
where
p - the power in watts, w- the energy in joules, and – the time in seconds.
Substitute in voltage and current definitions, the power equation is obtained as follows:
p=
=⎜ ⎟⎜ ⎟ dt ⎝ dq ⎠ ⎝ dt ⎠
= vi
where
p – the power in watts, v – the voltage in volts, and i – the current in amperes.
– Power is energy per unit of time and is equal to the product of terminal voltage and current
, able to tell whether power is being delivered to the pair of terminals or extracted from it (i.e. to/from element/s connected at the terminals).
Passive sign convention • an element is in the direction of the reference voltage drop across the element, use a positive sign in any expression that relates the voltage and the current. Otherwise use a ne ative si n.
•
Following passive sign convention: – If the power is positive (p > 0), power is being absorbed by the circuit inside the box. – I t e power is negat ve (p < ), power is eing del vered rom t e circuit insi e t e box.
•
Example 1.4: a
uppose t at we ave se ecte t e po ar ty re erences as s own n . and v = -10 V, then the power associated with the terminal pair 1,2 is
ven =
p = - (-10)(4) = 40 W Thus, the circuit inside the box is absorbing 40 W
.
i = -4 A and v = 10 V, then the power is
p = - (10)(-4) = 40 W Thus, the circuit inside the box is still absorbing 40 W
•
Example 1.5: – A high-voltage direct-current (HVDC) transmission line between Celilo, Oregon and y mar, a orn a s opera ng a an carry ng . a cu a e e power a Oregon end of the line and state the direction of power flow.
– At Celilo, it is non-passive sign convention
p = −vi = − ( 800 )(1.8 ) = −1440 MW – Power is negative, and hence power is delivered from Celilo, Oregon – Therefore, direction of power flow is from Celilo to Sylmar – Note: At Sylmar, it is passive sign convention
p = vi = ( 800 )(1.8) = 1440 MW •
•
Try Chapter Problems 1.12 to 1.15
Circuit elements •
. – An active element is one that is capable of generating energy. Examples are batteries, voltage sources and current sources. –
.
,
capacitors and inductors.
•
Independent and dependent Sources – An independent source establishes a voltage or current in a circuit without relying on voltages or currents elsewhere in the circuit. – A dependent source establishes a voltage or current whose value depends on the value of a voltage or current elsewhere in the circuit.
•
Ideal independent voltage and current sources – n ea vo age source s a c rcu t e ement t at ma nta ns a prescr e vo tage across its terminals regardless of the current flowing in those terminals. – An ideal current source is a circuit element that maintains a prescribed current .
•
Examples of ideal independent sources a) A time varying independent voltage source ) A constant in epen ent vo tage source
v(t)
V
i(t)
c) A time varying independent current source d) An independent voltage source e) An independent current source
•
c
a
Common practices but not always adhere to – Lower case for time varying (AC, square wave), e.g. v and i.
vs
is
– Upper case for constant (DC), e.g. V. – Subscripts used to differentiate one source from another
(d)
(e)
•
Ideal dependent (or controlled) voltage and current sources – Output is determined by a voltage or curren a a spec c oca on n a c rcu . – A dependent source is unilateral, i.e. the output variables have no influence on the in ut variables. a)
Voltage-controlled voltage source •
is dimensionless
b)
Current-controlled voltage source • ρ is in ohms (Ω)
c)
Voltage-controlled current source • α is in mhos ( Ω-1)
d)
Current-controlled current source •
β is dimensionless
Series and Parallel Connections • series and parallel connections – Those connected in series share the same current –
•
Example 1.6:
– Resistors 50 Ω and 10 Ω are in series – Resistors 20 Ω and 30 Ω are in arallel • They are also in parallel to the series combination of 50 Ω and 10 Ω resistors
– Resistors 40 Ω and 70 Ω are in series with the 60 V voltage source • They are also in series to the parallel combination of the remaining resistors
•
Note: A parallel connection of voltage sources or a series connection of current sources is forbidden unless the sources are pointing in the same direction and have .
Invalid
•
ry
Invalid
Valid
Valid
Invalid
Invalid
apter ro ems . an
. Valid
v1
v1+v2
v1
v2
v1=v2
2
v1
v1-v2
v
v
v2
Volta e sources in series i1
Volta e sources in arallel
i1=i2
i1
i2
i1+i2
i
i1
i2
i1-i2
i2
R i
EE2001 – L ECTURE 2 Ohm’s Law (2.2) Kirchhoff’s Laws (2.4, 2.5) . , . , .
Ohm’s L aw •
, specifically, the flow of electric charge. The circuit element used to model this behavior is the resistor .
•
’ and it is expressed as follows:
=
where
v - the voltage in volts, , R - the resistance in ohms.
– Express voltage as function of current
•
For the purpose of circuit analysis, we must reference the current in the resistor to the terminal voltage. – We can do so in two ways, according to passive sign convention or otherwise
•
Ohm’s law equivalent - express current as function of voltage
v = = R
where
v - the voltage in volts, - t e current n amperes, G - the conductance in mhos.
– Conductance is simply the inverse of resistance. • It is represented by the symbol G. • The unit of conductance is Siemens (S). Sometimes it is also known as mho ( Ω-1).
• • •
ower ca cu a on a
1 S = 1 A/V = 1 Ω-1
e erm na s o a res s or can e expresse
p = vi when v = iR
and
p = -vi when v = -iR
p = vi = (iR)i = i2R ,
and
p = -vi = -(-iR)i = i 2R
n severa ways:
or or
= v2/R which is inde endent of the olarit reference •
Note: regardless of voltage polarity and current direction, the power at the terminals of a resistor is always positive. –
,
.
•
Example 2.1: In each of the circuit, calculate the value of v or i and determine the power dissipated in each resistor. a
ass ve s gn conven on va = 1 × 8 = 8 V p = 12 × 8 = 8 W
b) Passive sign convention ib = 50 × 0.2 = 10 A p = 502 × 0.2 = 500 W c) Non passive sign convention vc = -1 × 20 = -20 V = (-20) 20 = 20 W d) Non-passive sign convention =d= p = 502 / 25 = 100 W
•
Example 2.2: a. What value of vg is required in order for e n erconnec on o e va
ib = - 8 A Æ vg = 0.25 (- 8) = - 2 V b. For this value of vg, find the power the 8A source.
p = vg i = -2 × 8 = -16 W (delivered)
•
Example 2.3: a) What value of α is required in order for the interconnection to be valid?
vx = -25 V and αvx = -15 A Æ α = 0.6 b) For the value of α calculated in part (a), find the power delivered/absorbed associated with the 25 V source.
p = vi = (vx)(αvx) or 25 × 15 = 375 W (absorbed)
•
ry
ssessment ro ems . an
. , an
apter ro ems . an
.
K irchhoff’s L aws •
’
.
– However, when it is coupled with Kirchhoff’s two laws, we have sufficient, powerful set of tools for analysing a large variety of electric circuits.
’
•
K irchhoff’s Voltage L aw (K VL ). •
Constructing a circuit model: – with each resistor – Label current of voltage source – of an individual element – A node is where two or more elements meet: e.g. a, b, c, and d
•
7 unknowns re uires 7 inde endent e uations – Ohm’s law gives 3 of the equations
v1 = i1 R1 vc = ic Rc l
=
l
l
The remaining 4 equations come from applying Kirchhoff’s law
node a
is − i1 = 0
node b i1 + ic = 0
− ic − il = 0 node d il − is = 0 node c
K irchhoff’s Current L aw (K CL ) •
. N
∑i
n
=0
where N is the number of branches connected to the node and in is the nth current entering (or .
– By this law, currents entering a node may be regarded as positive, while currents leaving the node may be taken as negative or vice versa.
•
Example 2.4: Consider the node in the figure below. – Applying KCL gives
i1 + (-i2) + i3 + i4 + (-i5) = 0
i2
i1
or -i
+i + -i + -i + i =
or
i4
i5 i3
i1 + i 3 + i 4 = i 2 + i 5 =
K irchhoff’s Voltage L aw (K VL ) • a circuit is zero M
v =0
m=1
•
where M is the number of voltages in the loop (or the number of branches in the loo and is the mth voltage.
Example 2.5: Consider a single-loop circuit shown below. – Applying KVL yields
+ v2 -
+ v3 -
-v1 + v2 + v3 – v4 + v5 = 0 v4
v1
or
v1 - v2 - v3 + v4 - v5 = 0 or
- v5 +
v2 + v3 + v5 = v1 + v4 – which may be interpreted as Sum of voltage drops = Sum of voltage rises
•
Example 2.6: – Applying KCL, node a node b node c node d
•
i1 + i4 – i2 – i5 = 0, i2 + i3 – i1 – i b – ia = 0, i b – i3 – i4 – ic = 0, i5 + ia + ic = 0.
Example 2.7: – Applying KVL, path a
-v1 + v2 + v4 – v b – v3 = 0,
path b
-va + v3 + v5 = 0, v b – v4 – vc – v6 – v5 = 0, -va – v1 + v2 – vc + v7 – vd = 0.
path c path d
•
Example 2.8: The currents i1 and i2 in the circuit shown are 20A and 15A respectively. a
n e power supp e voltage source.
2Ω
y eac
b) Show that the total power supplied
i5
i1 8Ω
230V
4 Ω i4
in the resistors. 260V (a) – V2 – V1 + V0 = 0 ⇒ V0 = V1 + V2 ⇒ V0 = 20(8) + 16(15) = 400V ⇒ i0 = 400/80 = 5A ⇒ i5 = 20 + 5 = 25A ⇒ i3 = 15 + 5 = 20A ⇒ i4 = 20 – 15 = 5A
16Ω 2Ω i2
+ V1
i0
80Ω
+ V0
+ V2 -
i3
==P260V = -260(20) = -5200W (delivering) (b) ΣPabs = (25)2(2) + (20)2(8) + (5)2(4) + (15)2(16) + (20)2(2) + (5)2(80) = 10950W ΣPdel = 5750 + 5200 = 10950W ∴ ΣPdel = ΣPabs = 10950W
.
.
-
Circuit with dependent sources •
0
– 3 unknowns of iΔ , i0, and v0 – Knowing any 2 of the unknowns rd
•
Solving for the currents –
- - - ,
– KCL at node b,
Δ
0
i0 = iΔ + 5iΔ = 6iΔ L ( 2 )
– Solving by substituting (2) into (1) for i0,
iΔ = 4A i0 = 24A
⎫ ⎭
0
0
•
Example 2.9: Find the total power developed in the circuit shown if Vo = 100V. App ying KVL aroun Loop A yie s -60 + V1 = 100 ⇒
V1 = 160V ig
Likewise around Loop B
-
2+
80 + 100 = 0
⇒
2=
180V 60V
Applying KCL at Node 1 yields
⇒ ig = 3iΔ = 12A
Pdel = 180(4) + 100(8) + 60(12) = 2240 W – CHECK:
•
Σ Pabs = 160(12) + 80(4) = 2240W
Try Chapter Problems 2.28 to 2.30
80V Loop A
and we know that iΔ= 4A.
-ig + iΔ +2iΔ = 0
+ V1 -
4A
iΔ
+ V2 -
2iΔ Loop B
+ V0 -
Operational amplifier (op-amp) •
-
-
– Output voltage is a function of the difference between the input voltages
– If a positive voltage is supplied to the non-inverting input, op-amp yields a positive output . –
,
-
This op-amp circuit is known as inverting op-amp. – Output voltage is constrained to a value between the negative and positive power supply voltage.
•
Our interest is on terminal behavior of the op-amp – Black-box approach to its operation as affected by external circuit connections – I nore the internal structure and volta e/current exce t that it im oses constraints on the terminals
.
Op-amp terminals ⎧−VCC A( vp − vn ) < −VCC ⎪ ⎪ v0 = ⎨ A( vp − vn ) −VCC ≤ A( vp − vn ) ≤ +VCC ⎪⎩+VCC
•
A( vp − vn ) > +VCC
Actual op-amp – Input resistance between positive and negative input terminals is very large (105 to 1012 Ω) – Amplification factor (gain) is very high, ranges from 105 to 107 Ω – Output resistance is very low, ranges from 1 to 50 Ω
•
Ideal o -am – Input resistance is infinite – Amplification factor is infinite – Out ut resistance is zero
Analyzing ideal op-amp •
– Since input resistance is infinite, –
i p = in = 0
, approaches infinity, the voltage across the input terminals must simultaneously become infinitesimally small,
vp − vn ≈ 0 or
•
vp ≈ vn
If the gain is infinite, it is impossible to control the output – In actual application, voltage control is accomplished through feedback via a resistor which feeds the out ut si nal back to the invertin in ut
•
Example 2.10: Given va = 1 V and vb = 0 V, find vo – Bot input termina s ave same vo tage,
vn = vp = vb = 0 V L (1) – Input currents are zero,
i25 + i100 = in = 0 A L 2 – Taking note that, 25
=
i100 =
( va − vn ) 25k
( vo − vn )
= =
1 25
vo
mA
– Substituting into (2),
1
•
+
vo
= 0 ⇒ vo = −4 V
Try Chapter Problems 5.3, 5.5 and 5.6
EE2001 – L ECTURE 3 Resistors in Series and Parallel (3.1, 3.2) Voltage Division and Current Division (3.4) - -
.
Resistors in Series • •
Applying KVL around the whole loop
− vs + is R1 + is R2 + is R3 + is R4 + is R5 + is R6 + is R7 = 0 or vs = is ( R1 + R2 + R3 + R4 + R5 + R6 + R7 ) = is Req eq
•
1
2
3
4
5
6
7
For k resistors connected in series, the equivalent single resistor has a resistance equal to the sum of the k resistances k
eq
i i =1
1
2
L
k
Resistors in Parallel • •
Applying KCL at node a, s =
•
=
vs R2
vs R3
vs = R4
s
⎛1
1
1
1 ⎞
⎝ R1
R2
R3
R4 ⎠
=
For k resistors connected in parallel,
1
Req •
vs R1
=
k i =1
1
Ri
=
1
R1
+
1
R2
+L +
1
Rk
It is more convenient to use conductance for parallel-connected resistors eq
=
k i i =1
=
1
2
L
k
vs Req
•
For two resistors in parallel,
1
Req •
=
1
1
R1
R2
=
R1 + R2 R1R2
eq =
R1R2 R1 + R2
Example 3.1: Find the equivalent resistance Rab 7Ω
a 15Ω
30Ω
24Ω
30Ω
Req1 =
20Ω
1
Req2
b
a
b
= eq2
7Ω
a eq1
(15)(30)
eq2
∴ Rab = eq1
b
eq2
1 24
+
1 30
= 10Ω +
1 20
=
1 8
= (10)(7 + 8) 10 + (7 + 8)
= 6Ω
•
Example 3.2: Find the equivalent resistance between terminals a-b. 10Ω 48Ω
15Ω
b
Req = 6Ω
18Ω
15Ω
b
.
Req
48 + (10 + 6)
∴ Rab =
= 12Ω
(15)(12 + 18)
18Ω
,
a) the voltage v, b) the power delivered to the , c) the power dissipated in the 10 Ω resistor (a) Work out the equivalent resistance, (64)(6 + 10) (30)(7.2 + 12.8) Req1 = = 12.8 Ω ⇒ Req = = 12 Ω 64 + 6 + 10 30 + 7.2 + 12.8
= 10Ω
a) With Req, v = 5 × 12 = 60 V b) For the 5A source, the 5A current and voltage v is of non-passive sign convention, =-
=-
=-
c) The power dissipated can be found by finding the current through the 10 Ω resistor,
i7.2Ω = 5 × i
= 3×
30 + 20 64
+
= 3A = 2.4A
p10Ω = 2.42 × 10 = 57.6W •
Try Chapter Problems 3.5 and 3.6
V olta tag ge Di Divi vis sion •
– direct proportion to their resistances – the larger the resistance, the larger the voltage drop across it.
v
•
R1
R2
+ v1 -
+ v2 -
v1 =
R1
R2
v
In ge gen ner eraal, if a vol volta tage ge di divi vide derr has has N resistors (R , R , …, R ) in series with the source voltage v, the nth resistor (Rn) will have a voltage drop of
Rn R vn = v= n v R1 + R2 + L + RN Req •
v ; v2 =
Req =
N
R i =1
Cond Co ndit itio ion n to ap appl ply y vo volt ltag agee di divi visi sion on - the resistors must have the same current
•
Example 3.4: Find v1, v2, and v3 in the circuit shown below. – F rst com ne t e 10Ω an 15Ω n para e
Req =
10(15)
+ v1 -
= 6Ω
40V
15Ω
+ v3 + -
– Then apply the voltage division,
v1 =
14
v2 = v3 = •
=
14 + 6
Common mistake
6
(40) = 12V
•
Example 3.5: a) Find Find no-l no-loa oad d vol volta tage ge v0 ) F n v0 w en RL s 150 Ω c) How How muc much h powe powerr is diss dissip ipat ated ed in in the 25 k Ω resistor if the load .
75
a) v0 =
(200) = 150 V
( 75) (150 ) b) Req = = 50 kΩ ( 75 + 150 ) v0 = c)
50 25 + 50
p25kΩ =
2
25k
(200) = 133.3 V
= 1.6 W
Current Division •
The principle of current division – the total current i is shared by the resistors in inverse proportion to their resistances – the lower the resistance, the higher the current through it R2 i v
•
i1
i2
i1 =
R1
R2 R1 + R2
i ; i2 =
R1 R1 + R2
i
In general, if a current divider has N resistors (R1, R2, …, RN) in parallel th , n
in =
v n
=
Req n
i ;
1 eq
N
=∑
1
i =1
i
•
Condition to apply current division - the resistors must have the same voltage
•
or using conductance,
in =
Gn G i = n i ; Geq = ∑ Gi G1 + G2 + L + GN Geq i =1
•
Example 3.6: Find i1 through i4 in the circuit shown below. 10Ω i4
i2 20Ω
We first combine resistors in parallel, 40Ω
30Ω
R
1
3
20A
=
Req2 =
10(40) 10 + 40 20(30) 20 + 30
Using the current division rule:
i +i =
8 8 + 12
20 = 8 A ⇒ i =
i2 =
20 50 30
8 = 3.2 A
(8) = 4.8 A 50 12 10 i3 + i4 = (20) = 12 A ⇒ i3 = (12) = 2.4 A 8 + 12 50 40 i4 = (12) = 9.6 50
= 8Ω = 12Ω
•
Practice problems
6Ω
3Ω
(a) Find the equivalent resistance R ab. .
(Ans: R ab = 10 Ω)
45Ω
40Ω
ns:
•
o=
; vo =
Try Chapter Problems 3.21, 3.22 and 3.23
12Ω
1Ω 15Ω
b For the circuit shown use current division to find i0, and use voltage division to find vo.
a b
5.2Ω
5Ω
Delta-to-WyeEquivalent Circuits •
, (delta) networks. – Delta interconnections • , , m • R3 , Rs , and Rm – Wye interconnections • R1 , R3 , and R • R2, Rs , and Rm
•
Delta is also called i π – Δ can be shaped into π without disturbing electrical equivalence
•
Star is also called tee (T) – Y can be shaped into T without disturbin electrical e uivalence
Delta-WyeTransformation •
, resistance at each respective pair of terminals must be equal while the third terminal is open-circuited.
Rab =
Rc ( Ra + Rb ) a
Rbc = Rca =
b
= R1 + R2
c
Ra ( Rb + Rc ) Ra + Rb + Rc Rb ( Rc + Ra ) Ra + Rb + Rc
= R2 + R3 = R3 + R1
•
Straightforward algebraic manipulation of previous equations gives values for Yconnected resistors in term of Δ-connected resistors and vice versa.
R1 =
RbRc Ra + Rb + Rc
Ra =
R2 =
c
Rb =
3
•
a
Ra + Rb + Rc Ra Rb Ra + Rb + Rc
c
R1R2 + R2 R3 + R3 R1 R1 1
2
2
3
3
1
R2 R1R2 + R2 R3 + R3 R1 R3
Generally, the equations used for star-delta transformation can be expressed in the o ow ng orms:
RY = RΔ =
Product of the two nearest Δ branches Sum of Δ resistances Sum of cyclic products of two Y branches Resistance of furthest Y branch
•
For the balanced case,
R1 = R2 = R3 = R Y R Y = •
1 3
RΔ
and
Ra = Rb = Rc = RΔ
and
RΔ = 3RY
Example 3.7: Consider the circuit shown , find the input resistance seen by the source. – Four transformations are possible for the abcd network: • Δabc, • Δbcd, • Y at b, and • Y at c.
a 2Ω
2Ω 1Ω
1Ω
2Ω
c 1Ω
a
(a) Δabc transformation to Y
Ran = Rbn = Rcn =
2× 2
= 0.8Ω
2 ×1 2 + 2 +1
× 2 + 2 +1
0.8Ω
n
= 0.4Ω
0.4Ω
v
c
= 0.4Ω
=
1Ω
(a)
(c) Combining resistances between nodes n and d
.
2Ω
1Ω
(b) Summing up resistances in paths
.
0.4Ω
a
2.4 × 1.4
2.4 + 1.4 = 0.884Ω
a 0.8Ω
0.8Ω
n
n
v
The input resistance,
v 2.4Ω
Rinput = 0.8 + 0.884 + 1
1.4Ω
0.884Ω
= 2.684Ω .
d
. (b)
(c)
EE2001 – L ECTURE 4 Node Voltage Method or Nodal Analysis (4.2, 4.3, 4.4)
Planar Circuits •
Planar circuit
Non-planar circuit
Descriptions of a circuit Name
Definition
Examples
node (n)
A point where 2 or more elements join
a, b, c, d, e, f, g
essential A node where 3 or more
b, c, e, g
(ne)
•
To solve be unknowns – neequa ons – Needs be - (ne-1) KVL equations −i1 + i2 + i6 − I = 0
− 3− 5 = i3 + i4 − i2 = 0 1
v1 = i1 R1 + i2 R5 + i3 ( R2 + R3 ) v2 = −i3 ( R2 + R3 ) + i4 R6 + i5 R4 0 = −i2 R5 + i6 R7 − i4 R6
path
A set of elements may be traversed in order without passing the same node twice
many
branch
A path that connects 2 nodes
v1, v2, R1, R2, R R R R R7, I
A path that connects 2 essential nodes without
v1-R1, R2-R3, v2-R4, R5, R6,
b essential branch
7,
e
essential node loop
A closed path
many
mesh
A loo that does not enclose any other loops
-R -R -R -R v2-R2-R3-R6-R4, R5-R7-R6, R7-I
Examples of circuit analysis solutions •
a
e=
– denoted by a, b, c
•
5 essential branches (be = 5) –
•
-
,
,
,
i1
b
i3
i2
i4
an
(ne-1) or 2 KCL equations at nodes a and b,
c
i1 − i2 − i3 = 0
i2 + 2 − i4 = 0 •
Since 1 branch current is already given (i.e. 2A), we need to have ((be-1)-(ne-1)) or 2 KVL equations,
10 − i1 − 5i3 = 0 5i3 − 2i2 − 10i4 = 0 • – Solution gives currents through all of the branches • i1 = 0.91A, i2 = -0.91A, i3 = 1.82A, i4 = 1.09A – •
vac = 9.09V, vbc = 10.91V, vab = -1.82V
Cramer’s method of solving set of equations • •
•
After substituting for i4 as i4 = i2+2, 3 equations are left for i1, i2 and i3 To find i1 1 =
•
•
Δ
=
and i2, i3
i2 =
•
N1
0
−1
−1
10
0
5
Δ
=
Determinant:
i1 + 5i3 = 10
−12i2 + 5i3 = 20 0 − 0 − (−50) + ( −100) + 120 − 0
70
1
−12 5 −1 −1
1
0
5
0
−12
5
0
−1
1
−1
0
1 10
5
1
0
10
0
20
5
1
−1
−1
1
0
5
0
−12
5
20
1
N2
i1 − i2 − i3 = 0
=
0 − ( −60) − ( −5) + 0 + 12 − 0
− 20 = 50−100 = −0.91 ; i3 = 77
N3
Δ
=
=
77
= .
−12 20 1 −1 −1 0 1
0
5
0
−12
5
= 12077+ 20 = 1.82
a1
a2
a3
b1
b2
b3 = ab 1 2 c3 − ab 1 3 c2 − a2 bc 1 3 + a2 b3 c1 + a3 bc 1 2 − a3 b2 c1
c1
c2
c3
Read Appendix A.2, A.3 and A.4
Nodal Analysis or Node-Voltage Method • on the circuit diagram. •
Select one node to be a reference node (e.g node connected to the greatest .
•
Define a voltage between each remaining node and the reference node ((N-1) voltages defined in an N-node circuit).
• •
pp y
a eac o
e no es, resu ng n a se o
-
no e equa ons.
Following the same method, formulate a set of independent equations. Solve the equations simultaneously. 1
•
With node 3 as the reference, voltages at nodes 1 and 2 are:
v1 = v13 v2 = v23
2
+
+
v13
v23 -
3
-
•
Example 4.1: For the circuit shown, compute the voltages across each . – 3 essential nodes – Setting one as the reference node
ig1 1A
– Apply KCL at 2 remaining nodes, giving two equations involving v1 and v2. – KCL @node 1:
ig1 − i1 − i2 = 0 ⇒ ig1 − – KCL @node 2:
i2 − i3 − ig2 = 0 ⇒
v1
Node 1
R2=8Ω
Node 2
i2 i1
R1=8Ω
R3=16Ω
i3
ig2 -2A
Node 3 (reference node)
v1 v1 − v2 − = 0 ⇒ − 2v1 + v2 = − 8 L (1) R1 R2
v1 − v2 v2 − − ig2 = 0 ⇒ 2v1 − 3v2 = −32 L ( 2 ) R2 R3
– Solve for v1 and v2 by first adding (1) and (2), – Substitute for v2 in (1),
v2
−2v2 = −40 ⇒ v2 = 20V
−2v1 + 20 = −8 ⇒ 2v1 = 28 ⇒ v1 = 14V
•
Example 4.2: Use node voltage method on similar circuit in previous chapter –
oose o om no e as re erence
– KCL @node 1:
v1 −
+
1
KCL @node 2:
v1 5
+
v1 − v2 2
=0
17v1 − 5v2 = 100 L (1) – From (2), –
.
v2 = ,
v2 − v1
+
v2
−2=0
2 10 −5v1 + 6v2 = 20 L ( 2 )
5v1 + 20 6
77 700 700 ⎛ 5v1 + 20 ⎞ v v 100 = ⇒ = ⇒ = = 9.09V 1 1 ⎟ 6 6 77 ⎝ 6 ⎠
17v1 − 5 ⎜
v2 =
.
= 10.91V
•
6 Node voltage method is a simpler solution with 2 instead of 3 equations
•
Try Chapter Problems 4.9, 4.10
Node voltage method with dependent sources •
,
-
supplemented with constraint equations – Constraint equations are imposed by the dependent sources
•
Example 4.3: Find power dissipated by the 5 Ω resistor in the circuit below.
– KCL equations:
− 2
v2 − v1
+ +
20
v2
+
+
−
5 v2 − 8iφ
= 0 ⇒ 0.75v1 − 0.2v2 = 10 L 1 = 0 ⇒ − 0.2v + 0.8v = 4i
L 2
– The constraint equation, –
u .
φ
no
– Subsequently,
iφ =
v1 − v2 5
an so ve or v1 an v2,
iφ =
16 − 10 5
=1.2A
v1 =
⇒
; v2 =
p5Ω = iφ 2 × 5 = 7.2W
•
Homework: Make constraint equation as the 3rd equation and use Cramer’s method to solve for iφ
•
Try Chapter Problems 4.19 and 4.20
Node voltage method with voltage sources •
, through voltage sources are not known. – Need to introduce a new variable i to denote the current throu h the unknown voltage source – Note that v1=50V and thus left with 2 KCL equations at nodes 2 and 3:
v2 − 50 5
+
v2 50
+i = 0 ; −i+
v3 100
−4=0
– Constraint equations due to the 10iφ voltage source and it being dependent:
v3 = v2 + 10iφ
and
iφ =
v2 − 50 5
– 4 equations to solve for the 4 unknowns
•
Introduction of a new variable requires an additional equation to solve
•
Instead of introducin a new variable we can a
l Su ernode conce t
Supernodeconcept • KCL, enclose the voltage sources and consider each enclosed surface as a generalized node or supernode. •
.
,
’
law and supernode to solve circuit analysis problem. •
Principle of supernode - a supernode is formed by enclosing a (dependent or elements connected in parallel with it.
•
A supernode may be regarded as a closed surface enclosing the voltage source and .
•
KCL @ supernode, v2 − 50 + v2 + v3 − 4 = 0 L 1 ()
5
50
100
•
Constraint equation due to supernode, v3 = v2 +
•
Constraint equation from dependent source control variable, iφ =
•
Simplify to a set of 3 equations and solve for v2, v3 and iφ
1400
= v2 =
50 1 1
iφ =
•
o e:
L v2 − 50
0
−1 10 0 −5
0
v2 − v3 + 10iφ = 0 v2 − 5iφ = 50
1
φ
=
7000 + 500
+ +
= 60V
−1 10 0 −5
60 − 50
= 2A ; v3 = 60 + 10 ( 2 ) = 80V
5 ewer equa on o so ve w en us ng uperno e concep
L ( 3)
•
Example 4.4: Analyze the circuit with a supernode as shown. –
4Ω
superno e,
i1 + i4 = i2 + i3 v1 − v2
v1 − v3
v2
v3
2
4
8
6
10V
– Constraint equation,
v2 = v3 + 5 L 2 •
Noting that v1 = 10V, (1) becomes
5v2 •
+
5v3
=
15
L ( 3)
Using (2) and (3),
5 ( v3 + 5)
5v3
i4
i1 v1
15
8 12 2 v2 = 4.2 + 5 = 9.2V
3
.
su ernode
v2
5V
v3
2Ω
i2
8Ω
6Ω
i3
Properties of a Supernode • solve for the node voltage. • •
A supernode has no voltage of its own. superno e requ res
e app ca on o
o
an
.
– KCL to relate to external circuit elements – KVL to relate the voltages of the two combined nodes
•
i 7V
(Ans: v = -0.2V, i = 1.4A)
•
Try Chapter Problems 4.26 and 4.27
3V
4Ω
Practice problem: Find v and i in the circuit shown.
3Ω
v -
2Ω
6Ω
Summary – Node Voltage Method or Nodal Analysis 1. Select one node to be a reference nod e 2. Define node voltages . Resulting in a set of (N-1) node equations. The equation variables are node voltages Three cases: 1. Normal circuits (without neither dependant sources nor supernode) 2. The circuit with dependant sources 3. Supernode Constraint equations (KVL)
EE2001 – L ECTURE 5 Mesh Current Method or Mesh Analysis (4.5, 4.6, 4.7) Source Transformation (4.9) .
Mesh Current Method or Mesh Analysis • element more than once as a mesh (or a loop). • •
Assume mesh current that flows around a mesh in a chosen direction (clockwise). c rcu e emen may e presen n w
. – The total current through the element is the algebraic sum of the mesh currents (taking direction into account).
•
pp y
an
m’s aw an o a n se o mes equa ons.
•
Solve the equations simultaneously to determine the unknowns.
•
Examples of meshes in a planar circuit – Each circuit element can be in two meshes at the maximum – Branch currents are defined in term of mesh , . .
iR = i1 1
iR = i1 − i2 6
R8
= 2−
3
•
To determine branch currents through all resistors. – be = 3, ne = 2 Î 1 KCL and 2 KVL equations
i1 = i2 + i3 v1 = i1 R1 + i3 R3
−
2
=
2
2
−
3
3
– Need to solve for a set of 3 equations
•
Applying mesh-current method – Identify the (be-(ne-1)) meshes and define mesh currents (with direction) –
or mes
a
v1 = ia R1 + ( ia − ib ) R3 – KVL for mesh ib
−v2 = ( ib − ia ) R3 + ibR2
i1 = ia
– Only need to solve a set of 2 equations for the mesh currents ia and ib •
e ranc curren s can e expresse
n erms o
e mes curren s,
i2 = ib i3 = ia − ib
•
Example 5.1: Use mesh-current method to find power delivered/absorbed by the 80 V source, and power dissipated . ic
a
•
Mesh-current equations: – Mesh ia,
80 = 5 ia − ic + 26 ia − ib
⇒ 31ia − 26ib − 5ic = 80 L (1) –
b,
b
−
a
+
b
−
+
c
b
=
⇒ − 26ia + 124ib − 90ic = 0 L ( 2 ) –
c,
c
−
a
c
c
−
b
=
⇒ − 5ia − 90ib + 125ic = 0 L ( 3)
b
– Using Cramer’s rule, 80
ia =
i =
•
−5 −
−90 125 31 −26 −5 −26 124 −90 −5 −90 125 0
31
80
−26 −5
0
−5 −90
0
125
−
− −26 124 −90 −5 −90 125 31
ic =
−26
− −5
=
592000 118400
= 5A
= −400W (delivered) p5Ω = ( ia − ic ) × 5 = 45W
=
296000
= 2.5A
p30 Ω = ic2 × 30 = 120W p26 Ω = ( ia − ib ) × 26 = 162.5W 2
p90 Ω = ( ib − ic ) × 90 = 22.5W
−26 80
−90 0 31 −26 −5 −26 124 −90 − −
p80V = −80 × ia
=
236800 118400
= 2A
Try Chapter Problems 4.31 and 4.32
p8Ω = ib × 8 = 50W
Mesh-current method with dependent sources •
sources and/or dependent sources? – If the circuit contains dependent sources, the mesh current equations must be su lemented with the constraint e uations im osed b the resence of the dependent sources.
•
Example 5.2: Use the mesh analysis to determine the power dissipated in the 4 Ω resistor in the circuit shown below. – Mesh-current equations,
50 = 5 ( i1 − i2 ) + 20 ( i1 − i3 ) 0 = 5 ( i2 − i1 ) + 1( i2 ) + 4 ( i2 − i3 ) 0 = 20 ( i3 − i1 ) + 4 ( i3 − i2 ) + 15iφ – Constraint equation imposed by presence of 15iφ source,
iφ = i1 − i3
•
Substitute iφ into the mesh-current equations,
25i − 5i − 20i = 50 L 1
−5i1 + 10i2 − 4i3 = 0 L ( 2 ) −5i1 − 4i2 + 9i3 = 0 L ( 3) •
To find power dissipated by 4Ω resistor, we compute mesh currents i2 and i3. 0
−20 −4
0
9
25 50
i2 =
−5 −5
−5 −20 −5 10 −4 −5 −4 9 25
25
i3 =
•
=
= 26A
125
2
p4 Ω = ( i3 − i2 ) × 4 = 16W
−5 50
−5 10 0 −5 −4 0 25 −5 −20 −5 10 −4 −5 −4 9
=
3500
= 28A
.
.
Mesh-current method with current sources •
, some additional manipulations.
•
Example 5.3: Find the mesh currents ia, ib, and ic. – either mesh ia or ic, we must introduce into the equations the unknown voltage across the 5 A current source. Thus, – Mesh ia: 100 = 3 ( ia − ib ) + v + 6ia – Mesh ic:
L (1)
v = 2 ( ic − ib ) + 50 + 4ic L ( 2 )
– Combine (1) and (2) to eliminate v, 50 = 9ia − 5ib + 6ic – – Taking note that:
,
=
b
−
a
b
b
−
c
−ia + ic = 5 L ( 5 )
– Solve (3), (4) and (5) Î ia = 1.75 A, ib = 1.25 A, ic = 6.75A
L( 3) L
Supermesh concept • common element; the current source is in the interior of the supermesh. •
Apply KVL and Ohm’s law in meshes without the current source. Then solve the .
•
In Example 5.3, we can derive equation (3) w ou n ro uc ng e un nown vo age variable v by using supermesh concept. –
emove e curren source rom equa on y simply avoiding it when writing the meshcurrent equations
– KVL around the su ermesh
100 = 3 ia − i
+ 2 ic − i + 50 + 4ic + 6ia
– This equation simplifies to equation (3), which is solved with (4) and (5)
•
.
.
•
Taking time to look carefully at a circuit to identify shortcuts such as the above supermesh provides a big payoff in simplifying the analysis.
•
In genera , we use no a or mes ana ys s met o , w c ever s s mp er an appropriate – what the question asks for: voltages or currents?)
•
Practice problems (a) Use the nodal analysis method to find the power delivered/absorbed y e vo age source. Ans: P50V = - 150 W (delivered) (b) Use the mesh analysis method to find the mesh current ia. Ans: 15 A
Summary – Mesh Current Method 1. Select meshes 2. Define mesh currents . Resulting in a set of (be-(ne-1)) equations. The equation variables are mesh currents Three cases: 1. Normal circuits (without neither dependant sources nor supermesh) 2. The circuit with dependant sources 3. Supermesh Constraint equations
Useful circuit analysis techniques •
voltage and mesh-current methods 3Ω
•
•
or examp e, you are g ven e following circuit, are there any other alternative(s) to determine the voltage
+
vo
2Ω
-
Some of the common techniques for circuit simplification –
er es-para e re uct ons
– Δ-to-Y transformations – Source transformations –
even n an
– Superposition
orton equ va ents
5Ω
8A
20V
Source transformations • parallel with the same resistor provided vs = is × R
•
Example 5.4: For the circuit shown, find the power associated with the 6 V source.
– There are several approaches to calculate the power associated with the 6V source. However, all the approaches will focus on finding just one branch current in the 6V source. As a result, we first simplify the circuit by using source transformations.
•
Note we must reduce the circuit in a way that preserves the identity of the branch containing the 6V source. There is no need to preserve the identity of the branch . – Starting with this branch, we can transform the 40V source in series with 5Ω resistor to 8A source in parallel with 5Ω resistor, and so on.
i6V =
19.2 − 6 4 + 12
= 0.825A ⇒
p6V = 6 × 0.825 = 4.95W (absorbed)
•
Example 5.5: Determine the equivalent circuit between the two terminals in figure below via source transformation.
1A
2Ω
2Ω
1A
3Ω
1Ω 2A
6V
2A 2Ω
3A
2Ω
3A
1Ω
1Ω
1A
0.75Ω
3V
1Ω 3V 0.75Ω
1.75Ω 3.75V
OR 2 1/7A
0.75V
•
Try Assessment Problem 4.15 and Chapter Problem 4.59
1.75Ω
Ignoring independent sources •
, sources. We can do that by turning them off (“killing” them). – Turning off independent voltage source, Vs = 0 Î replacing it with a short-circuit –
,
s=
1.6Ω
36A
1.6Ω 20Ω 120V
60V
36A
6Ω
8Ω
5Ω urn ng o a vo age sources
8Ω
1.6Ω
5Ω omp e e c rcu
6Ω
20Ω 120V
60V 5Ω
6Ω
8Ω
Thevenin and Norton equivalents •
, pair of terminals.
•
or examp e: en we p ug an app ance n o an e ec r c y ou e , we are interested primarily in the voltage and current at the terminals of the appliance. We have no interest in the effect that connecting the appliance has on voltages and . , focus on the behavior of the circuit supplying the outlet, but only at the outlet terminals.
• Thevenin and Norton Equivalents are circuit simplification techniques that
focus on terminal behavior and thus are extremely valuable aids in circuit .
Thevenin’s theorem • with a resistor such that the current-voltage relationship at the terminals is unchanged. •
, circuit (a) can be replaced by an equivalent circuit (b) consisting of a volta e source V in series with a resistor R Th,
I
+
Linear two-terminal circuit
V
– R Th is the input or equivalent resistance at the terminals when the internal n epen ent sources are turne o .
R Th
Load
-
(a)
– V is the o en-circuit volta e at the terminals
a
I
b
a +
Th
oa -
b
Norton’s theorem •
’ ’ circuit is an independent current source in parallel with a resistor.
•
It states that a linear two-terminal circuit of a current source I N in parallel with a resistor RN,
I
+
Linear two-terminal circuit
– I N is the short circuit current through the terminals
V
Load
-
(a)
b
I
– the terminals when the internal independent sources are turned off.
+
IN
•
a
Note: Thevenin’s and Norton’s equivalent circuits are related by source transformation. N
=
Th
;
N
=
V R Th
V
RN
-
(b)
b
Load
EE2001 – L ECTURE 6 Deriving equivalent circuits (4.10, 4.11) Maximum power transfer (4.12) .
Method A for determining equivalent circuit •
, (a) Thevenin's equivalent resistance RTh with independent sources killed (i.e. short-circuit all voltage sources and open-circuit all current sources). or the short-circuit current (isc = voc/RTh).
•
=
oc
sc
. ’ , the terminals a-b in the circuit shown below. 6Ω
12V
6Ω
2A
a 4Ω
1Ω
b
– Ignoring the 1Ω resistor, determine an equivalent circuit to represent the circuit to its left (i.e. simplifying the circuit while retaining the 1Ω resistor).
•
Solving for R Th (kill all the independent sources), 6Ω
6Ω
4Ω
•
Solving for V Th,
R Th = 12 || 4 =
R Th
6Ω
(12 )( 4 ) 12 + 4
6Ω +
2A
6Ω
+
4Ω V Th
2A
4A
4Ω V Th
6Ω
6Ω
-
6Ω
3Ω + 4Ω
4+6+6
a
Th
-
V Th =
= 3Ω
24 = 6V
b
•
Example 6.2: Consider the circuit below, find I o using Norton’s theorem. 6k Ω
6V
3k Ω
2k Ω
2mA
3k Ω
o
– Note: the 2k Ω resistor is considered as a load, which is ignored when deriving the e uivalent circuit.
•
Solving for RN 6k Ω
3k Ω
RN = 6kΩ || ( 3kΩ + 3kΩ ) RN
=
6kΩ + 6kΩ
= 3k Ω
•
Solving for I N 6k Ω
6V
•
3k Ω
IN
2mA
3k Ω
IN =
6V 6kΩ
+
3kΩ 3kΩ + 3kΩ
( 2mA )
Connect back the 2k Ω resistor to the Norton’s equivalent circuit and solve for I o – The Norton equivalent circuit is drawn to the left of the 2 k Ω resistor. Using current division, the current I o in the 2 k Ω resistor is
2mA
3kΩ
2k Ω o
Io =
3k Ω 3kΩ + 2kΩ
( 2mA ) = 1.2mA
Method B for determining equivalent circuits •
, equivalent by calculating: (a) Terminal open-circuit voltage voc and short-circuit current isc. Th
•
=
oc sc.
Example 6.3: Find Vo for the circuit shown below. 2Vx
x
2Ω
4Ω
3A
+
+ Vo -
– The 2Ω resistor is the load and not included in equivalent circuit.
•
Find the open-circuit terminal voltage after the 2 Ω load is removed.
V 4Ω +
24V
3A
VX = −3 × 4 = −12V Voc + 24 + 2VX + VX = 0 ⇒ Voc = 12V
+ Voc -
•
Find the short-circuit current. 2Vx
Vx 4Ω
24V
I I sc
24 + 2VX + VX = 0 ⇒ VX = −8V 3A
I=
VX 4
= −2A
I SC = I + 3 = 1A
•
Find R Th and form the Thevenin’s equivalent circuit Th =
V 12 = = I sc 1
;
Th
=
oc
=
– (Note: This is not equal to 4 Ω since the presence of the dependent source has altered e e ec ve res s ance o e g ven ne wor .
•
Connect the Thevenin equivalent circuit to the 2 Ω load resistor and solve for Vo
R Th = 12Ω Voc = 12V
2Ω
+ Vo -
Vo =
2 2 + 12
(12 ) =
12 7
V
Method C for determining equivalent circuit • (a) Apply a 1 A current source at the terminals and calculate the terminal voltage. Since = , = o o= (b) Apply a 1 V voltage source at the terminals and calculate the terminal current i. Take R = 1/i.
•
Example 6.4: Find the Thevenin’s equivalent of the circuit shown. – Since the rightmost terminals are already open-circuited, i = 0. Consequently, the dependent source is dead, so voc = 0.
3Ω
.
i
– We apply a 1 A source externally, then using the nodal analysis,
v− 1.5i 1.5i
2Ω
+ v -
3 1A
∴
v
+ − = 2
v v
+
an
=−
= 0.5 ⇒ v = 0.6V
∴ R Th = 0.6Ω
– We can also apply a current source of iin externally, then using the nodal analysis,
1.5i
2Ω
+ v -
iin
•
Summary: – Method to use depends on which is convenient and simpler – Method B can be generally applied to all circuits – Method C is only useful for determining the equivalent resistance, Methods A or C are necessary o e erm ne e even n’ s vo age or or on’ s curren – Only independent sources can be “killed” when determining equivalent equ ivalent resistance
• •
ea
xamp es .
an
.
Try Tr y Cha Chapt pter er Pr Prob oble lems ms 4. 4.63 63,, 4.6 4.67 7 and and 4. 4.71 71
M axi xim mum po powe werr tr tra ans nsffer •
, transfer takes place when the load resistance RL = R T Th h. – Proof:
PL = i 2 RL = ⎜
2
VTh
⎟ RL
⎝ R Th + RL ⎠
–
2
⎛ ( R Th + RL ) − RL ⋅ 2 ( RTh + RL ) ⎞ dPL 2 = V Th ⎜ ⎟ =0 4 ⎜ ⎟ dRL ( RTh + RL ) ⎝ ⎠ 2
– This results in the following, 2
( R Th + RL ) = 2 RL ( RTh + RL ) ⇒ RL = RTh – The maximum power is:
2
PL ,max
⎛ V Th ⎞ VTh2 =⎜ ⎟ RL = 2 RL
4 RL
•
Exampl Exam plee 7: 7: For For th thee circ circui uitt sho shown wn,, fin find d the the va valu luee of of RL that results in maximum power being transferred transferred to RL and the value of this maximum power. – Find Thevenin’s equivalent circuit,
( 30 ) (150) Th
V Th =
30 + 150 150 150 + 30
( 360 ) = 300V
– Power transfer is maximum when
R = R = 25Ω – The maximum power: 2
L ,max
=⎜ 25 = 900 W ⎟ ⎝ 50 ⎠
– What percentage does this maximum power correspond to the total power delivered by
vab =
V Th 2
= 150V ⇒ i360V = ∴
•
L ,max
P360V
=
360 × 7
360 − 150 30
= 7A
=35.71%
Try Assessment Problem 4.21 and Chapter Problem 4.79
Superposition •
proportional) to its input.
•
A linear circuit is defined as any circuit that has only independent sources, linear , . – Homogeneity (scaling) property: v = i R Î k v = k i R – Additive property: v1 = i1 R ; v2 = i2 R Î v = (i1 + i2) R = v1 + v2 , node may be calculated as the sum of the contributions of each independent source acting alone. – (p = i v).
•
Basic rules of applying superposition theorem – – An ideal voltage source is “killed” or removed by replacing it with a short circuit. – An ideal current source is “killed” or removed by replacing it with an open circuit. –
.
Steps to apply superposition principle .
. current) due to that active source using nodal or mesh analysis. – Independent voltage sources are replaced by 0 V (short circuit) –
2. 3.
Repeat step 1 for each of the other independent sources. Find the total contribution by adding algebraically all the contributions due to .
•
Note: Dependent sources are left intact because they are controlled by circuit var a es.
•
Superposition reduces a complicated problem to several easier problems, each containing an independent source. – There is no advantage in using superposition to solve a network with dependent sources because the dependent source is never killed.
•
Example 6.6: Using superposition principle, find the currents through the resistors i1, i2, i3 and i4.
•
First, find the currents resulting from the 120V voltage source – Kill 12A current source b re lacin it with an o en-circuit – Apply nodal analysis to solve for the currents,
v1 − 120 6
⇒ i1' = i2' =
v1
120 − 30 30 3 ' 4
+
v1
3 2+ 4 v1 = 30V
= 15A
= 10A
i =i = ' 3
+
30 2+4
= 5A
=0
•
Secondly, find currents resulting from 12A current source – Kill 120V voltage source by replacing it with a short-circuit – App y no a ana ys s to so ve or t e currents,
v3 6
+
v3 3
v4 − v3 2
+
v3 − v4
+
v4
2 4
v3
= 0 L (1)
v4
+ 12 = 0 L ( 2 )
– Solving equations (1) and (2),
v3 = −12V ; v4 = −24V i1 = − i" =
3
6
=−
v3 − v4 2
−
=
= 2A ; i2 =
6 −12 − ( −24 ) 2
= 6A
3
3
=
− 3
i" =
= −4A v4 4
=
−24 4
= −6A
