Units of Electricity

8/13/2019 Units of Electricity

1/36

Week 2

Units of Electricity


2/36

2

Electrical Units

a. Current

b. Quantity of Electricity

c. Electric Potentiald. Resistance

e. Electromotive Force

f. Ohms Lawg. Electric Energy


3/36

3

a) Current

current which, if maintained in two straight parallelconductors of infinite length, of negligible circularcross-section, and placed 1 meter apart in a

vacuum, would produce between theseconductors a force of 2 x 10-7newton per meter oflength.

The conductors are attracted towards each other ifthe currents are in the same direction, whereasthey repel each other if the currents are inopposite directions.


4/36

4

Symbol and Unit

Symbol: I

Unit: ampere (A)


5/36

5

Electric current

Electric current in a wire is defined as thenet amount of charge that passes throughthe wire per unit time at any point. Theaverage current is defined as:

I = Q 1A = 1C

t 1s

Electric current is called the ampere.

1 Ampere (A) = 1 coulomb per second (C/s)


6/36

6

Electric Current Flow

An electric current exists whenever electric charge flows

through a region like a light bulb circuit. The magnitude of

the current is measures in amperes (I).

Battery

(Flow of charge)


7/36

7

Example 1

A steady current of 2.5 A flows in a wirefor 4.0 min.How much charge passed through any point

in the circuit?Q = I t= (2.5 C/s)(240 s) = 600 C

How many electrons would this be?

1 e= 1.60 X 10-19C, so 600 C equals to

600 C = 3.8 X 1021electrons

1.6X10-19 C/electron


8/36

8

Example 2

Which light bulb will light up? Why?


9/36

9

b) Quantity of Electricity

The unit of electrical quantity is the coulomb,

namely the quantity of electricity passing a

given point in a circuit when a current of 1ampere is maintained for 1 second.

Q[coulombs] =I[amperes] X t[seconds]

Symbol: Q

Unit: coulomb (C)


10/36

10

c) Electric Potential

The electric potential at a

point aor Vaequals the

potential energy (PEa

) per

unit charge at that point.

Va= PEa/q


11/36

11

Potential Difference

Potential difference between point a and b

is

Vab= VaVb

The units of electric potential and potential

difference is Joules/Coulomb and giventhe name vol t.

1V =1 J/C


12/36

12

The positive plate has ahigher potential than thenegative plate. Thus a

positively charged objectmoves naturally (like chargerepels) from a high potentialto low potential. A negatively

charged object does theopposite. Potential differenceis often referred to as volt orvoltage.

Potential Difference (cont)


13/36

13

d) Resistance

resistance between two points of a conductorwhen a potential difference of 1 volt, appliedbetween these points, produces in this

conductor a current of 1 ampere, the conductornot being a source of any electromotive force.

OR, the resistance of a circuit in which a currentof 1 ampere generates heat at the rate of 1 watt.

Symbol: R

Unit: ohm ()


14/36

14

Resistor

Resistors are used to control theamount of current flow. In a

circuit, resistors are indicated by

the symbol ab

c

d

How to read resistor value.a b X 10c with tolerance.

In this example = 25 X 103 or25,000 with 10% tolerance


15/36

15

Resistor Coding

Digit Color

0 Black

1 Brown

2 Red3 Orange

4 Yellow

5 Green

6 Blue7 Violet

8 Grey

9 White

Tolerance Color

5% Gold

10% Silver

20% No color band


16/36

16

Resistivity

Resistance of a wire depends on some

factors like as length (L), cross-sectional

area (A) and resistivity of material ().R = L

A

Resistivity of a material also depends ontemperature. In general, the resistance of

metals increases with temperature.

T

=0

[ 1 + ( TT0

) ]


17/36

17

Example 3Speaker wires. Suppose you want to connect your stereo

to remote speakers. If each wire must be 20 m long,

what diameter of copper wire should you use to keep the

resistance less than 0.10per wire?

A = L/R

= (1.68 X 108 .m)(20m) =

(0.10 )

= 3.4 X 10

-6

m

2

A = d2/4 (Area calculation formula)

d = (4A)/ = 2.1 X 10-3m = 2.1 mm


18/36

18

Exercise

1. What is a simpler way of expressing 0.000 005 A?

2. What is a simpler way of expressing 3 000 000 V?

3.

A potential difference of 6 V causes a current of0.6 A to flow in a conductor. Calculate theresistance of the conductor.

4. Find the potential difference to pass a current of 5A through a conductor of resistance 8 .

5. A 960 lamp is connected to a 240 V supply.Calculate the current in the lamp.


19/36

19

e) Electromotive Force

An electromotive force is that which tends

to produce an electric current in a circuit,

and unit of e.m.f.is the volt. Symbol: E

Unit: volt (V)


20/36

20

f) Ohms Law To produce an electric current, a difference in

potential is required. The current in a metal wire

is proportional to the potential difference V

applied to its ends: I V The current flow through the wire is inversely

proportional to the resistance of the wire for a

given voltage.

I = V / R or V = I R (Ohms Law)

The unit for resistance is called the ohm ()


21/36

21

Current vs. Voltage

A metal conductor which obeys

the Ohms Law.

A nonohmic device, like

semiconductor


22/36

22

Example 4

Flashlight bulb resistance. A small

flashlight bulb draws 300 mA from its 1.5V

battery.What is the resistance of the bulb?

R = V / I = 1.5 V / 0.30 A = 5.0

If the voltage dropped to 1.2V, how much would the

current change?

I = V / R = 1.2 V / 5.0 = 0.24 A


23/36

23

Understanding Circuits situation Closed circuitcomplete circuit

Open circuitdisconnected wiring or incompletecircuit

Short Circuitwhen two wires crossed

Closed Circuit Open Circuit Short Circuit

No current flow

current


24/36

24

g) Electric Energy

Electric energy can be transformed into

other forms of energy like thermal energy,

light. Electric energy => Thermal energy => light

Example of applications are toasters

(thermal), light bulb (thermal and light)


25/36

25

Rate of electrical energy

transformation The energy transformed when a charge Q

moves through a potential difference V

and denoted by QV. Power (P) is the rate the energy is

transformed

P = Power = energy transformed (E) = QV

time (t) t


26/36

26

Power transformed by an electrical

device Power (P) = Current (I) * Voltage (V)

P = IV

Iis the current passing through the device

and Vis the potential difference across it.

The SI unit of electric power is watt.

1 W= 1Jou le per second (J/s)


27/36

27

Equations of Power

The Power equation can be rewritten as

P = IV (a)

= I (IR) = I2R (b)

= (V/R)V = V2/R (c)

Equation (a) applies to any device while

equation (b) & (c) only applied to resistors.


28/36

28

Equations of Electric Energy

Let us recall that

P = Power = energy transformed (E) = QV

time (t) t

Or E = P t and P = V I

= (V I ) t = V I t (i)

= (I R) I t = I2 R t (ii)

= (V / R)2R t = V2t / R (iii)


29/36

29

Example 5

Calculate the resistance of a 40W

automobile headlight designed for 12V.

Input

P = 40W

V = 12V

From equation P = V2/R, we

obtain R = V2/P

So, R = (12V)2/40W = 3.6


30/36

30

Kilowatt-hour (kWh)

It is energy and not power that you pay for

your electric bill. Usually energy in homes

is specified as kilowatt-hour (kWh) whichis equivalent to

1 kWh = (1000W)(3600s) = 3.60 X 106J


31/36

31

Example 6An electric heater draws 15A on a 120V line. How muchpower does it use and how much does it cost per month(30 days) if it operates 3.0 h per day and the electriccompany charges 10.5 cents per kWh?

Power (P) = IV = (15A)(120V) = 1800W = 1.8kW

Operation per month is = 3hr/day multiply by 30day = 90hr

The monthly power consumption per month for the electricheater is (1.8kW) X (90h) = 162kWh

So the cost would be 162kWh X RM0.105/kWh = RM 17


32/36

32

Summary of Important Formulae

Electric Charge Q = I t(coulombs)

Voltage V= P/I(volts)

V=I R

Power P = IV

P = I2R

P= V2/R


33/36

33

Summary of Terms and Concepts

Currentis the rate of flow of electric

charge in a circuit. The terms is often used

to describe the flow of electric charge, e.g.a current is flowing in a circuit; this isambiguous but is so common that we have

to accept it.


34/36

34

Contd

Electric chargemay be either +ve orve.ve electrons are free to move around a

circuit thus transporting energy fromsource to load.

To maintain a current, the source must

provide a driving force called theelectromotive force (e.m.f.).


35/36

35

Contd

The potential difference across a load

indicates in volts the energy lost per

coulomb of charge passing through theload.

Since the current is the rate flow, its

product with the voltage gives the rate ofenergy transmission, i.e. the power.


36/36

36

Contd

Resistanceis a measure of the opposition

to the flow of charge through a load.

Ohms Law states that the ratio of voltageto current is constant, provided other

physical factors such as temperature

remain unchanged.

Units of Electricity

Documents

Transcript of Units of Electricity