Chapter 3
Resistive Network Analysis
Jaesung Jang
Node Voltage/Mesh Current Method
Superposition
Thevenin and Norton Equivalent Circuits
Voltage and Current Source Transformation
Maximum Power Transfer
1
Network Analysis
• The analysis of electrical network consists of determining each
of unknown branch currents (currents flowing through each
branch) and node voltages (voltage at each node).
• Once the known (usually resistances and sources) and
unknown (branch currents and node voltages) variables
are identified, a set of equations relating these variables is
constructed.
• If the number of unknown variables and equations are
equal, the set of equations can be solved.
2
Node Voltage Method
• Variables: voltage at each node (node voltages).
• One node must be the reference point (usually ground)
for specifying the voltage at any other node.
• Solving procedure for a circuit containing n nodes
– Step1: Select the reference node and define the remaining n-1 node
voltages as the independent or dependent variables. Each of m
voltage sources is associated with dependent variables.
– Step2: Assume branch currents. (direction!)
– Step3: Use KCL on the n-1-m nodes labeled as an independent
variables and express the branch currents in terms of node voltages.
– Step4: Solve n-1-m linear equations where the number of unknowns
is n-1-m.
• The node-voltage analysis is especially useful when we
have current sources.
3
Examples
n=3, m=0, n-1-m=2
Ground node
At node a, i1 + i2 − iS = 0
 1
v a − vc v a − vb
1 
1
v a −
+
= iS →  +
vb = i S
R1
R2
R
R
R
2 
2
 1
At node b, i3 − i2 = 0
→
→
 1
vb − vc v a − vb
v
v −v
1
1 
 vb = 0
−
=0→ b − a b =0→−
va + 
+

R3
R2
R3
R2
R2
 R2 R3 
4
Cramer’s rule
This is to solve the linear equations using determinants without matrix inversion.
a
c

b  x   p 
 = 
d   y   q 
p b
a
c
q d
pd − bq
x=
=
,y=
a b
a
ad − bc
c
d
c
p
q
aq − pc
=
b
ad − bc
d
 3 − 1 I A   − 14 
− 2 3  I  =  7 

 B  

− 14 − 1
7
3
− 42 + 7
IA =
=
= −5 A,
3 −1
9−2
−2 3
3 − 14
−2 7
21 − 28
IB =
=
= −1A
3 −1
9−2
−2 3
5
Examples (cont.)
i1
i1
i3
i2
i3
i4
n=4, m=0, n-1-m=3
At node 1, − i1 − i2 + I1 = 0 → −
v3 − v1 v2 − v1
−
+ I1 = 0
R1
R2
At node 2, i3 + i2 − I 2 = 0 →
v2 v2 − v1
+
− I2 = 0
R3
R2
At node 3, i1 − i4 + I 2 = 0 →
v3 − v1 0 − v3
−
+ I2 = 0
R1
R4
i2
i4
n=4, m=1, n-1-m=2
At node b, − i1 + i2 + i3 = 0 → −
vS − vb vb vb − vc
+
+
=0
R1
R2
R3
At node c, − i3 + i4 − iS = 0 → −
vb − vc vc − 0
+
− iS = 0
R3
R4
6
Mesh Current Method
• Variables: current of meshes (mesh current).
• A mesh is the simplest possible loop that does not contain
any other loops.
• Solving procedure for a circuit containing n meshes
– Step1: Define the n mesh currents as the independent or dependent
variables. Each of m current sources is associated with dependent
variables. The number of unknown variables is n-m.
– Step2: Define mesh current direction. Unknown mesh currents will
always be defined in the clockwise direction except for known mesh
currents (i.e. current sources) .
– Step3: Use KVL on each mesh containing unknown mesh currents
and express each voltage in terms of the mesh currents.
– Step4: Solve n-m linear equations for the mesh currents.
– Step5: Find the branch currents and voltages from the mesh currents.
7
Examples
n=2, m=0, n-m=2
Reference direction:
pp 17 at Chap. 2.
Net current
8
Examples (cont.)
n=2, m=0, n-m=2
Branch current at the resistor 2
= (i2 − i1 ) : mesh currents
KVL at mesh 1, − V1 + v1 + V2 + v2 = 0
→ −V1 + i1 R1 + V2 + (i1 − i2 )R2 = 0
→ i1 (R1 + R2 ) − i2 R2 = V1 − V2
KVL at mesh 2,−V2 + v3 + V3 + v 4 + v2 = 0
_
→ −i1 R2 + i2 (R2 + R3 + R4 ) = V2 − V3
+
→ −V2 + i2 R3 + V3 + i2 R4 + (i2 − i1 )R2 = 0
The voltage across the resistor 2 has different polarity depending on the meshes.
9
Examples (cont.)
n=2, m=1, n-m=1
+
KVL at mesh 1, − VS + v1 + v2 = 0
+ +
→ −VS + i1 R1 + (i1 + I S )R2 = 0
→ i1 (R1 + R2 ) = VS − I S R2
→ i1 =
VS − I S R2
(R1 + R2 )
n=3, m=1, n-m=2
+
Mesh 2 is not considered
because it has a current
source (known).
i1 = I (the current source)
KVL at mesh 2, − V + v2 + v3 = 0
+
+
+
+
+
→ −V + (i2 − I )R2 + (i2 − i3 )R3 = 0
→ i2 (R2 + R3 ) − i3 R3 = V + IR2
Three of the mesh current directions on resistors 1, 2, and 3 are different.
KVL at mesh 3, v4 + v3 + v1 = 0
+
→ i3 R4 + (i3 − i2 )R3 + (i3 − I )R1 = 0
→ −i2 R3 + i3 (R4 + R3 + R1 ) = IR1
10
Node & Mesh Analysis w Controlled
Sources
• Dependent source generates a voltage or current that depends on the value
of another voltage or current in a circuit.
• Initially we treat them as ideal sources to set up the a set of equations and
substitute constraint equation if you need.
• Constraint equation: an equation relating the dependent source with one of
the circuit voltage or current.
n=3, m=0, n-1-m=2
At node 1, − iS +
v1 − 0 v1
+
=0
RS
Rb
 1
iS
1 
v1 = iS → v1 =
→ 
+

 1
1 
 RS Rb 


+
R
R
b 
 S
0 − v2
At node 2, β ib −
= 0 → v2 = − β ib RC
RC
Constraint equation
ib =
1
iS
Rb
 1
1 


+
R
R
b 
 S
=
RS i S
RS i S
& v2 = − β ib RC = − βRC
(Rb + RS )
(Rb + RS )
11
Examples
n=3, m=0, n-m=3
KVL at mesh 1, − v1 + i1 R1 + v = 0
+
+
+
+
→ −v1 + i1 R1 + (i1 − i2 )R2 = 0
→ i1 (R1 + R2 ) − i2 R2 = v1
Note : v = (i1 − i2 )R2
KVL at mesh 2, i2 R3 + (i2 − i3 )R4 − 2v − v = 0
→ i2 R3 + (i2 − i3 )R4 − 3(i1 − i2 )R2 = 0
→ −3R2i1 + i2 (R3 + R4 + 3R2 ) − i3 R4 = 0
(R1 + R2 )
 − 3R
2

 2 R2
− R2
(R3 + R4 + 3R2 )
− (R4 + 2 R2 )
  i1  v1 
   
− R4  i2  =  0 
(R4 + R5 ) i3   0 
0
KVL at mesh 3, (i3 − i2 )R4 + i3 R5 + 2v = 0
→ (i3 − i2 )R4 + i3 R5 + 2(i1 − i2 )R2 = 0
→ 2 R2 i1 − i2 (R4 + 2 R2 ) + i3 (R4 + R5 ) = 0
12
Principle of Superposition
• In a linear circuit containing more than one (voltage or current) source,
the current or voltage in any part of the circuit can be found by
algebraically adding the effect of each (voltage or current) source
separately.
• When we consider the effect of a single (voltage or current) source only in
the circuit, the other (voltage or current) sources can be removed (zeroing)
by
– shorting voltage source and opening current source.
_ +
Shorting: replace the voltage source with a short circuit
Opening: replace the current source with a open circuit
13
Examples
Determine the current flowing
through the resistor R
iR
= iR
+ iR
current source only
Current source only
Voltage source-> shorting
1
R
iR =
iB
 1
1
1

+
+ 
R
R
R
G
 B
voltage source only
Note: iR can have a negative
value depending on the
polarity of a source.
Voltage source only
Current source-> opening
iR =
1
R
 1 1

+ 
R
 B R
i1 =
1
R
VG
 1 1  (RG + R || RB )

+ 
R
 B R
i1 RG
i2 R
B
iR
R
+ VG
14
One-Port Networks & Equivalent Circuits
• The flow of energy from a source to a load can be described by showing
the connection of two “black boxes” labeled sources and load.
• It is always possible to view even a very complicated circuit in terms of
much simpler equivalent source and load circuits.
15
Thevenin & Norton Theorem
• Thevenin Theorem: When viewed from a load, any network composed of ideal
voltage and current sources, and of linear resistors can be represented by an
equivalent circuit consisting of an ideal voltage source (VT) in series with an
equivalent resistance (RT) .
• Norton Theorem: When viewed from a load, any network composed of ideal
voltage and current sources, and of linear resistors can be represented by an
equivalent circuit consisting of an ideal current source (IN) in parallel with an
equivalent resistance (RN) .
Ideal current and
voltage sources,
Linear resistors,
etc.
Note: RT= RN
16
Thevenin Equivalent Circuit
• Determining Thevenin Resistance and Voltage
– Step1: Disconnect RL to find the voltage between open terminals A and B (->
VT).
– Step2: Zero all the independent voltage and current sources (short the
voltage source and open the current source) and calculating the circuit’s
total resistance as seen from open terminals A and B. (-> RT). -> Thevenin
equivalent circuit
– Step3: Reconnect RL with terminals A and B and find VL or IL.
RT
+
A
+
network
VT
-
B
Load RL
A
A
network
-
+
B
-
B
17
Example
Example
Find VL
Procedure
Application of Thevenin’s theorem. (a) Actual circuit with terminals A and B across RL. (b) Disconnect RL to find that
VAB is 24V (= VT). (c) Short-circuit V to find that RAB is 2Ω. (d) Thevenin equivalent circuit. (e) Reconnect RL at
terminals A and B to find that VL is 12V.
18
Example (Cont.)
iL
Thevenizing a circuit with two voltage sources V1 and V2. (a) Original circuit with terminals A and B
across the middle resistor R3. (b) Disconnect R3 to find that VAB is −33.6V (= VT). (c) Short-circuit V1
and V2 to find that RAB is 2.4 Ω(= RT). (d) Thevenin equivalent with RL reconnected to terminals A and B.
19
Norton Equivalent Circuit
• Determining Norton Current and Resistance
– Step1: Disconnect RL, short the terminal A and B, and find the current
between terminals A and B (-> IN).
– Step2: Zero all independent voltage and current sources (short the voltage
source and open the current source) and calculating the circuit’s total
resistance as seen from open terminals A and B. (-> RN=RTH). -> Norton
equivalent circuit
– Step3: Reconnect RL with terminals A and B and find VL or IL.
+
+
A
+
network
A
IN
network
-
-
B
Load RL
A
RN
B
-
B
20
Example
iL
Same circuit as in Fig. 10-3, but solved by Norton’s theorem. (a) Original circuit. (b) Disconnect A and B and short
circuit across terminals A and B. (c) The short-circuit current IN is 36/3 = 12 A. (d) Open terminals A and B but shortcircuit V to find RAB is 2 Ω, the same as RN. (e) Norton equivalent circuit. (f) RL reconnected to terminals A and B to
21
find that IL is 6A.
Example (Thevenin & Norton Resistance)
Rab = RT =
{[(R1 || R2 ) + R3 ] || R4 } + R5
Rab = RT =
{[(R1 || R2 ) + R3 ] || R4 }
Same circuit as in Fig. 10-3, but solved by Norton’s theorem. (a) Original circuit. (b) Disconnect A and B and short
circuit across terminals A and B. (c) The short-circuit current IN is 36/3 = 12 A. (d) Open terminals A and B but shortcircuit V to find RAB is 2 Ω, the same as RN. (e) Norton equivalent circuit. (f) RL reconnected to terminals A and B to
22
find that IL is 6A.
Source Transformations
• Thevenin’s theorem says that any one-port network can be represented
by a voltage source and series resistance.
• Norton’s theorem says that the same one-port network can be
represented by a current source and shunt (parallel) resistance.
• Therefore, it is possible to convert directly from a Thevenin form to a
Norton form and vice versa.
• Thevenin-Norton conversions are often useful.
RT=RN
VT=IN x RN
23
Example
I1
I2
I1+ I2
24
Maximum Power Transfer Theorem
• How much power can be transferred to the load
from the source under the most ideal condition?
-> Answer: Maximum power transfer theorem.
PL = iL RL =
2
∂PL
∂RL
2
vT RL
(RT + RL )2
2
2 [(RT + RL ) − 2 R L (RT + RL )]
2 (RT + R L )(RT − R L )
=
v
= vT
T
(RT + RL )4
(RT + RL )4
• The power curve peaks where RL = RT.
At this point, the source transfers
maximum power to the load.
• 50% of the power generated is
dissipated in the source resistor and the
other 50% is in the load resistance.
25