Block A Unit 1 Outline
 Concepts
> Charge & Current; Voltage; Resistance; Power
 Terminology
> Branch, Mesh, Node
 Laws
> Kirchhoff’s current and voltage laws; Ohm’s law
 Resistive networks
> Parallel and series; current and voltage divider rule
 Measuring instruments (application)
> Ammeters and Voltmeters
Block A Unit 1
1
Electric Charge
 Charge is a fundamental electric quantity, measured by the unit Coulomb
(C)
 The smallest amount of charge that exists is the charge that is carried by 1
electron = -1.602 x 10-19 C
 Therefore charge quantities in real life occur in integral multiples of an
electron’s charge
 Typically denoted by the symbol Q
Block A Unit 1
2
Current: Free electrons on the move
+
_
_
_
_
_
_
_
_
_
_
_
_
-
_
_
_
_
Conventional current flows from + to (Opposite direction to electron flow)
Block A Unit 1
_
_
Currents must run in loops
_
Direction of current
_
_
Current in relation to Charge
 Currents arise from flow of charges
 Unit: Ampere (A)
 Typically denoted by the symbol, I
 Current = Rate of change of charge
 Mathematically, I = dQ/dt
3
Fundamental law for charge
 Current has to flow in closed loop
 No current flows if there is a break in the path
 Underlying physical law: Charge cannot be created or
destroyed
 This is the basis of Kirchhoff’s Current Law
i1
i2
i4
i3
Block A Unit 1
Kirchhoff’s current law
Sum of currents at a node must equal
to zero:
i 1 + i 2 + i3 + i 4 = 0
4
Kirchhoff’s current law
Does the direction of current matter?
I2 running into node
i1
YES!!
I2 running out of node
i1
Different
i2
i2
i3
i1 + i2 + i3 = 0
Block A Unit 1
i3
i1 + i2 + i3 = 0 (WRONG)
i1 + i3 = i2 (CORRECT)
5
Kirchhoff’s current law
Sign convention when applying KCL
Currents EXITING (-ve),
Currents ENTERING (+ve)
i1
Entering: I1 & I2 (+ve)
i2
i4
i3
Block A Unit 1
Exiting: I3 & I4 (-ve)
i1 + i2 – i3 – i4 = 0
6
KCL example
Problem 2.14 and 2.15
Find the unknown current using KCL
6A - 5A + 2A - i = 0
6A - 5A + 2A + i = 0
 i = 3A
 i = -3A
Block A Unit 1
7
Voltage/Potential difference
Current
A
_
_
_
_
B
_
Unit: Volt (V)
_
vx = v A - vB
+
-
vx
High (+)
Energy is required to move charges between
2 points
Direction
of Flow
Voltage/potential difference is always made
with reference to 2 points
Low (-)
Block A Unit 1
8
Sign convention
In a LOAD
i
Voltage DROPS in the direction of the current
Energy is dissipated (or consumed)
+
-
v
In a SOURCE
Voltage RISES in the direction of the current
Energy is generated
Block A Unit 1
i
+
-
v
9
Fundamental law on voltage





Energy is required to push electrons through a resistive element
That same energy needs to be generated by a source
Total energy generated in a circuit must equal total energy consumed in the circuit
Energy cannot be created or destroyed
Therefore, voltage rise = voltage drop
- V3 +
Kirchhoff’s voltage law
+
-
V2
V4
Net voltage around a closed
circuit is zero:
+
v1 + v2 + v3 + v4 = 0
-
Block A Unit 1
+ V1 -
10
Kirchhoff’s voltage law
load
source
load
 Reference voltage in a circuit set to 0V
 All other nodes on the circuit can then be conveniently
referenced to GROUND
+ V2 DEFINING signs
 Voltage gain (+ve)
 Voltage drop (-ve)
+
V1
-
v1 – v2 + v3 = 0
+ V3 -
load
Ground symbol
GROUND
Block A Unit 1
11
KVL example
Problem 2.16
Apply KVL to find voltage V1 and V2
Loop 1 (Clock-wise):
5V - 3V - V2 = 0
 V2 = 2V
Loop 2 (Anti-clock-wise):
V1 - 10V - V2 = 0
 V1 = 12V
Block A Unit 1
12
Resistance and Ohm’s Law
+
v
_
i
I
Ideal RESISTOR shows linear
resistance obeying Ohm’s law
1/R
Unit: Ohm (Ω)
V
When current flows through any circuit element, there will always be a
resistance to its flow which results in a voltage drop across that circuit
element
Ohm’s law: V = IR
IMPORTANT: Positive current is
defined here as flowing from higher
to lower voltage (Remember)
Block A Unit 1
R
L
A
A
L
ρ: resistivity (material property)
A: cross-sectional area
13
Ohm’s law + KCL example
Problem 2.17
Use Ohm’s law + KCL to find the current through the 15Ω resistor
KCL: I1 + I2 = 10A
Ohm’s: 15I1 = V15Ω (1); 30I2 = V30Ω (2)
(KVL) V15Ω = V30Ω
Therefore, 2I2 = I1
Solving for the variables:
I2 = 3.33 A, I1 = 6.67 A
Block A Unit 1
14
Electrical Power
Electrical power generated/dissipated in a given element is defined by the
product of the voltage across that element and the current through it
Unit: Watt (W)
+
1.5V
_
P = VI
+
0.2A
1.5V
_
0.2A
P = I2 R
P = V2/R
Source: P = 1.5V x 0.2A = 0.3W
Load: P = 1.5V x 0.2A = 0.3W
Power generated by source
MUST EQUAL
Power dissipated in the load
Block A Unit 1
 In a source, power is generated
 In a load (eg. resistor), power is
dissipated/consumed
15
Power example
Problem 2.22
Determine which components are absorbing power and which are delivering power
Is conservation of power observed in this example?
Block A Unit 1
16
Power example solution
Finally calculate power through each element
A: (12V)(5A) = 60W [generating]
B: (3V)(5A) = 15W [generating]
C: (-5V)(5A) = -25W [absorbing]
D: (-10V)(3A) = -30W [absorbing]
E: (-10V)(2A) = -20W [absorbing]
Last part: “Is power conserved?”
Generating: 60W + 15W = 75W
Absorbing: 25W + 30W + 20W = 75W
YES!
This slide is meant to be blank
Block A Unit 1
17
Terminology: Branch and Node
Branch
BRANCH: Any path of a circuit
with 2 TERMINALS connected to it
Node
NODE: Junction of
2 or more branches
Block A Unit 1
18
Terminology: Loop and Mesh
Vs
+
-
LOOP: Any closed connection of branches
In the above circuit, there are 6 loops in total
Vs
+
-
Mesh: Loop that does not contain other loops
In the above circuit, there are 3 meshes in total
Block A Unit 1
19
DC vs AC
 DC – Direct Current: Current is constant with time
 AC – Alternating Current: Current varies with time and
reverses direction periodically
Block A Unit 1
20
Independent Voltage Source
+
v
+
i
+
_
v
Circuit
i
_
_
Independent Voltage Source supplies a prescribed voltage across
its terminals irrespective of current flowing through it
Current supplied is determined by load circuit connected
Block A Unit 1
21
Independent Current Source
+
+
i
i
Circuit
i
_
v
_
Independent Current Source supplies a prescribed current to any
load circuit connected to it
Voltage supplied is determined by load circuit connected
Block A Unit 1
22
Dependent Sources
 DEPENDENT source generates v or i that is a function of
some other v or i in the circuit
+
 Symbol – diamond shape outline
+
5V
vx
-
-
vs
+
_
Independent
Voltage Controlled Voltage Source (VCVS): vs = μvx
Current Controlled Voltage Source (CCVS): vs = rix
Voltage Controlled Current Source (VCCS): is = gvx
Current Controlled Current Source (CCCS): is = βix
Block A Unit 1
is
23
Example of a VCVS
Ro
+
vi
-
Ri
+
-
Avi
Note: This is a general model for an amplifier
which shall be re-visited later in BLOCK C
Block A Unit 1
24
Short Circuit
A
0.1A
20Ω
B
Short circuit: Connect 2 or more terminals so that
the voltage between each of them is the same
Typically associated with current, eg. short-circuit
current
Short circuit A & B
A
0.1A
20Ω
20Ω resistor is now bypassed
Current from source flows through the shortcircuit to give a short-circuit current
Isc
B
Short circuit current Isc = 0.1A
Block A Unit 1
25
Open Circuit
A
0.1A
20Ω
Typically associated with voltage,
eg. open-circuit voltage
B
A
0.1A
Block A Unit 1
Ioc= 0A
+
20Ω
B
Open circuit: Leave 2 terminals
unconnected externally
Voc
-
Open circuit voltage
Voc = 0.1A × 20Ω = 2V
26
Self-contradictory circuits
A
Prefixes: Memorize and apply them!
+
vs B
What is the voltage across A and B?
A
Is
B
tera
T
1012
giga
G
109
mega
M
106
kilo
k
103
milli
m
10-3
micro
μ
10-6
nano
n
10-9
pico
p
10-12
femto
f
10-15
Write as 2.15mA instead of 0.00215A
What is the current arriving at A?
Block A Unit 1
27
Parallel network (Highlights)
I1
Is
I2
R1
IN
R2
RN
Is
RP
Equivalent Resistance
1/RP = 1/R1 + 1/R2 + …+ 1/RN
Current divider rule
1 RN
IN 
IS
1 RP
Block A Unit 1
28
Parallel Network: Proof
X I2
Apply Kirchhoff’s current law (KCL) at X:
I1
I s = I2 + I 1
R1
Is
R2
This can be seen as Is is split into the 2 branches
X
2 I2
1
3
Note that points 1, 2, 3 are all at the same voltage,
therefore same node (X)
I1
R1
Is
4
Y
Block A Unit 1
5
R2
6
Points 4, 5, 6 are all at the same voltage, therefore
same node (Y)
Voltage across R1 = Voltage across R2 = VXY
29
Parallel network: Proof
X
2 I2
1
Apply Ohm’s to both R1 and R2:
3
VXY = I1R1 = I2R2
I1
R1
Is
4
Y
5
R2 I1 = VXY/R1; I2 = VXY/R2
6
Adding I2 & I1 up according to KCL:
Is = VXY(1/R1 + 1/R2)
We can now find how much of Is is distributed between the 2 branches
 1 R1

 I s
I1  
 1 R1  1 R2 
 1 R2 
 I s
I 2  
 1 R1  1 R2 
This is referred to as the current divider rule
Block A Unit 1
30
Parallel network: Proof
I2
Is = VXY(1/R1 + 1/R2)
I1
R1
Is
R2
Replace the parallel network of
resistors with a single equivalent
resistor
Remember that the voltage across RP is still VXY!
X
VXY/RP = VXY(1/R1 + 1/R2)
RP
Is
1/RP = 1/R1 + 1/R2
Y
Block A Unit 1
31
Parallel network (seeing it)
Current splits at one node
I1
Is
I2
R1
IN
R2
RN
Current re-combine at the other node
Suggestion: Think about parallel resistors as the rungs on a ladder
Block A Unit 1
32
Current divider: Example 1
V1


I
i1
R1
i2
V2
R2

Find i1 and i2 in terms of I
If R1 = R2, find the ratio between
i1 and i2
If R2 = 3R1, find the ratio between
i1 and i2
i1 = [R2 / (R1 + R2)]I; i2 = [R1 / (R1 + R2)]I
If R1 = R2, then i1 = i2 = I/2
If R2 = 3R1, then i1 = 3I/4, i2 = I/4
Therefore i1 = 3i2
Block A Unit 1
33
Current divider: Example 2

V1

I
i1
R
i2
R
i3
R

Find the current through each
resistor in terms of I
Find the current through each
resistor in terms of I if a 4th
resistor was added in parallel
How many resistor are required
to reduce the current in each
resistor to 1% of I (ie 1/100)
V2
i1 = i2 = i3 = I/3
Consider that, in = I/n
i1 = i2 = i3 = i4 = I/4
Hence n = 100 for in = 0.01I
Block A Unit 1
34
Current divider: Example 3
V1
I
i1
R1
i2
R2 i3
R3
V2
I3 > I5 > I1 > I4 > I2
Block A Unit 1
i4
R4
i5
R5
If R2 > R4 > R1 > R5 > R3

Which current is the largest?

Which current is the smallest?

Rank the currents from largest to
smallest
35
Series network (Highlights)
+
R1
Vs
+
-
R2
V1
+
Vs
+
-
RS
V2
-
+
RN
VN
-
Equivalent Resistance
RS = R1 + R2 + …+ RN
Voltage divider rule
VN = VS(RN/RS)
Block A Unit 1
36
Series network: Proof
+
R1
Vs
+
Is
R2
V1
+
V2
-
Apply Ohm’s law to both resistors:
V1 = ISR1; V2 = ISR2
Adding up V1 and V2 according to KVL:
Vs = IS(R1+R2)
We can now find how much of Vs is distributed between the 2 resistors
 R1 
Vs
V1  
 R1  R2 
 R2 
Vs
V2  
 R1  R2 
This is referred to as the voltage divider rule
Block A Unit 1
37
Series network: Proof
R1
Vs
+
R2
Vs = IS(R1+R2)
Replace the series network of resistors
with a single equivalent resistor
Remember that the current going into
RS is still IS!
IsRS = IS(R1+R2)
Vs
+
-
Block A Unit 1
RS
RS = R1+R2
38
Series network (seeing it)
R1
Vs
+
-
R2
+
V1
+
V2
-
RN
+
VN
-
Suggestion: Think about series resistors as the rings forming a chain
Block A Unit 1
39
Voltage Divider: Example 1
2V
5V
10Ω
20Ω
30Ω
Node 1
Node 2
 Find the voltage across each resistor
 Find the voltage at nodes 1 and 2
Total resistance (in series) RT = 10 + 20 + 30 = 60Ω
V10Ω = (1/6)(3) = 0.5V; V20Ω = (1/3)(3) = 1V; V30Ω = (1/2)(3) = 1.5V
At node 1:
V1 = 2 + 0.5 = 2.5V
At node 2:
V2 = 5 – 1.5 = 3.5V
Block A Unit 1
40
Voltage Divider: Example 2
0V
9V
R1
If R3=3R1 and R2 = 2R1
R2
Node 1
R3
Node 2
• Find the voltage across each resistor
• Find the voltage at nodes 1 and 2
VR1 = (R1/RT)(VB - VA) = 9/6 = 1.5V; VR2 = (R2/RT)(VB - VA) = 9/3 = 3V
VR3 = (R3/RT)(VB - VA) = 9/2 = 4.5V
At node 1: V1 = VR1 + VA = VB/6
At node 2: V1 = VB - VR3 = VB/2
Block A Unit 1
41
Voltage Divider: Example 3
+ V1 -
+ V2 -
+ V3 -
+ V4 -
+ V5 -
VA
VB
R1
R2
R3
R4
R5
If R2 > R4 > R1 > R5 > R3

Which voltage difference is the largest?

Which voltage difference is the smallest?

Rank the voltage differences from largest to smallest
V2 > V4 > V1 > V5 > V3
Block A Unit 1
42
Series-Parallel combo: Example 1
Find the total resistance as seen across A and B
A
All resistors have value of 1Ω
A
RAB = (R1 || R2) + (R3 || R4)
R1
= 1||1 + 1||1 = 1Ω
R2
R3
R1
R2
R3
R4
R4
RAB = (R1 + R3) || (R2 + R4)
= 2||2 = 1Ω
B
B
Block A Unit 1
43
Combo example 2
Problem 2.48
Find the equivalent resistance seen by the source
How much power is delivered by the source?
Combine the 1Ω and 2Ω in series: R1 = 3Ω
Combine the R1 with 3Ω in parallel: R2 =
1.5Ω
Combine the R2 with 4Ω & 5Ω in series: R3
= 10.5Ω
Combine the R3 with 6Ω in parallel: R4 =
3.818Ω
Combine the R4 with 7Ω in series: RT =
10.818Ω
Power delivered by source: P = VS2/RT =
18.1W
Block A Unit 1
44
Combo example 2
This slide is meant to be blank
Block A Unit 1
45
Combo example 3
Problem 2.48
Find the equivalent resistance seen by the source
Find the current through the 90Ω resistor
Block A Unit 1
46
Combo example 3
This slide is meant to be blank
Block A Unit 1
47
Power dissipation
Power: P = V I
i
Power dissipated in resistor R
R
P = I2R = V2/R
+V -
Power dissipated: P = v22 / R2
R2
2
P
V
2 s
( R1  R2 )
R1
+
-
+
Vs
R2
v2
_
For Max Power Transfer:
R1 = R2
Find condition for dP/dR2 = 0 (max pt)
Block A Unit 1
48
Practical voltage source
Motivation: The ideal source does not consider internal resistance of sources so we
need to modify ideal model to describe the physical limitations in practical sources:
Ideal voltage source in series with a source resistance
Rs
Vs +
RL
-
Ideal
Vs +
RL
-
Practical
Terminal voltage becomes:
VRL = [RL/(RL+RS)]VS instead of VS
VRL is now lower due to some voltage drop across RS
Block A Unit 1
49
Practical current source
Motivation: The ideal source does not consider internal resistance of sources so we
need to modify ideal model to describe the physical limitations in practical sources:
Ideal current source in parallel with a source resistance
Is
RL
Ideal
Is
Rs
RL
Practical
Current through the load now becomes:
IRL = [Rs/(RL+RS)]IS instead of IS
IRL is now lower due to a fraction of the source current
flowing into RS
Block A Unit 1
50
Voltmeter
 Voltmeter measures voltage across a circuit element
 Connected in parallel with the element being measured
R1
+ v
s
_
R1
R2
V
IDEAL
+
_ vs
R2
RV
V
ACTUAL
Voltmeter should draw as little current away R2 in the
main circuit (application of current divider rule)
Block A Unit 1
51
Voltmeter Example
i
vs
+
_
im
R1
R2
RV
V
If R1 = R2 = 1kΩ, and RV = 1MΩ
Find VL without the voltmeter connected across R2
Find VL with the voltmeter connected across R2
What happens when R1 and R2 are now 500kΩ?
Block A Unit 1
52
Voltmeter example solution
Without voltmeter:
VL = 0.5VS
With voltmeter (When R2 = 1kΩ):
Substitute R2 with R2 || RV = 1k || 1M = 999Ω
VL = 0.4997VS
With voltmeter (When R2 = 500kΩ):
Substitute R2 with R2 || RV = 0.5M || 1M = 333kΩ
VL = 0.4VS
This slide is meant to be blank
Block A Unit 1
53
Ammeter
 Ammeter measures current flowing through an element
 Connected in series with the element being measured
vs
+
_
R1
i
A
vs
R1
+
_
i
A
IDEAL
RA
ACTUAL
RA should contribute as little as possible to the overall series
resistance in the circuit (application of voltage divider rule)
Block A Unit 1
54
Ammeter Example
i
Vs
R1
+
_
Vs = 5V, RA = 0.5Ω
RA
A
For R1 = 500Ω and then 2.5 Ω,
Find i without and with the ammeter included
Block A Unit 1
55
Ammeter Example solution
Without ammeter:
I = VS/R1 = 10 mA
With ammeter:
When R1 = 500Ω (RA << R1),
I = VS/(R1 + RA) = 9.99 mA
When RA = 0.5Ω and R1 = 2.5Ω (RA < R1)
I = VS/(R1 + RA) = 1.67 A
This slide is meant to be blank
Block A Unit 1
56