CHAPTER THREE
TECHNIQUES OF CIRCUIT ANALYSIS
Terminology
Node - Point where two or more circuit elements join e.g. a.
Essential node - Point where three or more circuit elements
join e.g. b.
Path – Trace fo adjioning basic elements with no elements
included more then once.
Branch – Path that connects two nodes
Essential branch – Path which connect
two essential node without passing
through an essential node
Loop – Path whose last node is the
same as the starting node
mesh – Loop does not enclose any
other loop
Planner circuit – circuit with no crossing
branch.
3.1 Node-Voltage Method
Based on Kirchoff current law. First make a neat layout of the
circuit so that no branches across and make clearly the
essential node and select one as a reference nodes,
normally node with most branches is usually a goods choice.
Node voltage is defined as the voltage rise from the
reference node to a nonreference node.
Step to determine Node Voltages :
• Select a node as the reference node or ground since it is
assumed to have zero potential. Assign voltage e.g. v1, the
voltage are referenced with respect to ref node.
• Apply KCL to each of the n-1 nonreference nodes. Assign i1,
i2.. to R1, R2,.. where current flow from ahigher potential to
lower potential in a resistor. i = (vH – vL)/R. Use ohm law to
express the branch currents in term of node voltages.
• Solve the resulting simultaneous equations to obtain the
unknown node voltages.
3.1 Node-Voltage Method
3.2 Mesh-Current Method
3.3 Source Transformations
3.4 Thevenin Equivalent Circuits
3.5 Norton Equivalent Circuits
3.6 Maximum Power Transfer
3.7 Superposition
From circuit 4.3 ;
Nodes are a,b,c,d,e,f and g.
Essential nodes are b,c,e and g
Branches are v1,v2, R1, R2,R3,R4,R5,R6,R7 and I.
Essential branch are v1-R1, R2-R3, v2-R4, R5,R6,R7 and I
Mesh are v1-R1-R5-R3-R2,V2-R2-R3-R6-R4,R5-R7-R6 and R7-I.
Path is R1-R5-R6 but it is not a loop.
Loop is v1-R1-R5-R6-R4-v2 but is not a mesh.
EX:
Here node 3 as reference, and by using KCL ;
node 1; (v1-10)/1 + (v1/5) + (V1-V2)/2 = 0
node 2 ; (V1-V2)/2 + v2/10 - 2 = 0
So from the two equation ;
v1 = 100/11 = 9.09v
v2 = 120/11 = 10.91v
1
Ex : a. Find ia, ib and ic.
b. Find power associated with each sources.
Here 2 essential node and lower node as reference, and
define v1 by using KCL ;
node 1; (v1-50)/5 + v1/10 + V1/40 - 3 = 0
so v1 = 40V
ia = (40-50)/5 = -2A; ib = 40/10 = 4 A; ic = 40/40 = 1A
b. P50V = -50ia = -100W
P3A = -3v1 = -120W
Ptotal = -100W – 120W = -220W
Node-voltage equations must be supplemented with the
constraint equation imposed by presence of dependent
sources.
Ex : Get the power dissipated in the 5Ω resistor ;
Here circuit have 3 node and need 2 node-voltage equation ;
node 1; (v1-20)/2 + (v1/20) + (V1-V2)/5 = 0
node 2; (v2-v1)/5 + (v2/10) + (V2- 8iΦ)/2 = 0
These two equation contains 3 unknowns, v1, v2 and iΦ, so to
eliminates, iΦ = (v1-v2)/5
0.75v1 – 0.2v2 = 10
-v1 + 1.6v2 = 0
V1 = 16 v and V2 = 10 v
5Ω
ia ib
50V
Node-Voltage Method and Dependent Sources
v1
Then, iΦ = (16–10)/5 = 1.2A
10Ω ic
3A
40Ω
P5Ω = i2R =1.44 x 5 = 7.2 watt
Ex : Calculate the node voltage v1 and v2.
Node 1; i1 = i2+ i3 → (v1- v2)/4 + (v1- 0)/2 = 5
X 4 : v1 - v2 + 2v1 = 20
Node 2 : i2 + i4 = i1 + i5 → (v1 – v2)/4+ 10 = 5 + (v2 – 0)/6
X 12 : 3v1 – 3v2 + 120 = 60 + 2v2
- 3v1 + 5v2 = 60
so 4v2 = 80, v2 = 20V
3v1 – 20 = 20, v1 = 40/3 = 13.33V
5A
i1 = 5A
i2 = (v1 – v2)/2 = -1.67A
i1
i3 = v1/2 = 6.66A
i4
4Ω
v2
v1
i4 = 10A
i2
i5 = v2/6 = 3.33A
6Ω
10A
i
2Ω
i
Ex : Calculate the node voltage v1 and v2.
Ex : Determine the power associated with each sources.
Node-Voltage Method : Some Special Case
3
5
20v
5Ω
20Ω
iΦ
X 4 : 3v1 - 2v2 - v3 = 12
(1)
Node 2 : i2 + i3 = ix → (v1 – v2)/2 = (v2 – v3)/8 + (v2 – 0)/4
(2)
X 8 : -4v1 + 7v2 – v3 = 0
Node 3 : i1 + i2 = 2ix → (v1 – v3)/4 + (v2 – v3)/8 = 2(v1 – v2)/2
(3)
X 8 and ÷ 3 : 2v1 - 3v2 + v3 = 0
(1) + (3) : 5v1 - 5v2 = 12 → v1- v2 = 12/5 = 2.4
4Ω
(2) + (3) : -2v1 + 4v2 = 0 → v1 = 2v2
i1
so 2v2 – v2 = 2.4 ;
v2 = 2.4V ; v1 = 2v2 = 4.8V
2Ω
8Ω
v2
v1
v3
v3 = 3v2 – 2v1 = -2.4V
3A
i2
ix
4Ω
i3
2ix
When a voltage source is the only element between 2
essential nodes, the node-voltage can be simplified. Here
circuit have 3 node and need 2 node-voltage equation. A ref
node has been chosen and two other nodes have been
labeled but 100V source constrains voltage between node 1
and the ref node to 100V and only one unknown node
voltage, v2.
node 2 ; (v2-v1)/10 + v2/50 – 5 = 0
But v1 = 100v, v2 = 125v
So each branch current can be
calculate.
3i1
2Ω
Node 1; i1+ ix = 3 → (v1- v3)/4 + (v1- v2)/2 = 3
10Ω
5A
2
Node-Voltage Method : Some Special Case
Ex: Analyze the circuit below.
Circuit contain 4 essential nodes, so 3 node voltage equ.
However, 2 essential nodes are connected by independent
Vs, and 2 other connected to current control dependent Vs.
Ref node flagged and node voltages defines. Hence, there
actually is only 1 unknown node voltage.
node 2 ; (v2-v1)/5 + v2/50 + i = 0
node 3 ; v3/100 - i – 4 = 0
so, (v2 – v1)/5 + v2/50 + V3/100 - 4 = 0
Node-Voltage Method : Super node
Super node is when two node can be combines into one
when a voltage source is between two essential source (two
ref nodes) and any element connected parallel to its.
node 2 ; (v2-v1)/5 + v2/50 + i = 0
node 3 ; v3/100 - i – 4 = 0
(1)
so, (v2 – v1)/5 + v2/50 + V3/100 - 4 = 0
Since v1 = 50v, so v3 = v2 + 10iΦ
Current controlling dependent
voltage sources; iΦ = (v2 – 50)/5
So if Vs is connected between ref
node and nonref node, so Vs equ
nonref voltage node.
v2(1/50 + 1/5 + 1/100 + 10/500) = 10 + 4 + 1
v2(0.25) = 15, v2 = 60
iΦ =(60 -50)/5 = 2A, v3 = 80V
3.2 Mesh-Current Method
Ex : Calculate the node voltage v1 and v2.
Mesh is a loop with no other loop inside it. A mesh current is
the current that exists only in the parameter of a mesh. Using
mesh current instead of element current is convenient and
reduces number of equation to solves.
Mesh analysis only apply to planar circuit with branch
crossing one another.
Step to determine mesh current ;
• Assign mesh current i1, i2,…
• Apply KVL to each mesh. Use Ohm law to express the
voltage in term of mesh current.
• Solve the resulting simultaneous equation to get mesh
current.
Supernode contain Vs 2V, node 1, node 2 and 10Ω.
KCL : i1+ i2 + 7 = 2 → (v1v- 0)/2 + (v2v- 0)/4 + 7 = 2
2v1 + v2 + 28 = 8
or
v2 = -20 – 2v1
→ v2 = v1 + 2
KVL : - v1 - 2 + v2 = 0
so 3v1 = -22, v1 = -7.33V
v2 = v1 + 2 = -5.33V
Note : 10Ω does not make any diff
10Ω
coz it is connected across supernode.
v1
v2
1
2A
2Ω
1
4Ω
i1
7A 2A
i2
EX :
Using Kirchoff laws; i1 = i2 + i3
v1 = i1R1 + i3R3
-v2 = i2R2 – i3R3
so from that and substitute i3;
v1 = i1(R1 + R3) – i2R3
-v2 = i2R2 – i3R3
By using Kirchoff voltage law
v1 = iaR1 + (ia-ib)R3
-v2 = (ib – ia)R3 + ibR2
by know ia and ib, we can calculate
i1, i2 and i3.
2v
v1
2
v2
2
2Ω
4Ω
i1
i2
7A
Ex :
a.Determine the power associated with each Vs.
b. Calculate v0 across 8Ω resistor.
-40 + 2ia + 8(ia – ib) = 0
8(ib – ia) + 6ib + 6(ib – ic) = 0
6(ic – ib) + 4ic + 20 = 0
Using Cramer method
10ia – 8ib +0ic = 40
-8ia + 20ib -6ic = 0
0ia – 6ib + 10ic = -20
ia = 5.6A; ib = 2.0A; ic = 0.8A
P40V = -40ia = -224W
P20V = 20ic = -16W
6Ω
2Ω
v0 = 8(ia – ib) = 8x3.6 = 28.8V
40V
8Ω
ia
v0
4Ω
20V
6Ω
ib
ic
3
Ex : Find branch current I1, I2 and I3.
Mesh-Current Method and Dependent Method
mesh 1 : -15 +5i1 + 10(i1 – i2) + 10 = 0
3i1 – 2i2 = 1
mesh 2 : 6i2 + 4i2 + 10(i2 –i1) – 10 = 0
i1 = 2i2 – 1
using substitution method, so 6i2 – 3 – 2i2 = 1
i2 = 1A
i1 = 2i2 – 1 = 1A
5Ω
I1 = 1A
I2 = 1A
I1
15V
10Ω
I3 = i1 – i2 = 0
i1
If the circuit contains dependent sources, the mesh-current
equations must be supplemented by appropriate constraint
equation. Ex: Determine the power dissipated in 4Ω resistor.
6Ω
I2
v0
4Ω
i2
10V
I3
Circuit has 6 branches where i is unknown
and 4 nodes, so needs 3 mesh-current.
50 = 5(i1 - i2) + 20(i1 - i3)
0 = 5(i1 - i2) + 1i2 + 4(i2 - i3)
0 = 20(i3 - i1) + 4(i3 - i2) + 15iΦ
Branch current controlling dependent
voltage source in term of i mesh, iΦ = i1 – i3
From the equation;
50 = 24i1 – 5i2 – 20i3
0 = -5i1 + 10i2 -4i3
0 = - 5i1 – 4i2 + 9i3
Ex : Find i0.
mesh 1 : -24 + 10(i1 – i2) + 12(i1 – i3) = 0
11i1 – 5i2 – 6i3 = 12
mesh 2 : 24i2 + 4(i2 – i3) + 10(i2 – i1) = 0
-5i1 + 19i2 – 2i3 = 0
mesh 3 : 4i0 + 12(i3 – i1) + 4(i3 – i2) = 0
but at A, i0 = i1 – i2, so
4(i1 – i2) + 12(i3 – i1) + 4(i3 – i2) = 0
or –i1 – i2 + 2i3 = 0
using cramer rules
i1 = ∆1/∆ = 432/192 = 2.25A
i2 = ∆2/∆ = 144/192 = 0.75A
24V
Using Cramer rules ;
i2 = 26A and i3 = 28A
Current in 4Ω = i3 – i2 = 2A
So, P4Ω = 22 x 4 = 16watt
i3 = ∆3/∆ = 288/192 = 1.5A
i0 = i1 – i2 = 1.5A
Mesh-Current Method – Special Cases
When a branch include current source, mesh current req
some add manipulation.
Circuit contain 5 branches where i is unknown and 4 nodes.
Here, mesh current equ = 5 – (4-1) = 2, coz current source
reduce mesh current fro 3 to 2, coz diff between ia and ic = 5A.
(1)
mesh a : 100 = 3(ia-ib) + v + 6ia
(2)
mesh c : -50 = 4ic – v + 2(ic –ib)
(1) + (2) so : 9ia – 5ib + 6ic = 50 (3)
mesh b : 3(ib – ia) + 10ib + 2(ib – ic)
so from (3) and (4) : ic – ia = 5
so ia = 1.75A, ib = 1.25A, ic = 6.75A
(4)
i1
i3
10Ω
i2
i2
4Ω
24Ω
i1
12Ω
i3
4i0
Super mesh results when two meshes have a dependent or
independent current source in common.
Using super mesh, remove the current source by avoid the
branch and voltage is express as original current mesh.
the sum of voltage around supermesh ;
100 = 3(ia-ib) + 2(ic-ib) + 50 + 4ic + 6ia =0
so, 50 = 9ia – 5ib + 6ic
It has eliminated the need for introducing the unknown voltage
across current source.
It use KVL and KCL.
It has no current of its own
The current source provide
the constraint equation necessary
to solve for the mesh current.
4
Node-Voltage Method verses Mesh-Current Method
Ex : Find i1 and i2.
KVL : -20 + 6i1 +10i2 + 4i2 = 0
(1)
so 6i1 + 14i2 = 20
(2)
KCL at 0 :i2 = i1 + 6
solving (1) and (2);
i1 = -3.2A; i2 = 2.8A
•
•
•
•
10Ω
6Ω
20V
2Ω
Both reduce the number of simultaneous equations that must
be manipulated. To identify more efficient method ;
Does one of the method result in fewer simultaneous
equations to solve.
Does the circuit contain super node ?. Is so, it much better.
Does the circuit contain super mesh ?. Is so, it much better.
Does it solve some portion of the circuit with requested
solution.
4Ω
i2
i1
6A
0
3.3 Source Transformations
Source Transformation Summary
Source transformation allows a voltage source in series with
a resistor to be replace with a current source in parallel with
the same resistor or vice versa, but its charac. is maintain.
From
TO
Method
Fix
R p = Rs
is = vs/Rs
Ex : Find the relationship
between is and vs of two circuit
to prove it is same.
If nodes a, b connect with RL
From
iL = vs/(R+RL),
To
Method
Using current division
Fix
vs = isRp
R s = Rp
iL = R/(R+RL) x is
If the circuit are equivalent, the
iL must be same, is = vs/R.
If polarity of vs change so do is.
EX: Find power associated with 6V source.
6Ω
4Ω
6V
30Ω
20Ω
5Ω
40V
3Ω
2Ω
Ex : Find V0.
i = 2/(2+8) x 2 = 0.4
v0 = 8i = 3.2V
3A
4Ω
10Ω
4Ω
12V
8Ω
2Ω
12V
3Ω
8Ω
4A
i
2A
6Ω
3Ω
8Ω
v0
4A
2Ω
8Ω
2A
v0
5
Ex : Find Vx.
Transform dependent current
Source and 6V but not 18V
Coz not in series with resistor.
KVL : -3 + 5i + vx + 18 = 0 (1)
Apply KVL to loop 3V and 1Ω
-3 + 1i + vx = 0
vx = 3 - i
(2)
3A
From (1) and (2)
15 + 5i +3 –i= 0
Apply KVL to loop 18V and 4Ω
-vx + 4i + vx + 18 = 0
i = -4.5A
So vx = 3 – i = 7.5V
4Ω
0.25vx
2Ω
6V
18V
2Ω
vx
4Ω
2Ω
4Ω
1Ω
3V
18V
2Ω vx
vx
i
3.4 Thevenin Equivalent Circuits
Name after M. L. Thevenin, a French engineers in 1883.
Thevenin equivalent circuit is an independent voltage
source, VTh in series with a resistor RTh which replaces an
interconnection of sources and resistors. If same load
connected across a,b terminal, it will generate same voltage
and current at load terminal.
If RL is infinitely large, it is a open-circuit condition, so voltage
at terminal a,b is VTh. To calculate VTh, used the original
open circuit .
If RL is zero, it is a short circuit
condition, so current from b to a;
vx
18V
isc = VTh/RTh so RTh = VTh/isc
isc Thevenin must equal with isc
of original circuit.
Ex : Find the Thevenin equivalent circuit.
First, calculate the open-circuit voltage of vab.
Lower node as ref node
Node voltage ;(v1-25)/5 + v1/20 – 3 = 0
v1 = 32V = VTh
a,b short-circuit current, isc.
Lower node as ref node, so
(v2-25)/5 + v2/20 – 3 + v2/4 = 0
v2 = 16V, so isc = 16/4 = 4A
Then , RTh = VTh/isc = 32/4 =8Ω
Ex : Find Thevenin equvalent circuit and current through RL =
1Ω
6, 16 and 36Ω.
4Ω
Find RTh by short circuit voltage
source and open circuit
32V
RL
2A
12Ω
current source.
RTh = 4||12 + 1 = 4Ω
Mesh : -32 + 4i1 + 12(i1-i2) = 0
4Ω
1Ω
i2 = -2A
RTh
12Ω
So i1 = 0.5A
VTh = 12(i1-i2) = 30V
Or node KCL : (32-VTh)/4 + 2
VTh
=VTh/12
1Ω
4Ω
96-3VTh+24 = VTh
32V
2A
VTh
12Ω
VTh = 30V
i1
IL = VTh/(RTh + RL)
RL = 6; IL = 3A
RL = 16; IL = 1.5A
RL = 36; IL = 0.75A
i2
Ex :
With ix = 0;
VTh = vab = -20i x 25 = -500i
i = (5 – 3v)/2000 = (5 – 3VTh)/2000 5V
VTh = control voltage = -5V
To calculate s/c, place s/c across a,b
So v = 0, so isc = -20i
So, i = 5/2000 = 2.5mA
Combine these two equation
5V
Isc =-20 x 2.5mA = -50mA
RTh = VTh/isc = -5/-50 x 103 = 100Ω
2KΩ
i
a
3v
ix
20i v
25Ω
b
2KΩ
i
20iv 25Ω
100Ω
a
5V
b
6
Ex :
5Ω
Deactive all independent source
Rab = RTh = 4 + 5||20 = 8Ω
4Ω
20Ω
25V
3A
a
i
b
5Ω
4Ω
20Ω
8Ω
a
32V
b
Ex : Find Thevenin equvalent circuit and current through RL =
1Ω
6, 16 and 36Ω.
4Ω
Find RTh by short circuit voltage
source and open circuit
32V
RL
2A
12Ω
current source.
RTh = 4||12 + 1 = 4Ω
Mesh : -32 + 4i1 + 12(i1-i2) = 0
4Ω
1Ω
i2 = -2A
RTh
12Ω
So i1 = 0.5A
VTh = 12(i1-i2) = 30V
Or node KCL : (32-VTh)/4 + 2
VTh
=VTh/12
1Ω
4Ω
96-3VTh+24 = VTh
32V
2A
VTh
12Ω
VTh = 30V
i1
3.5 Norton Equivalent Circuits
A linear two-terminal circuit can be replaced by an equivalent
circuit consist of current source IN parallel with RN.
Thevenin equivalent circuit consists of an independent
current source in parallel with the Norton equivalent
resistance. It can derive from Thevenin by making source
transformation. Norton current equal to isc terminal of interest
and Norton resistance is identical to RTh.
IN = VTh/RTh
i2
Step3 : Source transformation,
series resistor combined,
producing the Thevenin
equivalent circuit.
Step4 : Source transformation,
producing the Norton
equivalent circuit.
Ex : Derive Thevenin and Norton
equivalent circuit ;
Step1 : Source transformation
Step2 : Combine parallel source
and parallel resistor.
8Ω
Ex : Find Norton equ
RN = 5 || (8+4+8) = 4Ω
Find IN , apply mesh
i1 = 2A, 20i2 – 4i1 – 12 = 0
i2 = 1A = isc = IN
or
Mesh : i3 = 2A
25i4 – 4i3 – 12 = 0
i4 = 0.8A
Voc = VTh = 5i4 = 4V
iN = VTh/RTh = 4/4 = 1A
8Ω
2A
i
12V
2A
i
12V
3.6 Maximum Power Transfer
a
5Ω
4Ω
b
8Ω
a
i 5Ω
12V
2A
Ex : Determine RL value for maximum
power delivery.
4Ω
a
8Ω
5Ω
Circuit analysis is important in system designed to transfer
power from source to a load.
First, the efficiency of power transfer e.g. in power utility
systems, e.g. generation, transmission and distribution.
Second, the amount of power transferred, e.g. comm and
instrumentation on data transmission.
Maximum power transfer is the power that being transmitted
by a source that is represent by a Thevenin equivalent circuit
when the RL is equal to RTh.
b
Draw a Thevenin equivalent circuit.
Power dissipated in RL = i2RL
4Ω
b
=(VTh/(RTh+RL))2RL
7
Next, we recognize that for a given circuit, VTh and RTh will be
fixed. Therefore the power dissipated is a function of he
single variable RL. So ;
dP/dRL = V2Th[((RTh+RL)2 – RL.2(RTh+RL))/(RTh+RL)4]
The derivative is zero and p is maximized when
(RTh + RL)2 = 2RL(RTh + RL)
So RTh = RL
Pmax = i2RL = V2ThRL/(2RL)2 = V2Th/4RL
Ex : Find value RL for max power transfer
VTh = 150/180 x 360 = 300V
RTh = 150||30 = 25Ω
Pmax = (VTh/2RL)2 x RL
30Ω
360V
RL
b
=(300/50)2 x 25 = 900W
when RL = 25Ω
vab = 300/50 x 25 = 150V
a
150Ω
25Ω
a
RL
300V
b
3.7 Superposition
Superposition is whenever a linear system is excited, or
driven, by more than one independent source of energy, the
total response is the sum of the individual responses. It is
applied in analysis and design of circuit.
Superposition principle step ;
1. Switch off all independent source except one. Get the
output either voltage or current of the source.
2. Repeat step one for all independent source.
3. Get the total output of all the source to get the total output.
Ex :
Step 1 : switch off current source.
So v0 = 2kΩ/4kΩ x 10V = 5V
Step 2 : switch off voltage source.
So i0 = 2kΩ/4kΩ x 2mA = 1mA
V0 = 1mA x 2kΩ = 2V
So total v0 = 2V + 5V = 7V
Note :
1. Independent voltage source will become close circuit with
zero voltage.
2. Independent current source will become open circuit.
3. If there is a a dependent current source, it must active
during superposition process.
Ex : Find branch current.
Get v1.
(v1 – 120)/6 + v1/3 +v1/(2+4) = 0
120V
So v1 = 30V
Deactive current source
i1’ = (120-30)/6 = 15A
i2’ = 30/3 = 10A
i3’ = i4’ = 30/6 = 5A
120V
Two nodes voltage equation
v3/3 + v3/6 + (v3-v4)/2 = 0
(v4-v3)/2 + v4/4 + 12 = 0
So v3 = -12V, v4 = -24V
6Ω
i1
2Ω
i3
3Ω
i2
6Ω
i1’
v1
2Ω
i3’
3Ω
i2’
12A
4Ω
i4
v1
4Ω
i4’
i1” = -v3/6 = 12/6 = 2A
i2” = v3/3 = -12/3 = -4A
i3” = (v3-v4)/2 = (12+24)/2 = 6A
i4” = v4/4 = -24/4 = -6A
So
i1 = i1’ + i1” = 17A
i2 = i2’ + i2” = 6A
i3 = i3’ + i3” = 11A
i4 = i4’ + i4” = -1A
6Ω
i1”
2Ω
i3”
3Ω
i2”
v0
4Ω
12A
i4”
2Ω
6Ω
3Ω
12A
4Ω
v3
v4
8