SECTION 2:
RESISTIVE CIRCUIT ANALYSIS I
MAE 2055 – Mechetronics I
2
K. Webb
Resistance & Conductance
MAE 2055 – Mechetronics I
Resistance
3


Electrical resistance is the degree to which a circuit
element opposes the flow of electrical current
Shown as a resistor in a schematic diagram
R


Resistance is measured in ohms (Ω)
Resistors may be discrete, intentional circuit
components, or parasitic resistance of wires,
cables, interconnects, etc.
K. Webb
MAE 2055 – Mechetronics I
Resistance – fluid analogy
4

Electrical resistance is analogous to the resistance
of a pipe to fluid flow due to friction
small-diameter pipe
large resistor
d1
d1 < d2
R1 > R 2
large-diameter pipe
small resistor
d2
K. Webb
MAE 2055 – Mechetronics I
Resistance – thermal analogy
5

Electrical resistance is analogous to the resistance
of a solid to the conduction of heat
thick slab
large resistor
w1
R1 > R2
w1 > w2
thin slab
small resistor
w2
K. Webb
MAE 2055 – Mechetronics I
Conductance
6




Electrical conductance is the degree to which a
circuit element allows the flow of electrical current
Conductance is the inverse of resistance
Denoted as G
G
Conductance is measured in siemens or mhos
(S or Ω-1)
K. Webb
MAE 2055 – Mechetronics I
Real Resistors
7


Resistors for use in electronic circuits come in
many shapes and sizes depending on their target
application
Physical size primarily determined by the resistor’s
power handling capability
 Larger
resistors can generally dissipate more power
before burning up

Two primary form factors of discrete resistors are
axial lead and chip resistors
K. Webb
MAE 2055 – Mechetronics I
Axial Lead resistors
8


Cylindrical resistive component with
wire leads extending from each end
Used with through-hole technology
printed circuit boards (PCB’s)


Useful for prototyping
Size varies with power handling capacity
http://www.doctronics.co.uk/resistor.htm
K. Webb
http://www.doctronics.co.uk/resistor.htm
MAE 2055 – Mechetronics I
Resistor Color Code
9
http://www.elexp.com/t_resist.htm
K. Webb
MAE 2055 – Mechetronics I
Chip Resistors
10

Small rectangular
footprint
0805 – 0.080” x 0.050”
 0603 – 0.060” x 0.030”
 0402 – 0.040” x 0.020”



Used with surfacemount technology
PCB’s
More common than
axial lead in modern
electronics
K. Webb
http://www.mini-systemsinc.com/msithick/chipanat.asp
MAE 2055 – Mechetronics I
11
K. Webb
Ohm’s Law
MAE 2055 – Mechetronics I
Ohm’s Law
12
V
I
R
Georg Simon Ohm, 1789 – 1854
“The current through a resistor is proportional to the voltage across
the resistor and inversely proportional to the resistance.”
K. Webb
MAE 2055 – Mechetronics I
Ohm’s Law – said differently
13
V  I R
Georg Simon Ohm, 1789 – 1854
“The voltage across a resistor is proportional to the current through the
resistor and proportional to the resistance.”
K. Webb
MAE 2055 – Mechetronics I
Ohm’s Law – fluid analogy
14

Voltage is analogous to pressure

Driving potentials

Electrical current is analogous to flow rate

A pipe carrying fluid has some resistance determined by physical
characteristics (length, diameter, roughness, etc.)
resistor
v1
v2
Section of pipe
p1
K. Webb
Rpipe
p2
MAE 2055 – Mechetronics I
Ohm’s Law – thermal analogy
15

Voltage is analogous to temperature

They are both driving potentials

Electrical current is analogous to heat flux

A solid slab or wall has some thermal resistance determined by physical
characteristics (thickness, material properties, etc.)
resistor
v1
v2
solid wall
T1
K. Webb
T2
MAE 2055 – Mechetronics I
16
K. Webb
Series & Parallel Circuits
MAE 2055 – Mechetronics I
Series Circuits
17

Components connected in series share one
common node – connected end-to-end
 Equal
current flows through each component
I1
Is
K. Webb
I2
Resistors, R1 and R2, and
voltage source, Vs, are all
connected in series
Is = I1 = I2
MAE 2055 – Mechetronics I
Parallel Circuits
18

Components connected in parallel share two
common nodes – connected side-by-side
 Equal
voltage across each component
Is
K. Webb
I1 +
V1
-
I2
+
V2
-
Resistors, R1 and R2, and
voltage source, Vs, are all
connected in parallel
Vs = V1 = V2
MAE 2055 – Mechetronics I
Series Resistances
19

Resistances in series add


Can be replaced with an equivalent resistance whose value is the sum
of the series-connected resistances
Equivalent resistance will be greater than either individual resistance

All current must flow through both resistances
Rs = R1 + R2
K. Webb
MAE 2055 – Mechetronics I
Parallel Resistances
20

Conductances in parallel add


Can be replaced with an equivalent resistance whose conductance is
the sum of the parallel-connected conductances
Equivalent resistance will be less than either individual resistance

K. Webb
Some current flows through each resistor (amount determined by Ohm’s Law)
MAE 2055 – Mechetronics I
21
K. Webb
Voltage & Current Dividers
MAE 2055 – Mechetronics I
Voltage Divider
22


Series-connected resistors divide the voltage across them
Voltage across each resistor given by Ohm’s Law
K. Webb
MAE 2055 – Mechetronics I
Voltage Divider
23



Equal current through each resistor
Voltage drop across each resistor is
proportional to the value of that resistor
Use a voltage divider to generate
desired voltages from a single reference
voltage or supply
If R1 = R2
K. Webb
MAE 2055 – Mechetronics I
Current Divider
24



Parallel-connected resistors split the flow of current
Equal voltage across each resistor
Current through each resistor given by Ohm’s Law
K. Webb
MAE 2055 – Mechetronics I
25
K. Webb
Kirchhoff’s Laws
MAE 2055 – Mechetronics I
Kirchhoff’s Current Law – (KCL)
26
“The algebraic sum of currents entering
any node must be zero.”


Charge cannot accumulate in a node
What flows in, must flow out
Gustav Kirchhoff, 1824 – 1887
K. Webb
MAE 2055 – Mechetronics I
KCL & the Conservation of Mass
27




Imagine fluid-carrying pipes connected in a Tee
Tee connector is analogous to electrical node
Pipes analogous to branches
According to the conservation of mass, what flows in must flow out

The sum of the mass flow rates must be zero
node
branches
K. Webb
MAE 2055 – Mechetronics I
KCL – an example
28



KCL says the sum of the currents flowing into a node is zero
Same is true of the sum of currents flowing out of a node
Sign of the current value gives the direction

When summing currents into a node, negative current flows out
Determine the current, I3, flowing
through the 1 KΩ resistor
K. Webb
MAE 2055 – Mechetronics I
Kirchhoff’s Voltage Law – (KVL)
29
“The algebraic sum of voltage drops taken
around any loop in a network is equal to zero.”
KVL around Loop 1
KVL around Loop 2
Gustav Kirchhoff, 1824 – 1887
K. Webb
MAE 2055 – Mechetronics I
KVL & the Conservation of Energy
30



Voltage in a circuit is analogous to
potential energy while traversing a
mountainous loop
PE varies with elevation
 Increases with each climb
 Decreases with each descent
Initial/final elevation & PE are the same
 Sum of PE rises/drops around the loop
is zero - just like voltage drops around
a loop in a circuit
Elevation
K. Webb
MAE 2055 – Mechetronics I
KVL – an example
31



KVL says the sum of the voltage drops around a loop is zero
Same is true of the sum of voltage rises around a loop
Sign of the voltage value gives polarity

Negative voltage has opposite polarity from that assumed when
applying KVL
Determine the voltage, V2, across
the 3.3 KΩ resistor
K. Webb
MAE 2055 – Mechetronics I
32
K. Webb
Resistive Network Analysis
MAE 2055 – Mechetronics I
Circuit Analysis Methods
33



Ohm’s Law and Kirchhoff’s Laws can be used to
determine unknown currents and voltages in a
circuit
Formal circuit analysis methods allow us to derive
a set of simultaneous equations that completely
characterize all circuit operating conditions
Mesh (Loop) and nodal analysis systematically
employ Kirchhoff’s Laws and Ohm’s Law to
determine all currents and voltages in a circuit
K. Webb
MAE 2055 – Mechetronics I
Mesh/Loop Analysis
34





Systematic application of KVL to obtain a system of
equations with mesh currents as unknown variables
Mesh currents are fictitious circulating currents – not
necessarily equal to branch currents
Number of simultaneous equations equals the number
of unknown mesh currents
Solve the system of equations using a convenient
method to obtain mesh currents
Once mesh currents are determined, Ohm’s Law then
used to obtain node voltages
K. Webb
MAE 2055 – Mechetronics I
Meshes and Mesh Currents
35

What is a mesh?



A mesh is a loop that does not contain any other loops
Loop 1 and Loop 2 are meshes, Loop 3 is not
What is a mesh current?



Fictitious circulating current in a mesh
Mesh current in Loop 1 is I1, mesh current in Loop 2 is I2
Not necessarily equal to branch currents

Current in branch 1 is I1, but current in branch 2 is I1 – I2
I1
K. Webb
I2
MAE 2055 – Mechetronics I
Mesh Analysis – step-by-step procedure
36
1)
If a reference (ground) node is not specified, select one – this
node is defined to be 0 V.
2)
Identify all meshes in the circuit and label all mesh currents and
branch currents, clearly indicating polarity. Identify and label all
unknown node voltages.
3)
Apply KVL around each mesh, using Ohm’s Law to express voltage
drops in terms of mesh currents.
4)
Solve the resulting simultaneous system of equations using
substitution, Cramer’s Rule, MATLAB, etc.
5)
Determine branch currents from calculated mesh currents.
6)
Use Ohm’s Law and branch currents to determine node voltages.
K. Webb
MAE 2055 – Mechetronics I
Mesh Analysis – an example
37

K. Webb
Use mesh analysis
to determine all
node voltages and
branch currents in
the circuit shown
MAE 2055 – Mechetronics I
Mesh Analysis – step 1
38


Step 1: Select a
reference node if
one is not specified
Select a node that:
 Connects
the
maximum number
of elements
 Is connected to the
maximum number
of voltage sources
K. Webb
MAE 2055 – Mechetronics I
Mesh Analysis – step 2
39



K. Webb
Step 2: Identify all
meshes in the circuit
and label all mesh
currents, clearly
indicating polarity
Two meshes and
unknown mesh
currents in this circuit
Choice of direction of
currents is arbitrary –
sign will determine
actual direction
MAE 2055 – Mechetronics I
Mesh Analysis – step 3
40



K. Webb
Step 3: Apply KVL
around each mesh,
using Ohm’s Law to
express voltage drops
in terms of mesh
currents
Pay careful attention
to polarities
Note that some
voltage drops will be
functions of two mesh
currents (e.g. the
voltage across R2)
MAE 2055 – Mechetronics I
Mesh Analysis – step 3, cont’d
41

Step 3: Apply KVL around each mesh, using Ohm’s Law
to express voltage drops in terms of mesh currents
KVL around mesh 1
KVL around mesh 2
K. Webb
MAE 2055 – Mechetronics I
Mesh Analysis – step 4
42

Step 4: Solve the resulting simultaneous system of equations
Rearrange system of equations in matrix form:
K. Webb
MAE 2055 – Mechetronics I
Mesh Analysis – step 4, cont’d
43

Can solve by simple substitution or by using Cramer’s Rule
1.5𝑘Ω
−1𝑘Ω
−1𝑘Ω 𝐼1
5𝑉
=
=𝑹∙𝑰=𝑽
1.6𝑘Ω 𝐼2
0
According to Cramer’s Rule:
det 𝑹𝟏
𝐼1 =
,
det 𝑹
det 𝑹𝟐
𝐼2 =
det 𝑹
where 𝑹𝟏 is obtained by replacing the first column of 𝑹 with 𝑽, and 𝑹𝟐 is
obtained by replacing the second column of 𝑹 with 𝑽.
5𝑉 −1𝑘Ω
0 1.6𝑘Ω = 𝟓. 𝟕𝟏𝟒𝒎𝑨,
𝐼1 =
1.5𝑘Ω −1𝑘Ω
−1𝑘Ω 1.6𝑘Ω
K. Webb
1.5𝑘Ω 5𝑉
𝐼2 = −1𝑘Ω 0
= 𝟑. 𝟓𝟕𝟏𝒎𝑨
1.5𝑘Ω −1𝑘Ω
−1𝑘Ω 1.6𝑘Ω
MAE 2055 – Mechetronics I
Mesh Analysis – step 4, cont’d
44

Can also solve with the help of MATLAB – useful for large circuits
I1
I2
K. Webb
MAE 2055 – Mechetronics I
Mesh Analysis – step 5
45

Step 5: Determine branch currents from calculated
mesh currents

Denote branch currents as in
to distinguish them from
mesh currents, In
𝑖1 = 𝐼1 = 𝟓. 𝟕𝟒𝒎𝑨
𝑖2 = 𝐼1 − 𝐼2 = 𝟐. 𝟏𝟒𝟑𝒎𝑨
𝑖3 = 𝐼2 = 𝟑. 𝟓𝟕𝟏𝒎𝑨
K. Webb
MAE 2055 – Mechetronics I
Mesh Analysis – step 6
46

Step 6: Use Ohm’s Law and branch currents to
determine node voltages
𝑉1 = 𝑉𝑠 − 𝑖1 𝑅1
𝑉1 = 5𝑉 − 5.714𝑚𝐴 ∙ 500Ω = 𝟐. 𝟏𝟒𝟑𝑽
or
𝑉1 = 𝑖2 𝑅2 = 2.143𝑚𝐴 ∙ 1𝑘Ω = 𝟐. 𝟏𝟒𝟑𝑽
𝑉2 = 𝑉1 − 𝑖3 𝑅3
𝑉2 = 2.143𝑉 − 3.571𝑚𝐴 ∙ 200Ω = 𝟏. 𝟒𝟐𝟗𝑽
or
𝑉2 = 𝑖3 𝑅4 = 3.571𝑚𝐴 ∙ 400Ω = 𝟏. 𝟒𝟐𝟗𝑽
K. Webb
MAE 2055 – Mechetronics I
Nodal Analysis
47




Systematic application of KCL to obtain a system of
equations with node voltages as unknown
variables
Number of simultaneous equations equals the
number of unknown node voltages
Solve the system of equations using a convenient
method to obtain node voltages
Once node voltages are determined, Ohm’s Law
then used to obtain branch currents
K. Webb
MAE 2055 – Mechetronics I
Nodal Analysis – step-by-step procedure
48
1)
2)
3)
4)
5)
6)
K. Webb
If a reference (ground) node is not specified, select
one – this node is defined to be 0 V
Identify and label all nodes in the circuit – distinguish
known from unknown node voltages
Assign and label polarities of voltages across, and the
corresponding currents through, all branches
Apply KCL at each node, using Ohm’s Law to express
branch currents in terms of node voltages
Solve the resulting simultaneous system of equations
using substitution, Cramer’s Rule, MATLAB, etc.
Use Ohm’s Law and node voltages to determine
branch currents
MAE 2055 – Mechetronics I
Nodal Analysis – an example
49

K. Webb
Use nodal analysis
to determine all
node voltages and
branch currents in
the circuit shown
MAE 2055 – Mechetronics I
Nodal Analysis – step 1
50


Step 1: Select a
reference node if
one is not specified
Select a node that:
 Connects
the
maximum number
of elements
 Is connected to the
maximum number
of voltage sources
K. Webb
MAE 2055 – Mechetronics I
Nodal Analysis – step 2
51


K. Webb
Step 2: Identify and
label all nodes in
the circuit –
distinguish known
from unknown
node voltages
Vs is a known node
voltage (5 V) – V1
and V2 are unknown
MAE 2055 – Mechetronics I
Nodal Analysis – step 3
52


Step 3: Assign and label
polarities of voltages
across, and
corresponding currents
through, all branches
Assumed polarities
needn’t be correct


K. Webb
Correct polarity given
by the sign of the
determined quantity
Current and voltage
polarities should be in
agreement
MAE 2055 – Mechetronics I
Nodal Analysis – step 4
53

Step 4: Apply KCL at each node, using Ohm’s Law to
express branch currents in terms of node voltages
KCL at node 1
𝐼1 − 𝐼2 − 𝐼3 = 0
5𝑉 − 𝑉1 𝑉1 𝑉1 − 𝑉2
−
−
=0
𝑅1
𝑅2
𝑅3
KCL at node 2
𝐼3 − 𝐼3 = 0
𝑉1 − 𝑉2 𝑉2
−
=0
𝑅3
𝑅4
K. Webb
MAE 2055 – Mechetronics I
Nodal Analysis – step 5
54

Step 5: Solve the resulting simultaneous system of equations
5𝑉 − 𝑉1 𝑉1 𝑉1 − 𝑉2
−
−
=0
𝑅1
𝑅2
𝑅3
𝑉1 − 𝑉2 𝑉2
−
=0
𝑅3
𝑅4
𝑉1 −
𝑉1
1
1
1
1
5𝑉
−
−
+ 𝑉2
=−
𝑅1 𝑅2 𝑅3
𝑅3
𝑅1
1
1
1
+ 𝑉2 − −
=0
𝑅3
𝑅3 𝑅4
Rearrange system of equations in matrix form:
−
1
1
1
−
−
𝑅1 𝑅2 𝑅3
1
𝑅3
K. Webb
1
5𝑉
𝑅3
𝑉1
−
=
𝑅1
1
1 𝑉2
− −
0
𝑅3 𝑅4
−8𝑚𝑆
5𝑚𝑆
𝑉1
−10𝑚𝐴
5𝑚𝑆
=
0
−7.5𝑚𝑆 𝑉2
MAE 2055 – Mechetronics I
Nodal Analysis – step 5, cont’d
55

Can solve by simple substitution or by using Cramer’s Rule
−8𝑚𝑆
5𝑚𝑆
𝑉1
−10𝑚𝐴
5𝑚𝑆
=
=𝑮∙𝑽=𝑰
0
−7.5𝑚𝑆 𝑉2
According to Cramer’s Rule:
det 𝑮𝟏
𝑉1 =
,
det 𝑮
det 𝑮𝟐
𝑉2 =
det 𝑮
where 𝑮𝟏 is obtained by replacing the first column of 𝑮 with 𝑰, and 𝑮𝟐 is
obtained by replacing the second column of 𝑮 with 𝑰.
−10𝑚𝐴
5𝑚𝑆
0
−7.5𝑚𝑆 = 𝟐. 𝟏𝟒𝟑𝑽,
𝑉1 =
−8𝑚𝑆
5𝑚𝑆
5𝑚𝑆 −7.5𝑚𝑆
K. Webb
−8𝑚𝑆 −10𝑚𝐴
0
𝑉2 = 5𝑚𝑆
= 𝟏. 𝟒𝟐𝟗𝑽
−8𝑚𝑆
5𝑚𝑆
5𝑚𝑆 −7.5𝑚𝑆
MAE 2055 – Mechetronics I
Nodal Analysis – step 5, cont’d
56

Can also solve with the help of MATLAB – useful for large circuits
𝑉1
−8𝑚𝑆
=
𝑉2
5𝑚𝑆
5𝑚𝑆
−7.5𝑚𝑆
−1
−10𝑚𝐴
= 𝑽 = 𝑮−𝟏 𝑰
0
V1
V2
K. Webb
MAE 2055 – Mechetronics I
Nodal Analysis – step 6
57

Step 6: Use Ohm’s Law and node voltages to
determine branch currents
𝑉𝑠 − 𝑉1 5𝑉 − 2.143𝑉
𝐼1 =
=
= 𝟓. 𝟕𝟏𝟒𝒎𝑨
𝑅1
500Ω
𝐼2 =
𝑉1 2.143𝑉
=
= 𝟐. 𝟏𝟒𝟑𝒎𝑨
𝑅2
1𝑘Ω
𝑉2 1.429𝑉
𝐼3 =
=
= 𝟑. 𝟓𝟕𝟏𝒎𝑨
𝑅4
400Ω
K. Webb
MAE 2055 – Mechetronics I
Nodal & Mesh Analysis – results
58

Both nodal and mesh analyses yield all the branch
currents and node voltages in the circuit
 Note
that in the preceding example, each method
yields identical results

Choose method based on desired information – do
you want to determine node voltages or branch
currents?
 Node
method yields node voltages first
 Mesh method yields branch currents first
K. Webb
MAE 2055 – Mechetronics I