Unformatted text preview: ELEC 101 DC Circuits 1 DC Circuits
1 Basic Definitions A Circuit diagram
is a graphical representation of a circuit
(closed connection of elements). Node a (wire, or
contains no element) Example
+ v1 − Branch
voltage Node b R1 Node c i4
Branch
current R2 R3 R6 branch Physical crossing
(no connection,
not a node) 7 nodes, 9 branches (elements) 2009/10 Spring ELEC 101 DC Circuits 2
Node a (wire, or
contains no element) + v1 − Branch
voltage Node b R1 C 2009/10 Spring Node is the electrical joint connecting the terminals of two
or more elements. Node c i4
Branch
current a node
R2 R3 R6 branch Physical crossing
(no connection,
not a node) Not a node
(no electrical contact) Two (or more) nodes can be reduced to one node. 7 nodes, 9 branches (elements)
D B Branch Loop is any closed path through the circuit. No
node is crossed more than once, and the
beginning node is also the end node. One branch is one twoterminals element.
E.g. E Mesh is a loop that does not contain other loop.
It is the simplest loop. ELEC 101 DC Circuits 3 Example F 2009/10 Spring TwoTerminals (or OnePort) Network How many nodes ? i B 2 nodes: node A
and node B
5 branches (5 elements) Example i B Example
B A
A
B 5 elements are in parallel +
v
− Example How many branches ? +
v
− B
A G Resistive Circuit A circuit containing only resistances R,
voltage sources V, and current sources I. How many loops ?
How many meshes ? 4Ω Example
2A 4Ω 3 loops
2 meshes 3 nodes, 4 elements 4V ELEC 101 DC Circuits 4 2 Circuit Theorems A 2009/10 Spring Equivalence Example Two resistive oneport networks are
equivalent if they have the same currentvoltage (IV) curve across the two
terminals for ALL loads (or sources).
I2 I1
Network
A V1 load R Network
B V2
Example
load R If I1 = I2 and V1 = V2 for ALL load R,
then network A and network B are
equivalent. ( A ≡ B )
Hence a complex network (Network A) can be
replaced by a simple equivalent (Network B) and
network analysis can be simplified. Is network A equivalent to network B ?? network A 2Ω
4V network B I1 = 1A
V1 = 2V
R = 2Ω 6Ω I2 = 1A 8V
V2 =2V
R = 2Ω NO. I1 ≠I2 and V1 ≠V2 for other R.
For example, when R = 6 Ω,
I1 = 0.5A, V1 = 3V ; I2 = 2/3A, V2 = 4V ELEC 101 Example
2Ω DC Circuits 5
Network A equivalent to network B ? I1 network A 2Ω V2
4Ω 8V 4V When R = 0Ω (short) a I2 4Ω V1 2009/10 Spring R R 4V I1 I1
V1
8V R For example, when R = 2 Ω, R When R = ∞Ω (open) b 2Ω I1 = 1A, V1 = 2V ;
4V I2 = ?, not easy to calculate. 4Ω ∵ V 2 = 0V
8V
∴ I2 =
= 2A
4Ω ∵ V1 = 0V
4V
∴ I1 =
= 2A
2Ω ( YES. For ALL R , I1 = I2, V1= V2 ) I2 V2 network B network A
network B I2 4Ω I1 I2 4Ω V1
8V V2
4Ω V1 network A ∵ I1 = 0 A
∴ V1 = 4V R R network B
∵ I 2 = 0A ∴ V2 = 8V
× 4Ω = 4V
8Ω ELEC 101 DC Circuits 6 For network A c 2009/10 Spring d The I1V1 curve for all R is the line joining
R = 0 and R = ∞. The curve is a straight line
as the IV curve for a resistive circuit is
linear (can be proved later). For network B Similarly the I2V2 curve is a straight line. I1 I2
R=0 R=0 2A 2A
R=∞ 0 e 4V V1 Hence I1V1 curve = I2V2 curve.
Hence network A and network B are equivalent. R=∞ 0 4V V2 ELEC 101 DC Circuits 7 Series and Parallel B Example a. Series Connection
i +
v R1
R2 − is equivalent to
+
v
− v = i (R1 + R2 + ... + RN) = v = iR i i
R1 R2 RN − i= + ∴ = 50Ω 50Ω 60Ω v R i= Hence R = 25Ω 1
1
1
1
1
+
+
+ ..... +
=
R1 R 2 R 3
RN R v
R 50Ω 20Ω 30Ω + 20Ω = 50Ω 50Ω// 50Ω = 25Ω − v
v
v
v
+
+
+ ..... +
R1 R 2 R 3
RN 30Ω 30Ω 30Ω / /60Ω
30 × 60 1800
=
=
= 20Ω
30 + 60
90 b. Parallel Connection is equivalent
to 50Ω R ∴R1 + R2 + ... + RN = R + 30Ω Find the equivalent
resistance R. i RN hence v 2009/10 Spring 25Ω ELEC 101 Example DC Circuits 8 Example Find the equivalent resistance R. 1
S
50 1
S
50 1
S
50 1
S
50 2009/10 Spring Find the equivalent conductance G.
1
S
30
1
S
20 1
S = 0.02 S = 0.02 Ω −1 ≡ 50 Ω
50 is equivalent to 50 Ω 50 Ω
30Ω is equivalent to 25Ω Hence
R = 50Ω // 50Ω
= 25Ω 1
S
25 or G = 1 / R
= 0.02S + 0.02S
= 0.04S Hence
R = 30Ω + 20Ω
= 50Ω 1
S
50 20Ω or G = 1 / R
= 0.033S // 0.05S
= 0.02S ELEC 101 C DC Circuits 9 2009/10 Spring Kirchhoff’s Current Law Kirchhoff's Current Law (KCL) : at any time
instant, the algebraic sum of all currents
entering (or leaving) a node must be zero (or
total current flowing into a node = total
current flowing out of a node). a Example
Find i6 . i1 = 5A R1 R4 i8 = −2 A Gustav Robert Kirchhoff
(18241887) i6
a Apply KCL to node a
I flow into node a = I flow out of node a
Σ Iin = Σ Iout 5A = i6 – 2A
∴i6 = 7A
Proof of KCL b or algebraic sum of I = 0 Iin Σ (Iin + Iout ) = 0 Iout i1 − i6 − i8 = 0 5A − i6 − (− 2A) = 0
∴i6 = 5 + 2 = 7A Under conservation of charge and lumped
circuit approximation, charge can not be
stored in a node or element. Hence charge
flowing in = charge flowing out of lumped
device. ELEC 101 DC Circuits 10 I
Example Example a Find I. 5mA 10A Find I. 2009/10 Spring lumped element I 5A
10A
Circuit is not valid (violates
KCL), since at node a, Iin (5A)
is not equal to Iout (10A). Find I. Iout = Iin a
5A 10A 5A I
Example Apply KCL to node a, Iout = Iin
I = 5A + 10A = 15A a
10A
5A I 5mA Hence I = 5mA KCL can also be applied to
ANY lumped element
(the element is also called a super node). a Example Apply KCL to the lumped element, A B i
Questions:
1. Is node A = node B ?
2. Is i = 0 ?
3. VAB = ?
4. Is ohm’s law valid between AB ?
yes, yes or no, 0, yes I ELEC 101 D DC Circuits 11 2009/10 Spring Kirchhoff’s Voltage Law (KVL) Kirchhoff's Voltage Law (KVL) : at any time
instant, the algebraic sum of the branch
voltages around a loop (any closed path) is zero.
Example 2V + 4V − Find v6 . 3V − 5V + +
6V
−
+
v6
− + 4V − +
+
6V
2V
−
−
+
+
3V
v
− − 5V + −
6 Short Proof of KVL
Under conservation of energy and lumped
circuit approximation, electron gains zero
energy (ΣE = 0) or zero voltage.
(ΣV = ΣE / Q = 0) when moving back to A. Apply algebraic sum of V = 0 (Σ V = 0)
add clockwise voltages (or add
counterclockwise voltages) 3V + 2V – 4V – 6V – v6 – 5V = 0
Hence v6 = − 10V electron A ELEC 101 DC Circuits 12 10V Example
Find v. Example v Find VO. 5V + 10V − v = 0
Hence v = 15V 10V Find v. 5V v mesh a Voltage Divider Circuit
i 2 × R1 VIN i2 i2 R1 V = i2 ×RO
O
V
= IN ×RO
R1 + RO VO RO ∴VO = VIN × 36V ∴VO = 36V × Circuit is not valid (violates KVL), since
at mesh a, Σ V ≠ 0V (= 10V – 5V = 5V).
E 2Ω Example 5V ΣV=0 2009/10 Spring RO
R1 + RO 4Ω VO 4Ω
= 24V
2Ω + 4Ω ELEC 101 DC Circuits 13 2009/10 Spring Current Divider Circuit F 16V
2Ω IIN − IO IIN R1 IO VO RO R1
R1 + R O Example I1 = ?
I2 = ? 8A 2Ω
4A I1 4A −4A
24V 4A 8A 4Ω 2Ω
0V VO = IO RO = (IIN − IO ) R1 ∴ IO = I IN × 8A 12A I2 I3
2Ω 4Ω I3 = ? I1 = − 4 A (− current from 4 A source)
2Ω
2Ω
= 12 A ×
= 4A
2Ω + 4Ω
6Ω
I 3 = 12 A − 4 A = 8A
I 2 = (8A + 4 A ) × Check: 4A×4Ω = 2Ω×8A ELEC 101 Example DC Circuits 14 V1 i=?
V1 = ?
V2 = ? G i 2A 2009/10 Spring Dependent Source 2Ω 4V V2 i = 2A ( = current from 2A current source)
V2 = − i × 2Ω = − 4V (Ohm’s law) For dependent voltage or current source, the
voltage and current are linear function of the
variable v and i (a, b, c, d are constants). 4V + V1 + V2 = 0V (KVL) 2Ω Find Va,
i1 and i2. i2
4V 1Ω Example hence V1 = 0V Example I = ci + dv V = ai + bv 10V 3A i 2i 0V
i1 2Ω 2A
Va ( −4V, 2A, 0A) Can use KVL method to find out the answer (discuss later)
10V
10V 2Ω
1.4A
0V 7.2V 1Ω
3A 4.4A
2×1.4=2.8V ELEC 101 H DC Circuits 15 Ground (or Earth) (0V)
Symbol Voltage is the potential difference between two
nodes. To define node voltage and simplify
calculation, a common reference node (usually the
Earth) is defined as the 0V.
Example Find all V and I.
4Ω 2A 4V 0V
2A 2A 2A
4Ω −8V
4V −12V 4Ω
4V 12V
2A 2A
4Ω 4V
4V 0V 2009/10 Spring ELEC 101 I DC Circuits 16 2009/10 Spring Maximum Power Transfer
for maximum power
delivered to load by
source when RL = RS
maximum
power
PL ( max ) source
resistance
RS
source
voltage
VS ( VS / 2) 2 VS 2
=
=
RL
4R for maximum current
flow into load, RL = 0, VS
IL = VS/RS
IL
+ VL RS IL = VS/RS RL = 0 will study more in the loading effect and
op amp circuits −
load
RL IL
VS
RS If RL = 0, VL = 0, PL = VL×IL = 0.
If RL = ∞, IL = 0, PL = VL×IL = 0. RL = 0 RL = ∞ RS
for maximum voltage
across load, RL = ∞,
V L = VS VS 0
+ V L = VS − RL = ∞ VS VL ELEC 101 J DC Circuits 17 Nodal Analysis (or Node Voltage
Method, or KCL Method)
3Ω Example
Find Va,
Vb. 2009/10 Spring A general circuit can be very complicated
need to learn a systematic way to analyze circuit 2Ω 2A
Va 1Ω + Vb −1A − rear windshield defroster circuit
Method usually applied to circuit with more
current sources. It is usually a fast method.
a Define 0V node and other node voltages 0V node usually is the node connected to
most elements or negative terminal of V or I
sources.
b Find element (branch) currents in terms
of the node voltages (using ohm’s law)
can arbitrary assign the current directions c Apply KCL to the nodes d Find the unknown node voltages ELEC 101 DC Circuits 18 2Ω 3Ω
2A Va a Node a c −1A + 1Ω Vb Σ Iin = Σ Iout Vb Va − Vb Va − 0
+
2Ω
3Ω
V − Vb
V −0
= 1A + b
node b ∴ a
2Ω
1Ω
node a ∴ 2A = Node b
−1A 3Ω 1Ω
d 0V Va Va − Vb
2Ω Va − 0
3Ω 3Ω 0V
2A Va − 0
3Ω Vb − Va
2Ω 2Ω Vb 3Ω ∴ Va − 3Vb = 2V 5
∴ Va = V
2 1A
Va ∴ 5Va − 3Vb = 12V (2) Vb Vb − 0
1Ω or Solve for Va and Vb
(1) 2Ω 1Ω
2A Apply KCL to node a and b − 2Ω Va 2A b 2009/10 Spring 1Ω 2A 5
V
2 7A
6 1
Vb = V
6
2Ω 3Ω
1A 0 − Vb
1Ω 2A 5
A
6 1
V
6 1
A
6 1A −1A
1Ω 0V (1)
(2) ELEC 101 DC Circuits 19 2009/10 Spring Can solve the equations using 2 Substitution Method
Substitute Va = 2 + 3Vb into (1),
5(2 + 3Vb) − 3Vb = 12, 12Vb = 2, ∴ Vb = Va = 1
V
6 −3 12 −3 2 × (−3) − 12 × (−3)
=
1 −3
1 × (−3) − 5 × (−3)
5 −3 = (−6 + 36) / (−3 + 15) = 30 / 12 = 2.5V Can also solve for Va and Vb using Cramer’s
Rule (determinant method) ax + by = E
cx + dy = F Not mandatory in exam
V1 V2 V3 Eb
x= F d Ed − bF
=
,
ab
ad − bc
cd
aE y= c F aF − cE
=
a b ad − bc
cd 4 nodes, 3 unknown node voltages:
3 equations, 3 unknowns ELEC 101 DC Circuits 20 4Ω Va Example
Find Va. The 12V source can be considered as a node
(a supernode) and
I flowing into node = I flowing out of node. a 2A 4Ω 2009/10 Spring 4V 1. assign voltages to the nodes
2. express element currents in terms of node
voltages supernode Use nodal analysis (apply KCL to node a)
6 V − Va
2kΩ Iin = Iout
Va − 4V Va
V
+
= a − 1A
4Ω
4Ω 2Ω
∴ Va = 6V
∴ 2A = 2kΩ 2kΩ
6V Va
1k Ω Find IO.
12V 1kΩ −4V 2kΩ 1kΩ
Example Va+12V 12V Va 6V Va + 12 V − ( − 4 V )
2 kΩ IO 2kΩ 0V
3. apply KCL to the ‘supernode’ 2kΩ I
IO 2kΩ Iin = Iout −4V 6V − Va Va Va + 12V Va + 16V
=
+
+
2kΩ
1kΩ
2kΩ
2kΩ
22
Va + 12V
∴ Va = − V
∴ IO =
= 3.8mA
5
2kΩ
∴ The current (I) flowing into the 12V source
can not be expressed by the node voltages. ELEC 101 K DC Circuits 21 Mesh Analysis (or Mesh or Loop
Current Method , or KVL Method)
3kΩ Example
Find IO . 12V IO
3kΩ 2kΩ
6kΩ 3kΩ I1
12V
3V IO
mesh A I1 I1 I2 2kΩ 6kΩ
I2 3V mesh B
1kΩ 3kΩ I2 1kΩ Method usually applied to circuit with more
voltage sources.
a 2009/10 Spring b Find element voltages of mesh A I1 × 3kΩ I1 3kΩ Assign all element (branch) currents
can arbitrary assign the current directions b Apply KVL to the loops (meshes) d a (I1 − I2) ×6 kΩ 3kΩ Find the unknown element currents
Assign I1 and I2 to mesh A and B,
and therefore IO ( = I1 – I2 ) 6kΩ I1 Find element voltage in terms of
element current (from Ohm’s law) c 12V I1 − I2 = IO c I1 × 3kΩ Apply KVL to mesh A (clockwise) 12V − I1 × 3kΩ − (I1 − I 2 ) × 6kΩ − I1 × 3kΩ = 0
∴12V = I1 × 12kΩ − I 2 × 6kΩ ELEC 101 DC Circuits 22 I2 Similarly for
mesh B
I1 − I2 3V 1kΩ I1 − I2 +
(I − I ) ×6 kΩ
−
1 2 I2 I2 1kΩ − I ×1kΩ +
2 +
3V
− ∴ 3V = I1 × 6kΩ − I 2 × 9kΩ Solve the equations to find the currents
12V = I1 × 12kΩ − I 2 × 6kΩ
3V = I1 × 6kΩ − I 2 × 9kΩ 3− I 2Ω I 3A 5V Can not apply KVL to the 3A source ! Apply KVL to outer loop
(7 − I) × 3Ω (4 − I) × 2Ω I2 (I1 − I 2 ) × 6kΩ − I 2 × 2kΩ − 3V − I 2 × 1kΩ = 0 d 3Ω 2Ω 4V − 2kΩ 6kΩ 4A 4A 1Ω I 2 × 2kΩ + 4A Example
Find I. I2 ×1kΩ and − 1
3
I2 = mA ∴IO = I1 − I2 = mA
2
4 2kΩ 6kΩ or use + 5
∴I1 = mA
4 I 2 × 2kΩ (I1 − I2) ×6 kΩ 2009/10 Spring I ×1Ω
I
4V 4−I 7−I 3A (3 − I) × 2Ω
5V 4V + I ×1Ω − (4 − I) × 2Ω − (7 − I) × 3Ω − (3 − I) × 2Ω − 5V = 0
∴−36V = −8Ω× I ∴I = 4.5A ELEC 101 L DC Circuits 23 2009/10 Spring Principle of Superposition In a linear network, the current or voltage
at any point is equal to the sum of currents or
voltages contributed by each source (with the
other sources killed or inactive).
Linear
network (DC
and AC)
Noted that in DC network (network with constant
voltage or constant current sources), inductance
L behaves like a short circuit and capacitance C
behaves like an open circuit.
1 Linear network and superposition x2(t) a y1(t)
output input
Linear
Network A y2(t) then input x1(t) + x2(t) has output
y1(t) + y2(t) b Superposition Rule
x1(t) +
x2(t) c Linear
Network A y1(t)
+ y2(t) and input ax1(t) has output ay1(t). Linear Rule Linear networks have superposition properties.
If y1(t) is the output (response) of the
network due to an input x1(t) , and y2(t)
is the output due to an input x2(t), Linear
Network A x1(t) ax1(t) Linear
Network A ay1(t) Linear rule can be derived from superposition rule. ELEC 101 Kill
sources
(or
sources
inactive) DC Circuits 24 Replace current
source by open circuit Replace voltage
source by short circuit By definition, ideal current source with I = 0
is an open circuit 2009/10 Spring Example
Find all node
voltages. 4Ω VA = 6V , VB = VA ×
VC = VB × 2Ω VB 6V VC 2Ω
VD 4Ω / /4Ω
= 3V
4Ω / /4Ω + 2Ω 2Ω
= 1.5V , VD = 0 V
2Ω + 2Ω and voltage source with V = 0 is a short circuit Dependent voltage and current
sources can NOT be killed. 2Ω VA VA Example
Find all node
voltages. 2Ω 2Ω VB 12V From Linear Rule 4Ω 2x1(t) VC 2Ω 2y1(t) V source (input) × 2 (6V to 12V)
Hence all node voltages (output) × 2
In circuit analysis, superposition method is mainly
used to find the contribution of each source.
It is also used to reduce the number of sources and
hence simplify circuit analysis. VA = 12 V , VB = 6V
VC = 3V , VD = 0V
(and all element currents × 2) VD ELEC 101 Example x1 DC Circuits 25 superposition Linear
Network A y1 2Ω 18V 2Ω
4Ω x1 = 18V Linear
Network A y2 6V 2Ω
4Ω x2 = 6V
x1 + x2
Linear
Network A 4Ω 4.5V
2Ω y1 = 4.5V
2Ω x2 2009/10 Spring 1.5V 3Ω y2 = 1.5V 2Ω 36V
x1 = 36V y1 = 8V 4Ω 6A
3Ω 2Ω 8V −4V
2Ω y2 = −4V x2 = 6A y1 + y2
2Ω 24V 2Ω
4Ω 6V 4Ω 6A
3Ω 2Ω 36V
x1 + x2 = 24V y1 + y2 = 6V x1 + x2 = 36V+6A 4V
2Ω y1 + y2 = 4V ELEC 101 DC Circuits 26 Example 3Ω 4Ω Kill the 36V source VO Find VO. 3Ω
36V a VO = VO due to 36V alone +
VO due to 6A alone b Find VO due to 36V alone (VO,36V)
(kill other source or other source
inactive) 4Ω
VO,36V 36V ∴ VO,36 V = 36V ×
c 2Ω 2Ω
=8V
9 Find VO due to 6A alone (VO,6A)
(hence kill other source). 4A 6A
2A 2A VO,6A 2Ω ∴ VO,6 A = − 2 A × 2Ω = −4V
From Superposition Rule
∴VO total = VO ,36V + VO,6A d = 8V – 4V = 4 V Kill the 6A source
3Ω 4Ω 2Ω 6A 36V 2009/10 Spring 2 Nonlinear circuits does not have
superposition and linearity properties
Example diode is nonlinear element
VO V1
V2 R VO due to V1+V2 may
not necessary equal to
VO due to V1 + VO due
to V2 . ELEC 101 3 DC Circuits 27 Superposition also applies for time varying voltage. Power does not obey superposition Power is not a linear function (P ∝ V2 or I2)
and does not obey superposition.
Example 8V 16V Example 2V VB V2 2Ω 0V
−2V Find all node
voltages. 2Ω VO 2Ω
2
O from Superposition Rule
2 V
(4V )
=
= 8W
R
2Ω Find power absorbed by 2Ω.
2Ω
2Ω VO 1 VB = VB due to V1 alone + VB due to V2 alone
= VB1 + VB2 = V1 + V2 VB1 VB 2 V2 + V1 V2 14V
12V
10V V1 2
VO 82
=
= 32 W
Power absorbed by 2Ω = P =
R
2 2 VA = VB
2 VB VB = V1 + V2
0 Voltage × 2, Power × 4 VA
2Ω 12V
V1 Find power absorbed by 2Ω. Power absorbed by 2Ω = P =
Example 2009/10 Spring VA 7V
6V
5V
0 ELEC 101 DC Circuits 28 3kΩ Example
Find IO. 12V 2kΩ Va
IO 2009/10 Spring Example
3V 6kΩ Show that network A and network B
are equivalent. 7Ω
1kΩ 3kΩ Since the two networks are equivalent.
3kΩ
12V
3kΩ 12V 18V 3Ω I1 36V R 4Ω 6A I2
R V1 V2 3kΩ 3kΩ ≡ 1. when R = 0Ω (short), 12V ∵ V1 = 0V Va ∴I2 = 18V 18
=A
7Ω 7 2. when R = ∞Ω (open), ∵ I 2 = 0 A
∵ I1 = 0 A
∴ 36V = 3Ω × (6 A + I 2) + 4Ω × I 2 2kΩ 1kΩ 6kΩ ∵ V2 = 0V
∴ 36V = 3Ω × (6 A + I 2 ) + 4Ω × I 2 18V 18
∴ I1 =
=A
7Ω 7 Hence the circuit can be converted to :
3kΩ 3kΩ Network B Network A 3V ∴ V1 = 18V ∴ V2 = 36V − 6A × 3Ω = 18V 3. I1V1 curve = I2V2 curve Hence Va can be found using node voltage
(KCL) method . 12V − Va
V
V − 3V
= a+ a
3kΩ + 3kΩ 6kΩ 2kΩ + 1kΩ Hence network A and network B are equivalent.
3Ω
4Ω
7Ω
18V ≡ 36V 6A ELEC 101 M DC Circuits 29 a Thevenin’s Theorem A linear network can be replaced by an
equivalent network containing only an
independent voltage source in series with
a resistance.
a linear
network RTH ≡ VTH b Load is the element with unknown I.
Other parts of the circuit is then the network.
I flows from node a to node b.
3Ω a 6A 4Ω b b 6A I 2Ω b Load Find the Thevenin equivalent for
the Network . VTH
2Ω a
I RTH
a 4Ω Network b Find I.
3Ω 36V Define the Network and the Load 36V The theorem is mainly used to find the simple
equivalent (model) of a complex network.
Hence simplify network analysis. The theorem
normally is not a fast calculation method.
Example 2009/10 Spring a I
2Ω b
Load Thevenin equivalent for the Network ELEC 101 DC Circuits 30 ii Find VTH (or VOC) i VTH is called Thevenin voltage. VOC
is called opencircuit voltage. It is the
voltage across the terminals when the
port is open circuit (load disconnected). 3Ω
36V 4Ω a 6A VOC 6A
3Ω
36V c 4Ω 6A 3Ω 4Ω RTH a ≡ b ≡ b
d = 0 ) in the network (this method cannot be
used if the network contains dependent sources
which depend on the parameters outside the
network). b a RTH is called Thevenin
equivalent resistance. Find RTH RTH is found by killing all sources (or VTH Load is
disconnected
(open circuit)
RTH Network 2009/10 Spring VTH a VOC b No current flows
in 4Ω since ab is
open
∴ VOC = Vab = Vcb VOC b ∴ R TH = 3Ω + 4Ω = 7Ω
c Replace Network with Thevenin
equivalent and solve for I
3Ω 36V 4Ω 6A = 36V − 6A × 3Ω
= 18V a Network R TH = 7Ω a
b ≡ a VTH = 18V b
Thevenin equivalent ELEC 101 DC Circuits 31 R TH = 7Ω I VTH
= 18V b 6Ω Find Thevenin
equivalent of
network A . a Hence Thevenin
equivalent of A is or 36V 6Ω Since i = 0 ∴ VTH VTH ≡ b a ≡ 6Ω 24V 2Ω a 6Ω 36V a a 24V a
36V Example
3Ω ≡ 6Ω 2Ω 3Ω 3Ω 3Ω RTH ∴ R TH = 3Ω / /6Ω = 2Ω Network A
Example ≡ 2Ω
Load 18V
∴I =
= 2A
2Ω + 7 Ω Thevenin equivalent
of the Network a a 3Ω a 2009/10 Spring a
RTH
VTH Va
VTH 36V
= Va =
× 6Ω = 24V
6Ω + 3Ω 2Ω 24V Application : can save one voltage
supply in BJT circuit (the BJT
circuit will discuss later).
36V
36V
RC a
RE 3Ω ≡
36V 6Ω a RC RE ELEC 101 DC Circuits 32 Example 8V Is network A equivalent to network B ? 2Ω 2009/10 Spring Example Unknown
Network A Find Thevenin equivalent for Network A
8V 2Ω ∴ VTH = 8V 2Ω 1kΩ 1V Unknown
Network A Network B VTH 4V Find V1. 8V Network A Unknown
Network A 3kΩ V1 Find Thevenin equivalent for unknown network A.
Then find V1. ∴ R TH = 0Ω
VTH RTH 4V ∴ VTH = 4V 0Ω Thevenin
equivalent of A is 8V RTH
1kΩ 4V 1V ∴ R TH = 3kΩ Hence Network A is equivalent to Network B
R in parallel to an ideal V source can be neglected. 3kΩ
4V 3kΩ V1 ∴ V1 = 2V ELEC 101 N DC Circuits 33 2009/10 Spring
Bell Labs engineer for whom
the Norton equivalent circuit is
named (Photograph taken
October 13,1925 and
reproduced courtesy of the
AT&T Archives) Norton’s Theorem A linear network can be replaced by an
equivalent network containing only an
independent current source in parallel
with a resistor.
a linear
network 3Ω a ≡ b Define the Network and the Load a RN IN 36V b Similar to Thevenin’s Theorem (dual for
Thevenin equivalent). It can be derived using
Thevenin’s Theorem.
Find I.
3Ω
36V 6A 4Ω b 6A b 2Ω Load a IN
2Ω I Find the Norton equivalent for
the Network. Same question as
Thevenin equivalent. a a Network
b Example 4Ω RN I
Norton equivalent
for the Network I
2Ω b
Load ELEC 101 DC Circuits 34 2009/10 Spring ∴ 36 V − 18V − 3I SC − 4I SC = 0 Find IN (or ISC) i ∴18V − 7I SC = 0 IN is called Norton current, ISC is called
shortcircuit current. It is the current
across the terminals when the port is short
circuit (load shorted).
3Ω
36V 4Ω a 6A I SC I SC ≡ b ∴ I SC = IN RN a 18
A
7 Find RN ii RN is called Norton resistance kill all sources in the network
b Load is shorted
(replaced by a short circuit) Network 6A + Isc 3Ω
36V 4Ω 6A 3Ω 4Ω a ≡ a RN b
a
b I SC Apply KVL to outer loop 36V − (6A + ISC ) × 3Ω − ISC × 4Ω = 0 ∴ R N = 3Ω + 4Ω = 7Ω
Norton resistance is the same as
Thevenin resistance. b ELEC 101 DC Circuits 35 Replace Network with Norton
equivalent and solve for I b I SC 18
= IN = A
7 R N = 7Ω Example I I1 IO R b 2Ω
When R = 0, I1 = 4A = Isc Load Norton equivalent
for the Network 4Ω a 2Ω If R = 0, I1 = 4A.
V
If R = 4Ω , find I1. O
IO and VO are
unknowns. a
b ∴I = 2009/10 Spring Open IO and short VO, hence RN = 4Ω
4Ω a 2Ω 18
7Ω
= 2A
A×
7
2Ω + 7 Ω a ≡ RN
b b a
c Hence Norton
equivalent is Norton equivalent is equivalent
to Thevenin equivalent linear
network ≡ 7Ω
18V ≡ 18V
7Ω 7Ω R N = 4Ω Isc = 4A
a Norton
equivalent 4Ω
4A b
R = 4Ω ∴ I1 = 2A
b ELEC 101 O DC Circuits 36 2009/10 Spring Source Transformation I2 Thevenin equivalent and Norton equivalent
are equivalent. Hence any series VR circuit is
equivalent to a parallel IR circuit. V2
IN RN V2 R 0 Can be used to simplify circuit and simplify circuit
calculation. It is usually a fast calculation method. ≡ R TH
VTH Thevenin equivalent VTH = I N × R N VTH VTH VOC
=
IN
ISC 20Ω
40V ≡ 40V
= 2A
20Ω 20Ω b b Source transformation can be
applied to dependent sources also. When R = 0
I1 = VTH/RN Thevenin equivalent
with dependent source V1
0 R a
a Norton equivalent V1 R TH RN IN I1 I1 When R = ∞
V2 = IN×RN Example R TH = R N = Proof I2 When R = 0
I2 = IN When R = ∞
V1 = VTH RT
VOC ≡ Norton equivalent
with dependent source ISC RT ELEC 101 DC Circuits 37
20Ω 8Ω Example 2009/10 Spring
I Find I. 20 5A −16V ∴ I= 12Ω − 16 V
= − 0.5 A
32 Ω I
30Ω 40V 12Ω Use source transformation method
8Ω
20Ω 40V 30Ω Find Va. 4Ω Va Example 4Ω 2A 4V 12Ω 2A Use source transformation method :
8Ω 4Ω 40V
8V 2A 12Ω 12Ω ∴I =
12Ω
24V 8Ω 40V 12Ω 4Ω Va
I 8V − 4 V
= 0.5A
4Ω + 4Ω ∴ Va = 0.5A × 4Ω + 4V = 6V 4V ELEC 101 DC Circuits 38 Superposition method for network with
more than two sources Example Is network A equivalent to network B ? 2Ω I Example 2Ω 6V
Find I . 2A 2Ω a 4V 2Ω 2Ω
2Ω 6V Find I due to 6V and 2A 2Ω 2Ω
−4V
6V 2Ω ∴ I1 = 6V + 4V
= 2.5A
4Ω 2Ω 2Ω 4V Network B 2Ω b Find I due to 4V 6V 8A Find Norton Equivalent for Network B I1 Find Thevenin
equivalent 8A
Network A 2A I1 I2 2009/10 Spring ∴ I 2 = 0A 2Ω I SC 2Ω 8A ∴ I SC = 8A
Hence Norton
equivalent is ∴ R N = ∞Ω I SC = 8A Hence Network A is equivalent to Network B
c ∴ I = I1 + I 2 = 2.5A R in series to an ideal I source can be neglected ELEC 101 DC Circuits 39 2009/10 Spring Method 2 Example (past paper)
(a) Show that the Thevenin resistance (RTH) at
terminals ab is 2kΩ .
(b) If IO = 20mA/3, find the Thevenin voltage (VTH)
at terminals ab.
(c) If the load is changed from 1kΩ to 2kΩ , find
the new IO.
(d) Find the short circuit current ISC at terminals
ab (from a to b). (a) Disconnect load, kill all sources.
1kΩ 1kΩ a 1kΩ b R between node a
and node b is 2kΩ 2kΩ (b) 12V ∴ R TH = 2kΩ 20/3 mA a IO = 20/3 mA 2kΩ
4mA 6V 1kΩ
1kΩ b 1kΩ
load
IO
2kΩ a
2mA Method 1
4mA 6V 1kΩ
1kΩ b 1kΩ 2mA 6mA V1
2kΩ a ∴ VTH = V1 + 6V + 12V + 2V
2mA ∴VOC = Va − Vb = VTH
∴VTH = V + 20V − V = 20V
1
1 b 20
mA × (2kΩ + 1kΩ) = 20V
3 (c) V1 + 6V 12V 1kΩ load VTH 1kΩ a
2kΩ
20V
Thevenin
equivalent (d) ∴ ISC = b IO
2kΩ
load ∴ new IO =
= 5mA VTH 20V
=
= 10mA
R TH 2kΩ 20V
4kΩ ELEC 101 DC Circuits 40 Source transformation can ONLY be used to
find current (or voltage) outside the Thevenin
equivalent (or Norton equivalent) circuit. Example 2009/10 Spring 3Ω
9A
I1 = 9A × Show that source transformation
can not be used to find I1. 9A 3A 6Ω
= 6A
3Ω + 6Ω Current through
the 3Ω resistor
changed after the
source
transformation. Can apply source transformation
to find I2 (which is outside the
equivalent circuit). Example
6Ω 3Ω 6Ω I1=
6A I1
6Ω 3Ω
If source transformation
is used, 9A
I1 27V
I1 =
= 3A
3Ω + 6Ω 6Ω
27V 3Ω I2 6Ω 27V
But I1 is actually equal to 6A. I2 ∴ I2 = 3Ω 27V
= 3A
9Ω I2 ELEC 101 DC Circuits 41 Apply 10V − 2Ω × ia − 1Ω × ia − 2Ω × ia = 0
KVL
10V Dependent Source Examples Q ∴ ia = Dependent source cannot be killed
Example Use superposition to find i .
2Ω 10V a b 2i 10V = 2A Kill 10V, can not kill dependent source ib 1Ω
3A 2ib (3A+ib) × 1Ω V 2Ω × ib V Kill 3A, can not kill dependent source
2Ω 5Ω 2Ω 1Ω
3A i 2009/10 Spring 1Ω
ib 2ia ia 3A 2Ω × ib V Apply KVL 2Ω × ib + (3A + ib) × 1Ω + 2Ω × ib = 0
2Ω × ia V
10V ia 5Ω × ib + 3V = 0 1Ω × ia V ∴ ib = −0.6A 2Ω × ia V
c ∴ i = ia + ib = 2A − 0.6A = 1.4A ELEC 101 DC Circuits 42 4.4A Check 2Ω 10V 7.2V 1Ω a ia×2Ω V 2.8V
2i 0V 10V 1Ω c 2ia 3A ∴ ia = −3A ∴ V1 = 10V − ia × 2Ω = 10V − ( − 3A) × 2Ω = 16V I
d 3A i 2ia V 0V Use Thevenin’s Theorem to find I.
2Ω d c 10V
ia Example load disconnected
(open circuit)
VTH Find VTH
2Ω V1 3A 1.4A 2009/10 Spring ∴ VTH = V1 − 2ia = 16V − 2Ω × ( − 3A) = 22V
2i
Can not kill dependent source to find RTH 1Ω Find Thevenin Equivalent
2Ω
i
10V c 1Ω c I
d 3A
2i ≡ RTH
VTH I b d Find RTH using VTH/ISC or VOC/ISC
2Ω
10V ib c ISC
3A d
2ib ELEC 101 2Ω×ib V
10V DC Circuits 43 Isc 2Ω 2009/10 Spring Example
2Ω×ib V ib kill V and I sources, apply 1A a 3A Another way to find RTH Find Va, then RTH = Va/1A 0V Va i×2Ω Apply KVL 10V − 2Ω × ib − 2Ω × ib = 0 2Ω 10V
∴ ib =
= 2.5A
4Ω c ∴ R TH = VTH 22V
=
= 4Ω
ISC 5.5A Replace network with Thevenin Equivalent
c
4Ω i 22V d 0V
∴ i= − 1A
Apply KVL ∴ i × 2Ω + Va + 2i = 0 ( − 1A × 2Ω) + Va + (2Ω × −1A) = 0
∴ Va = 4V b ∴ R TH 1Ω 22V
= 4 .4 A
I ∴I =
5Ω d
2i V ∴ ISC = 3A + ib = 3A + 2.5A = 5.5A
Find RTH using
VTH/ISC 1A c = V
=a
1A 4V
= 4Ω
1A c
RTH 1A 4V
d ELEC 101 DC Circuits 44 VO 4Ω Find Vo. Thevenin equivalent of a pure dependent source
network is only a single resistance
4Ω 4Ω 16V 4Vo Example VO
4Ω 8Ω 16Vo VO ∴ i = −1A
8Ω 2A 2Vo 4Ω 1.5V ∴VO = (2A − 2VO ) × 2Ω = 4V − 4VO
8Ω 3Ω 2Ω ∴ Va = 0.6V 0.6V 0.3A
c 1.5A 4Ω 1A 2Ω 1A 0.8V 4Ω − 6V 0.2A i 0V 1.7A 1.9A 16V 0.6V 0.7A Rth
d Apply KCL ∴ − 1.5V − Va = Va + −1A
3Ω
2Ω ∴VO = (2A − 2VO ) × (8Ω / /4Ω / /8Ω)
∴VO = 0.8V 3Ω 1.5i c d
Va  1.5V a apply 1A 8Ω 2Ω 4Ω
4Ω 2A 3Ω 1.5i Find Rth . c i ≡ 8Ω Example 2009/10 Spring 3.2A 4Ω b ∴ Rth = Va 0.6V
=
= 0.6Ω
1A
1A Rth 1A d ELEC 101 DC Circuits 45 ∆−Y (or π T) Transform P a a
R3 ≡ R2 Ra R1 Rb Rc b c b Given in Exam c R3 (R1 + R 2 )
R1 + R 2 + R3
R (R + R 3 )
R bc = R b + R c = R1 / /(R 2 + R 3 ) = 1 2
R1 + R 2 + R 3
R (R + R 3 )
R ac = R a + R c = R 2 / /(R1 + R 3 ) = 2 1
R1 + R 2 + R 3
R ab = R a + R b = R3 / /(R1 + R 2 ) = ∴Ra = R 2R 3
R1 + R 2 + R 3
∴Rc = ∴Rb = R1R 3
R1 + R 2 + R 3 R1R 2
R1 + R 2 + R 3 Ra Rb + Rb Rc + Ra Rc
Ra
R R + Rb Rc + Ra Rc
∴R2 = a b
Rb
R R + Rb Rc + Ra Rc
∴R3 = a b
Rc
∴ R1 = 2009/10 Spring ELEC 101 DC Circuits 46 Example Example
10Ω
b a
30Ω
20Ω c ≡ Ra
Rb
b a a a
Rc
c R 2 R3
30Ω ×10Ω
=
= 5Ω
Ra =
R1 + R 2 + R3 20Ω + 30Ω + 10Ω R1 R3
20Ω ×10Ω
10
=
=Ω
Rb =
R1 + R 2 + R3 20Ω + 30Ω + 10Ω 3
Rc = 2009/10 Spring R1 R 2
20Ω × 30Ω
=
= 10Ω
R1 + R 2 + R 3 20Ω + 30Ω + 10Ω 30Ω 10Ω
b
50Ω/3 c 20Ω
d ≡ 5Ω
10Ω/3 c b 10Ω 50Ω/3 10Ω a ≡ 10Ω d
5Ω
20Ω 20Ω d
a ≡ 15Ω
d ELEC 101 DC Circuits 47 Example Example Find the equivalent
resistance between
terminals ab
(REQ). The
resistance in each
branch is R. b a R I R R
R
R a to ∞ 1Ω a ≡
1Ω 1Ω 1Ω 1Ω I I/3 b R equivalent R = r R R R I/3 I
I
I
5I
∴ Vab = × R + × R + × R = × R = I × R EQ
3
6
3
6
5
∴ R EQ = R
6 equivalent R also = r r
1+r
∴ r(1 + r) = 1 + r + r = r + r 2
∴ r = 1 + 1//r = 1 + ∴r 2 − r − 1 = 0
∴r = r
b can not be solved using
normal circuit theorems R
R R Find r. 1Ω 1Ω a 1Ω b I/6 I/3 2009/10 Spring 1+ 5
Ω
2 ...
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 Fall '09
 CHAN
 Volt, Thévenin's theorem, Voltage source, Current Source, DC Circuits

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