Contents

Technical Mathematics For Electronics
1. Algebra Review
1.1. Quadratic Equations
1.2. Equation of a Straight Line
1.3. Graphical and Algebraic Representation of Voltage and Current Waveforms

2. Systems of Linear Equations and DC Networks

2.1. Solve Systems of Equations using Graphical Methods
2.2. Solve Systems of Equations using Substitution Methods
2.3. Solve Systems of Equations using Elimination Methods
2.4. Gauss and Gauss-Jordan Elimination Methods of Solving Linear System of Equations
2.4.1. Solve Systems of Equations using Gauss Elimination 
2.4.2. Solve Systems of Equations using Gauss-Jordan Elimination
2.5. Use Cramer's Rule to express the solution of a system in Determinant Form
2.5.1. Solving a 2x2 system of equations using determinants
2.5.2. Solving a 3x3 system of equations using determinant
2.6. Evaluate a Determinant of any Order
2.7. Redundant and Inconsistent Systems
 



Physics

1. Measurement and Units

1.1. Scientific Method

1.2. Measurement

Circuit Analysis

1. Introduction
1.1. Force between Charges
1.2. Voltaic Cell
1.3. Electromagnetism

1.4. Magnetic Effects

1.5. Ohm's Law

1.6. Electromagnetic Induction

1.7. Capacitor

1.8. Maxwell's Equations

1.9. Electromagnetic Theory of Light

1.10. Microwaves

1.11. Laws of Voltages and Currents

1.12. X-Rays

1.13. Electricity

1.14. Electronics

1.15. Power Generation and Distribution

1.16. Communication Systems

1.17. Edison Effect - Transmission of Radio Waves

1.18. Transmission of Radio Waves

1.19. Electromagnetic Induction

1.20. Electromagnetic Induction

3. Resistance

3.1. Memristor

DC Circuit Basics

DC Circuit Basics

v - voltage or potential difference
SI unit: volt ; symbol: V
named for Alessandro Volta (voltaic pile  ~ first battery)

I - current
SI unit: ampere or amp ; symbol: A
named for André-Marie Ampère (Electromagnetism)

R - resistance
SI unit: ohm ; symbol: W
named for Georg Simon Ohm (extented Volta's work, math prof too!!)

Ohm's Law: Ohm demonstrated experimentally that over a wide range of operating conditions, that v, I and R are related by:

v = I R

v = R I
y = m x + b

Ohm's Law (version 1):

v = I . R  or  I = v / R or  R = v / I

The end of the resistance at which the assumed current enters is called the assumed high potential terminal (+);

The end of the resistance at which the assumed current leaves is called the assumed low potential terminal (-);

Thus, in a resistor, current is said to flow from high potential to low potential. This also referred to as voltage drop.

Example:

Calculate v_R.

Solution:

v_R = R I = (12 W) (2 A)

v_R = 24 V

Example:

Calculate I.

Solution:

I = v / R = (16 V) / (4 W)

I = 4 A

Example:

Add caption

Calculate v.

Solution:

v = R I = (5.2 x 10⁶ W) (2 x 10^-3 A)

v = 10 400 V

Ohm's Law (version 2):

Note: Useful in Nodal analysis.

Clearly, v = V_A - V_B

\ I =  (V_A - V_B) / R   or    (V_A - V_B) / = I R

\ I =  V_AB / R            or    V_AB = I R

V_A = potential at A relative to a Common or Ground

V_B = potential at B relative to Same Common or Ground

We define  V_AB = (V_A - V_B)  to be the potential at A with respect to B.

Comments:

Generally,

I = [(Potential at end of reistor where assumed current enters) - (Potential at end of reistor where assumed current leaves)] / R

Example:

Calculate I (an assumed current)

Calculate V_AB

Solution:

I = (V_A - V_B) / R = (16 V - 5 V) / (10 W) = 15 V / 10 W

I = 1.5 A

\ Therefore, our true physical current is 1.5A ¯

Example:

In each case below, calculate I.

Solution:

I = (V_A - V_B) / R = (20 V - 0 V) / (10 W) = 20 V / 10 W

I = 2 A

\ Therefore, our true physical current is 2A ¯

Example:

Solution:

I = (V_A - V_B) / R = (160 V - 140 V) / (10 W) = 20 V / 10 W

I = 2 A

\ Therefore, our true physical current is 2A ¯

Example:

Solution:

I = (V_A - V_B) / R = (-20 V - (-40 V)) / (10 W) = 20 V / 10 W

I = 2 A

\ Therefore, our true physical current is 2A ¯

What does the example above illustrate?

The current is determined by the difference in potentials, not the absolute potentials themselves.

Example:
(i) Calculate I₁ and I₂
(ii) what is the magnitude and direction of the actual physical current in each case?

Solution:

Assume I₁ & I₂

I₁ = (V_A - V_B) / R = (40 V - 100 V) / (10 W) = -60 V / 10 W

\ I₁ = -6 A

==> Interpretation:

\ True physical current is 6 ↑ (upwards direction)

I₂ = (V_A - V_B) / R = (100 V - 40 V) / (10 W) = 60 V / 10 W

\ I₂ = 6 A

==> Interpretation:

\ True physical current is 6A ↑ (upwards direction)

­

Example:
Find formulas for I₁ ,  I₂ ,  I₃  in terms of the given potentials and resistances.

Solution:

I₁ = (V_A - 10 V) / R₁ = (40V - 10V) / 10W = 30V / 10W = 3A;

I₂ = (10 V - V_B) / R₂ = (10V - 5V) / 10W = 5V / 10W = 0.5A;

I₃ = (10 V - V_C) / R₃ = (10V - 5V) / 10W = 5V / 10W = 0.5A

or
Redraw the schematic:

==>

(10W + 10W) I₁ -  (10W)            I₂  - (10W) I₃  = (40 V - 5 V)

-(10W)           I₁ + (10W + 10W) I₂ + (10W) I₃  = (5 V - 5 V)
-(10W)           I₁ + (10W)             I₂ + (10W) I₃  = (5 V - 10 V)

==>

 (20W) I₁ - (10W)   I₂ - (10W) I₃ = (35 V)

-(10W) I₁ + (20W) I₂ + (10W) I₃ = (0 V)
-(10W) I₁ + (10W) I₂ + (10W) I₃ = (-5 V)==>
I₁ = 3A

I₂ = 0.5A

I₃ = 2A

==>

I₁ = I₁ = 3A

I₂ = I₂ = 0.5A

I_10V = I₃ = 2A

I₃ = I₁ - I₂ - I₃ = 3A - 0.5A - 2A

I₃   = 0.5A

Example:
Calculate  I₁ , I₂ and  V_AB   in the (partial) circuit below.

Solution:

I₁ = (V_B - V_A) / 5WW

\hysical current is 2A ®

I₂ = (V_B - V_C) / 10W = (10V - 2V) / 10W = 8V / 10W = 0.8A

\ Physical current is 0.8A ®

V_AB = V_A - V_B = 20V - 10V 

V_AB = 10V

Node:
In electronics, a node is a common joining point for three or more components (batteries, resistors, capacitors, etc...).

All points connected to a given node are at the same potential (relative to some Common or Ground)

Ex. Identify (circle) all nodes in the complete or partial circuits shown below

Aside: Some books refer to sometimes like as a node... not done here...

Kirchhoff's Current Law (KCL):

Basically a statement of charge conservation.

KCL (version A):

The sum of all the currents entering a node = The sum of all the currents leaving same node.

"Sum of":                  å (currents In)  =  å (currents Out)

Note:  å - Sigma, upper case greak letter for Sum

KCL (version B):

The algebraic sum (taking currents directed into a node as '+'; currents directed out as '-') of all currents entering (or leaving) any node equals zero.

å (currents In) - å (currents Out) = 0

å [(currents In) + (- currents Out)] = 0

alg. å (currents In and Out) = 0

alg. å I_in/out = 0

Example:

Find I

Solution:

KCL: version A: å I_in = å I_out

==> 3A = I + 2A ==> I = 3A - 2A = 1A

KCL: version B: alg. å I = 0

==> 3A + (-2A) + (-I) = 0

or   3A - 2A - I = 0 ==> I = 3A - 2A = 1A

Example:

Find I₁

Solution:

KCL: version A: å I_in  = å I_out

==> I₁ + 2A = 3A + 6A + 10A 

\  I₁ = 3A + 6A + 10A - 2A = 17A

KCL: version B: alg. å I  = 0

==> 2A + (-10A) + I₁ + (-6A) + (-3A) = 0

 \ I₁ = 10A - 2A + 6A + 3A = 17A

same answer!

Example:

Find I₁ and I₂

Solution:
Node 2: 
26A + I₁ = 2A + 4A + 12A
 \ I₁ = -8A
or
26A - 2A + I₁ - 4A - 12A = 0
I₁ = 12A + 4A + 2A -26A
\ I₁ = -8A

either way, physical current is 8A ¯

Node 2:
12A =  I₂ + 3A
\ I₂ = 12A - 3A
\ I₂ = 9A

either way, physical current is 9A ®


Kirchhoff's Voltage Law (KVL):

• Basically a statement of energy conservation.

Conventions (terminology)


travelling from '-' to '+' through a component referred to as as a voltage rise (potential rise)

 travelling from '+' to '-' through a component referred to as as a voltage drop (potential drop)

KVL (version A):

The sum of all the voltages rises (potential rises) around any closed path

 =

The sum of all the voltages drops (potential drops) around the same closed path.

"Sum of":                 

å V_rises  = å V_drops

KVL (version B):

The algebraic sum (taking voltage rises as '+'; voltage drops as '-') of voltage rises and drops around any closed path equals zero.

å (voltage rises) - å (voltage drops) = 0

å [(voltage rises) + (- voltage drops)] = 0

alg. å (voltage rises and drops) = 0

alg. å V_rises/drops = 0

Example:

Find V₁ 

Solution:

KVL version A: start at  x P (clockwise):  å V_rises = å V_drops

20V = V₁ + 15V

\ V₁ = 5 volts

KVL version B: start at x P  (clockwise): alg. å V_rises/drops = 0

20V + (-V₁) + (-15V) = 0

20V - V₁ - 15V = 0

\ V₁ = 20V - 15V = 5 volts

Example:

Find V₂

Solution:

KVL version A: start at y P (clockwise): å V_rises = å V_drops

20V + V₂ = 15V

\ V₂ = 15V - 20V
\ V₂ = -5 volts

KVL version B: start at y P  (clockwise): alg. å V_rises/drops = 0

20V + V₂ - 15V = 0

\ V₂ = 15V - 20V = -5 volts

KVL version B: start at y Q  (counterclockwise): alg. å V_rises/drops = 0

15V - V₂ - 20V = 0

\ V₂ = 15V - 20V = -5 volts

Example:

Find V₂

Solution:

KVL version A: start at  P  Q (clockwise): 

25V + 25V + 5V = 20V + 6V + 16V + V₂ 

55V = 42V + V₂
\ V₂ = 13 volts

KVL version B: start at P Q (clockwise): 

25V + 25V - 20V - 6V - 16V - V₂ + 5V = 0

\ V₂ = 13 volts

Example:

Find V_ab

Solution:

Note: when dealing with notation like

Assume:

 V_ab  (a: + (high); b: - (low)) or V_xy  (x: + ; y: -) or  V_cd (c: + ; d: -)

KVL version A: start at x P (clockwise):

20V + 12V + V_ab = 15V + 50V

\ V_ab = 33 volts

KVL version B:  start at x P (clockwise):

20V - 15V + 12V - 50V + V_ab =  0 

\ V_ab = 33 volts
==> means terminal ''a" is 33 volts higher in potential than terminal "b".

Example:

Find V_ab , V_xy

Solution:

Assume:

V_ab (a: high; b: low) , V_xy (x: + ; y: -)

Apply KVL in loop 1, start at P P (clockwise):

4V + 10V - V_ab - 5V = 0

\ V_ab = 9 volts

Apply KVL in loop 2, start at P P (clockwise):

+20V + V_xy - 4V = 0

\ V_xy = -16 volts

remember: by definition V_xy = V_x - V_y = -16V
\ V_y = 16V + V_x

\ terminal "y" is 16 volts higher in potential than terminal "x".

Example:

(i) Find V₁ and V₂

(ii) If  R₁  = 10 Wk and R₂ = 2 Wk , Find I₁ and I₂.

_Solution:

(i)

loop 1:  start at Q P (clockwise):  alg. å V_rises/drops = 0

40V - V₁ + 10V - 18V + 10V = 0
\ V₁ = 42 volts

loop 2: start at P P (clockwise): alg. å V_rises/drops = 0
18V - 10V +17V - V₂ - 12V - 15V = 0
\ V₂ = -2 volts

(ii) 

V₁ = R₁ . I₁

--> I₁ = V₁ / R₁ 

\ I₁ = 42 volts / 10x10³ ohms

\ I₁ = 4.2x10^-3 amps = 4.2 mA

V₂ = R₂ . I₂

--> I₂ = V₂ / R₂

\ I₂ = -2 volts / 2x10³ ohms

\ I₂ = -1x10^-3 A = -1 mA





In this course we will look at the following applications of systems of Linear Equations.

1) Loop Analysis of DC Circuits
2) Design of a Ring-Shunt Ammeter
3) Curve-fitting (Interpolating Polynomials)
4) Nodal Analysis of DC Circuits
5) Series / Parallel Circuits and Circuit Reduction Formulas
6) Power Dissipation in DC Circuit
7) Source Conversions


Loop Analysis (Background... Preamble)

" _ " one or more components connected...

• Define a loop current I₁ , I₂ , ...

  - A loop current is a continuously cycling current travelling around a closed path
  - A loop current relates to actual physical currents in ways to be discussed below

• Need as many loop currents as we have indivisible small loops
• In setting up loop currents, need to ensure that every component gets included in at least one loop...
• We will arrange our direction of travel around each loop so that we can treat all resistors as "voltage drops" in our equations
• In situations where only a single loop current flows in a part of circuit, it will be easy to find actual physical current... see examples to follow...
• In situations where several loop currents flow thru a component, need to be careful...

net downward physical current is  I₁ -  I₂ = 5a - 3A = 2A

net downward physical current is  I₁ +  I₂ = 2A + 3A = 5A

• We will generate our loop equations using KVL...

Loop Analysis Examples

Example:
(Two-Loop Circuit)

a) Set up and solve the equations for the loop currents I₁ and I₂
b) Determine the physical currents (magnitude & direction) in each resistor
c) Repeat problem using a different definition for loop currents (as will be described in class...)

Solution:
a) write equation...
alg. å V_rises/drops = 0
loop 1 P :  20V - 3 I₁ - 4 (I₁ - I₂) = 0
loop 2 P :  0V - 4 (I₂ - I₁) - 5 I₂ = 0

Simplifying:
 7 I₁ - 4 I₂ = 20
-4 I₁ + 9 I₂ = 0

∆ =

7	-4
-4	9

= (7) (9) - (-4) (-4) = 63 - 16 = 47
∆ = 47

∆₁ =

20	-4
0	9

= (20) (9) - (0) (-4) = 180 - 0 = 180
∆₁ = 180

∆₂ =

7	20
-4	0

= (7) (0) - (-4) (20) = 0 + 80 = 80
∆₂ = 80
\ I₁ = ∆₁ / ∆ = 180 / 47 =  3.83A

\ I₂ = ∆₂ / ∆ = 80 / 47 = 1.70A

b) The physical currents (magnitude & direction) in each resistor
I₃W = I₁ = 3.83A ®
I₄W = I₁ - I₂ = 3.83A - 1.70A = 2.13A ¯
I₅W = I₂ = 1.70A ¯

c) write equation...

alg. å V_rises/drops = 0
loop 1 P : 20V - 3 I₁ - 4 (I₁ + I₂) = 0
loop 2 Q:   0V - 4 (I₂ + I₁) - 5 I₂ = 0

Simplifying:
7 I₁ + 4 I₂ = 20
4 I₁ + 9 I₂ = 0

∆ =

7	4
4	9

= (7) (9) - (4) (4) = 63 - 16 = 47
∆ = 47

∆₁ =

20	4
0	9

= (20) (9) - (0) (4) = 180 - 0 = 180
∆₁ = 180

∆₂ =

7	20
4	0

= (7) (0) - (4) (20) = 0 - 80 = -80
∆₂ = -80

\ I₁ = ∆₁ / ∆ = 180 / 47 = 3.83A

\ I₂ = ∆₂ / ∆ = -80 / 47 = -1.70A

The physical currents (magnitude & direction) in each resistor
I₃W = I₁ = 3.83A ®
I₄W = I₁ + I₂ = 3.83A  + (-1.70A) = 2.13A ¯
I₅W = I₂ = 1.70A ¯



Example:
(Two-Loop Circuit)

a) Set up and solve the equations for the loop currents I₁ and I₂
b) Determine the physical currents (magnitude & direction) in each resistor
c) Find V_ab
Solution:
a) write equation...
alg. å V_rises/drops = 0
loop 1 P: 1V - 1 I₁ - 3 (I₁ + I₂) = 0
loop 2 Q: 2V - 3 (I₂ + I₁) - 2 I₂ = 0

Simplifying:
4 I₁ + 3 I₂ = 1
3 I₁ + 5 I₂ = 2

∆ =

4	3
3	5

= (4) (5) - (3) (3) = 20 - 9 = 11
∆ = 11

∆₁ =

1	3
2	5

= (1) (5) - (2) (3) = 5 - 6 = -1
∆₁ = -1

∆₂ =

4	1
3	2

= (4) (2) - (1) (3) = 8 - 3 = 5
∆₂ = 5

\ I₁ = ∆₁ / ∆ = -1 / 11 = -0.091A

\ I₂ = ∆₂ / ∆ = 5 / 11 = 0.455A

b) The physical currents (magnitude & direction) in each resistor
I₁W = I₁ = 0.091A ¬
I₃W = I₁ + I₂ =  (-0.091A) + 0.455A = 0.364A ¯
I₂W = I₂ = 0.455A ¬

c) The physical currents (magnitude & direction) in each resistor
V_ab = (1W) I₁ =  (1W)  (-0.091A) = -0.091V



Example:
(Three-Loop Circuit)

a) Set up the loop equations which allow you to solve for the loop currents I₁ , I₂ and I₃
b) Find V_ab

Solution:
a) write equation...
alg. å V_rises/drops = 0
loop 1 P: 10V - 14 I₁ - 6 (I₁ - I₂) = 0
loop 2 P:  0V - 6 (I₂ - I₁) - 2 I₂ - 4 (I₂ - I₃) = 0
loop 3 P:  - 4 (I₃ - I₂) - 4 I₃ - 20 - 5 I₃ = 0

Simplifying:
20 I₁ - 6 I₂ + 0 I₃ = 10
-6 I₁ + 12 I₂ - 4 I₃ = 0
0 I₁ - 4 I₂ + 13 I₃ = -20

Need I₃ ,... than V_ab  = 5W . I₃
I₃ , use determinants ... Cramer's...

∆ =

20	-6	0	20	-6
-6	12	-4	-6	12
0	-4	13	0	-4

= (20) (12) (13) + (-6) (-4) (0) + (0) (-6) (-4)
- (-6) (-6) (13) - (20) (-4) (-4) - (0) (12) (0)
= (3120) + (0) + (0) 
- (468) - (320) - (0)
= 3120 - 788 = 2332
∆ = 2332

∆₃ =

20	-6	10	20	-6
-6	12	0	-6	12
0	-4	-20	0	-4

= (20) (12) (-20) + (-6) (0) (0) + (10) (-6) (-4)
- (-6) (-6) (-20) - (20) (0) (-4) - (10) (12) (0)
= (-4800) + (0) + (240)
- (-720) - (0) - (0)
= -4560 + 720 = -3840
∆₃ = -3840

Finally: I₃ = ∆₃ / ∆ = -3840 / 2332 = -1.65A

\ V_ab = (5W) I₃ = (5W) (-1.65A) = -8.25 volts

(V_ba = -V_ab = 8.25 volts)

Example:
For the circuit given, set up but do not solve the KVL equations for loop currents I₁ , I₂ and I₃

Solution:
Apply KVL around each loop...
in form alg. å V_rises/drops = 0
loop 1 P: 10 - 10 I₁ - 30 - 5 (I₁ - I₂) - 15 = 0
loop 2 P: 15 - 5 (I₂ - I₁) - 15 I₂ + 50 - 20 I₂ - 25 (I₂ - I₃) = 0
loop 3 P:  - 30 I₃ - 25 (I₃ - I₂) - 60 - 40 I₃ = 0

...

15 I₁ - 5 I₂ + 0 I₃ = -35
-5 I₁ + 65 I₂ - 25 I₃ = 65
0 I₁ - 25 I₂ + 95 I₃ = -60


Loop Analysis Continued:

Example:
Wheatstone Bridge Circuit

Solve for the circuit in Rs.

Solution:
Good choice for I₁ , I₂ , I₃  since only need to find  in order to find current in R₅.

R₅ is variable.
(KVL)  å V_rises  =   å V_drops
loop 1 P: 10 = 1 (I₁ + I₃) + 1 I₁ + 3 (I₁ - I₂)
loop 2 P:   0 = 3 (I₂ - I₁) + 1 I₂ + 1 (I₂ + I₃)
loop 3 P: 10 = 1 (I₁ + I₃) + 1 I₃ + 1 (I₂ + I₃)

5 I₁ - 3 I₂ + 1 I₃ = 10
-3 I₁ + 5 I₂ + 1 I₃ = 0
1 I₁ + 1 I₂ + 3 I₃ = 10

... Need only find I₂ ...

∆ =

5	-3	1	5	-3
-3	5	1	-3	1
1	1	3	1	1

= (5) (5) (3) + (-3) (1) (1) + (1) (-3) (1)
- (1) (5) (1) - (1) (1) (5) - (3) (-3) (-3)
= (75) + (-3) + (-3)
- (5) - (5) - (27)
= 69 - 37 = 32
∆ = 32

Also need
∆₂ =

5	10	1
-3	<><><><><><><><>	0	1
1	10	3

= (-1) (-3)

10	1
10	3

+ (+1) (0)

5	1
1	3

 + (-1) (1)

5	10
1	10

\∆₂ = 3 (30 - 10) + 0 (15 - 1) - 1 (50 - 10)
\∆₂ = 20


\ I₂ = ∆₂ / ∆ = 20 / 32 = 0.625A
\ V_ab = R₅ . I₂ = 1W . (0.625) = 0.625 volts


Note: If the bridge is "balanced" then there will be no current flow in R₅
In this case:
                                R₁ R₄ = R₂ R₃  or  R₁ / R₃ = R₂ / R₄
\ if three of these resistors are known, the fourth can be determined.













Loop analysis method may be used to find values other than currents in a circuit, as in the following examples.

Example:
Solve for V and R in the following circuit

Solution:
Normally we are trying to find loop currents. But in certain design problems, we may know the current demands and want to infer other circuit component and source requirements...
two unknowns: V, R ...

Loop 1: V = 2 . I₁ + R (I₁ - I₂)
or        : V = 2 . (1) + R (1 - 0.5)
             V = 2 + 0.5 R                                             (1)


Loop 2: 0 = R (I₂ - I₁) + 1 . I₂
or :        0 = R (0.5 - 1) + 1 . (0.5)
              0 = - 0.5 R + 0.5                                         (2)

Any method could be employed to solve system (1) & (2):
go with substitution (2) ==>  0.5 R = 0.5  ==> R = 1 W
then sub in (1): V = 2 + (0.5) (1)
\ V = 2.5 volts