Block A Unit 1 1

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 2

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 3

Current: Free electrons on the move

_

_

__

__

__ _

__

_ _

_

+ -

Conventional current flows from + to - (Opposite direction to electron flow)

_ _ _ _

Direction of current

_ _ _

Currents must run in loops

Current in relation to ChargeCurrents arise from flow of chargesUnit: Ampere (A)Typically denoted by the symbol, ICurrent = Rate of change of chargeMathematically, I = dQ/dt

Block A Unit 1 4

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

i4

i3

i2Kirchhoff’s current law

Sum of currents at a node must equal to zero:

i1 + i2 + i3 + i4 = 0

Block A Unit 1 5

Kirchhoff’s current law

Does the direction of current matter?

Differenti1

i2

i3

i1

i2

i3

i1 + i2 + i3 = 0 i1 + i2 + i3 = 0 (WRONG)

I2 running into node I2 running out of node

i1 + i3 = i2 (CORRECT)

YES!!

Block A Unit 1 6

Kirchhoff’s current law

Currents EXITING (-ve),Currents ENTERING (+ve)

i1

i2

i3

i4

Entering: I1 & I2 (+ve)

Exiting: I3 & I4 (-ve)

i1 + i2 – i3 – i4 = 0

Sign convention when applying KCL

Block A Unit 1 7

KCL exampleProblem 2.14 and 2.15

Find the unknown current using KCL

6A - 5A + 2A - i = 0

i = 3A

6A - 5A + 2A + i = 0

i = -3A

Block A Unit 1 8

Voltage/Potential difference

_ _ _ _ _ _

A BCurrent

+ -

Energy is required to move charges between 2 points

Voltage/potential difference is always made with reference to 2 points

vx vx = vA - vB

Unit: Volt (V)

High (+)

Low (-)

Direction of Flow

Block A Unit 1 9

Sign convention

i

v+ -

In a LOAD

Voltage DROPS in the direction of the current

Energy is dissipated (or consumed)

i

v+-

In a SOURCE

Voltage RISES in the direction of the current

Energy is generated

Block A Unit 1 10

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 +

+ V1 -

-

V4

+

+

V2

-

Kirchhoff’s voltage law

Net voltage around a closed circuit is zero:

v1 + v2 + v3 + v4 = 0

Block A Unit 1 11

Kirchhoff’s voltage law

source

load

load

load

GROUND

Reference voltage in a circuit set to 0V All other nodes on the circuit can then be conveniently

referenced to GROUND

Ground symbol

+ V1

-

+ V2 -

+ V3 -

DEFINING signs Voltage gain (+ve) Voltage drop (-ve) v1 – v2 + v3 = 0

Block A Unit 1 12

KVL exampleProblem 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 13

Resistance and Ohm’s Law

i

v

V

I

1/R

Ohm’s law: V = IR

Ideal RESISTOR shows linear resistance obeying Ohm’s law

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

+

_

IMPORTANT: Positive current is defined here as flowing from higher to lower voltage (Remember)

Unit: Ohm (Ω)

A

L

ρ: resistivity (material property)

A: cross-sectional area

ALR

Block A Unit 1 14

Ohm’s law + KCL exampleProblem 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 15

Electrical Power

In a source, power is generated In a load (eg. resistor), power is

dissipated/consumed

Electrical power generated/dissipated in a given element is defined by the product of the voltage across that element and the current through it

P = VI

0.2A0.2A1.5V 1.5V+

_

+

_Source: P = 1.5V x 0.2A = 0.3W

Load: P = 1.5V x 0.2A = 0.3W

Unit: Watt (W)

P = I2R P = V2/R

Power generated by source

MUST EQUAL

Power dissipated in the load

Block A Unit 1 16

Power exampleProblem 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 17

Power example solution

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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!

Block A Unit 1 18

Terminology: Branch and Node

Branch

BRANCH: Any path of a circuit with 2 TERMINALS connected to it

NodeNODE: Junction of 2 or more branches

Block A Unit 1 19

Terminology: Loop and Mesh

LOOP: Any closed connection of branches

In the above circuit, there are 6 loops in total

+-Vs

+-Vs

Mesh: Loop that does not contain other loops

In the above circuit, there are 3 meshes in total

Block A Unit 1 20

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 21

Independent Voltage Source

+_ Circuit

i

i

+

_

v

+

_

v

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 22

Independent Current Source

Circuit

i

i

+

_

i

+

_

v

Independent Current Source supplies a prescribed current to any load circuit connected to it

Voltage supplied is determined by load circuit connected

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

Block A Unit 1 23

+-5V

+vx

-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

_+

is

vs

Block A Unit 1 24

Example of a VCVS

+vi

-Avi

+-

Note: This is a general model for an amplifier which shall be re-visited later in BLOCK C

Ri

Ro

Block A Unit 1 25

Short Circuit

0.1A

A

B

0.1A

A

B

Short circuit A & 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

Isc

Short circuit current Isc = 0.1A

20Ω resistor is now bypassed

Current from source flows through the short-circuit to give a short-circuit current20Ω

20Ω

Block A Unit 1 26

Open Circuit

A

B

Open circuit: Leave 2 terminals unconnected externally

Typically associated with voltage, eg. open-circuit voltage

0.1A

0.1A

A

B

+

Voc

-

Ioc= 0A

20Ω

20Ω

Open circuit voltage

Voc = 0.1A × 20Ω = 2V

Block A Unit 1 27

Self-contradictory circuits

+-vs

A

BWhat is the voltage across A and B?

Is

A

BWhat is the current arriving at A?

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

Prefixes: Memorize and apply them!

Write as 2.15mA instead of 0.00215A

Block A Unit 1 28

Parallel network (Highlights)

R1 R2IsRN

I1 I2 IN

RPIs

Equivalent Resistance

1/RP = 1/R1 + 1/R2 + …+ 1/RN

Current divider rule

SP

NN I

RRI

11

Block A Unit 1 29

Parallel Network: Proof

R1 R2Is

I2

I1

Apply Kirchhoff’s current law (KCL) at X:

Is = I2 + I1

This can be seen as Is is split into the 2 branches

R1 R2Is

I2

I1

X

X

Y

1 2 3

4 56

Note that points 1, 2, 3 are all at the same voltage, therefore same node (X)

Points 4, 5, 6 are all at the same voltage, therefore same node (Y)

Voltage across R1 = Voltage across R2 = VXY

Block A Unit 1 30

Parallel network: Proof

R1 R2Is

I2

I1

X

Y

1 2 3

4 56

Apply Ohm’s to both R1 and R2:

VXY = I1R1 = I2R2

I1 = VXY/R1; I2 = VXY/R2

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

sIRR

RI

21

11 11

1sI

RRRI

21

22 11

1

This is referred to as the current divider rule

Block A Unit 1 31

Parallel network: Proof

R1 R2Is

I1

I2

Is = VXY(1/R1 + 1/R2)

Replace the parallel network of resistors with a single equivalent resistor

RPIs

Remember that the voltage across RP is still VXY!

VXY/RP = VXY(1/R1 + 1/R2)

1/RP = 1/R1 + 1/R2

X

Y

Block A Unit 1 32

Parallel network (seeing it)

R1 R2IsRN

I1 I2 IN

Current splits at one node

Current re-combine at the other node

Suggestion: Think about parallel resistors as the rungs on a ladder

Block A Unit 1 33

Current divider: Example 1

i1

V2

R1 i2 R2

V1

I

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 34

Current divider: Example 2 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)

i1

V2

R i2 R

V1

I

i3 R

i1 = i2 = i3 = I/3

i1 = i2 = i3 = i4 = I/4

Consider that, in = I/n

Hence n = 100 for in = 0.01I

Block A Unit 1 35

Current divider: Example 3

If R2 > R4 > R1 > R5 > R3

Which current is the largest?Which current is the smallest?Rank the currents from largest to smallest

i1

V2

R1 i2 R2

V1

I

i3 R3 i5 R5i4 R4

I3 > I5 > I1 > I4 > I2

Block A Unit 1 36

Series network (Highlights)

R1

R2

Vs

+-

RSVs+-

Equivalent Resistance

RS = R1 + R2 + …+ RN

RN

Voltage divider rule

VN = VS(RN/RS)

+

-V1

+

-V2

+

-VN

Block A Unit 1 37

Series network: Proof

R1

R2

Vs+-

+

-V1

+

-V2Is

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

sVRR

RV

21

11

This is referred to as the voltage divider rule

sVRR

RV

21

22

Block A Unit 1 38

Series network: Proof

R1

R2

Vs+-

Vs = IS(R1+R2)

RSVs+-

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)

RS = R1+R2

Block A Unit 1 39

Series network (seeing it)

R1

R2

Vs

+-

RN

+

-V1

+

-V2

+

-VN

Suggestion: Think about series resistors as the rings forming a chain

Block A Unit 1 40

Voltage Divider: Example 1

Find the voltage across each resistor Find the voltage at nodes 1 and 2

2V 5V

10Ω

Node 1 Node 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

20Ω 30Ω

Block A Unit 1 41

Voltage Divider: Example 20V 9V

R1 R2 R3

If R3=3R1 and R2 = 2R1

• Find the voltage across each resistor

• Find the voltage at nodes 1 and 2

Node 1 Node 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 42

Voltage Divider: Example 3

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

VA VB

R1 R2 R3 R4 R5

+ V1 - + V2 - + V3 - + V4 - + V5 -

V2 > V4 > V1 > V5 > V3

Block A Unit 1 43

Series-Parallel combo: Example 1

R1 R2

R3 R4

Find the total resistance as seen across A and B

All resistors have value of 1ΩA

B

RAB = (R1 || R2) + (R3 || R4)

= 1||1 + 1||1 = 1ΩR1 R2

R3 R4

A

B

RAB = (R1 + R3) || (R2 + R4)

= 2||2 = 1Ω

Block A Unit 1 44

Combo example 2Problem 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 45

Combo example 2

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Block A Unit 1 46

Combo example 3Problem 2.48

Find the equivalent resistance seen by the source

Find the current through the 90Ω resistor

Block A Unit 1 47

Combo example 3

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Block A Unit 1 48

Power dissipationPower: P = V I

+ V -

iPower dissipated in resistor R

P = I2R = V2/RR

+- Vs R2

R1

v2

+

_

Power dissipated: P = v22 / R2

22

21

2

)( sVRR

RP

For Max Power Transfer:

R1 = R2

Find condition for dP/dR2 = 0 (max pt)

Block A Unit 1 49

Practical voltage source

+-Vs RL

Terminal voltage becomes:VRL = [RL/(RL+RS)]VS instead of VS

VRL is now lower due to some voltage drop across RS

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

+-Vs

Rs

RL

Ideal Practical

Block A Unit 1 50

Practical current source

RLIs Rs

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

RLIs

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

Ideal Practical

Block A Unit 1 51

Voltmeter Voltmeter measures voltage across a circuit element Connected in parallel with the element being measured

+_ vs R2

R1

V+_ vs

R1

R2VRV

IDEAL ACTUAL

Voltmeter should draw as little current away R2 in the main circuit (application of current divider rule)

Block A Unit 1 52

Voltmeter Example

+_vs R2

R1

V

i im

RV

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 53

Voltmeter example solution

This slide is meant to be blank

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

Block A Unit 1 54

Ammeter Ammeter measures current flowing through an element Connected in series with the element being measured

+_vs

R1

i

A

+_vs

R1

i

ARA

RA should contribute as little as possible to the overall series resistance in the circuit (application of voltage divider rule)

IDEAL ACTUAL

Block A Unit 1 55

Ammeter Example

+_Vs

R1

i

ARA

For R1 = 500Ω and then 2.5 Ω,

Find i without and with the ammeter included

Vs = 5V, RA = 0.5Ω

Block A Unit 1 56

Ammeter Example solution

This slide is meant to be blank

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