All matters made of atoms which consist of electrons, protons

and neutrons.Protons and neutrons are in the nucleas and electrons are moving

around them.Each electron carries a negative charge of 1.6 x 10-19 C, each

proton carries positive charge of the same value and neutron

carries no charge.Atom having a number of electron not same as proton is called

ion.

A positive ion is an atom having a number of electron less than

number of proton.A negative ion is an atom having a number of electron more

than number of proton.Coulomb is a unit for electric charge.

1 coulomb = 6.25 x 1018 electronUnits for electric charge are microcoulomb (C = 10-6C) ,

nanocoulomb(nC= 10-9, picocoulomb (pC = 10-12 C),

Current is a net flow of electrical charges passing a point and given as i = dq/dt

Unit for current is ampere (A) i.e a rate of 1 C per sec.

E.g 230 mA = 230 x 10-3 A = 0.23 A

0.015 A = 0.015 x 103 mA = 15 mA125 A = 125 x 10-3 mA = 0.125 mA125 A = 125 x 10-6 A = 0.000125 A

The figure shows the flow of electric current. In 2 ms, the values of negative charge q1 and positive charge q2 flowing from X to Y crossing a cross-section at point A are10 C and -4 C respectively. (i)What is the current if both flowing in the same direction(ii)What is the current if both flowing in the different directions

(i)The charge net flow isQ = q1 + q2 = 10 C + (-4) C = 6 C.Time t = 2 msI = Q/t = (6 x 10-6)/(2 x 10-3) = 3 mA

(ii)The charge net flow isQ = q1 – q2 = 10 C - (-4) C = 14 C.I = Q/t = (14 x 10-6)/(2 x 10-3) = 7 mA

A

X

Y

q2

q1

Same charge repels . Different charge attracts Energy is required to bring a positive charge near to negative

charge. Energy is required to bring a positive charge away from another

positive charge. Potential different is a measurement of energy required to bring 1

C charge near to another charge or to bring away same. The unit of potential different is volt (V) and the formula is

v = dw/dq If W = 1 J is required to bring a charge q2 = 1 C from point A to

point B, the potential different between A and B is 1 V.A B

q1q2 = 1 C

The moving of charges in conductor caused collision and friction among them and causing losses of energy. Thus the moving of charges are said to have resistance.

Unit for resistance ohm (). 1 ohm if 1A of current flowing in a conductor to produce 1V between two points

The resistance value depends on material and some other parameter such as temperature

Resistance is important element to control the current in the circuit.. Resistance value is dependent on the following parmeter. length (l), cross-section (A) and resistivity (), Thus the resistance

an be writen asR = l/A or R = 1/(A) = 1/ = conductivity

Resistance also depends on temperature and can be fomulated as

R1 = temperature coef. Of resistance at t1

R2 = temperature coef. Of resistance at t2

0 = pekali suhu rintangan pada suhu 0 C Most insulator such as rubber , the resistance decreases with the

increases of temperature.

20

10

2

1

tα1

tα1

R

R

Material (m) at 0oC

Aluminium 2.7 X 10-8

Brass 7.2 X 10-8

Copper 1.59 X 10-8

Eureka 49.00 X 10-8

Manganin 42.00 X 10-8

Carbon 6500.00 X 10-8

Tungsten 5.35 X 10-8

Zinc 5.37 X 10-8

Material o(/oC) at 0oC

Aluminium 0.00381

Copper 0.00428

Silver 0.00408

Nickel 0.00618

Tin 0.0044

Zinc 0.00385

Carbon -0.00048

Manganin 0.00002

Constantan 0

Eureka 0.00001

Brass 0.001

For cuprum: = 0.0173 -m. If the length of the cuprum wire is 10 m and its cross-section is 0.5 mm2 . What is its resistance?

R = l/A = (0.0173 x 10-6 x 10)/(0.5 x 10-6)

= 0.346 = 346 m

A cable consists of two conductors which , for the purposes of a test , are connected together at one end of the cable. The combined loop resistance measured from the other end is found to be 100 when the cable is 700m long. Calculate the resistance of 8 km of similar cable.

2

1

2

1

R

R

R

1143700

8000100RR

1

212

A conductor of 0.5 mm diameter wire has a resistance of 300 . Find the resistance of the same length of wire if its diameter were double.

21

22

1

2

2

1

R

R

d

d

A

A

2

2

2 5.0

0.1

R

300

AR

1

752R

A coil of copper wire has a resistance of 200 when its mean temperature is 0oC. Calculate the resistance of the coil when its mean temperature is 80oC

5.2688000428.012001 11 ooRR

When a potential difference of 10V is supplied to a coil of copper wire of mean temperature 20oC, a current of 1 A flows in the coil. After some time the current falls to 0.95 A yet the supply voltage remains unaltered. Determine the mean temperature of the coil given that the temperature coefficient of resistance of copper is 4.28 x 10-3 / oC at 0oC

2

1

2

1

1

1

R

R

o

o

200428.01

2000428.01

10.53

10.0

53.1095.0

10

2

22 I

VR

101

10

1

11 I

VR

Co4.332

At 20oC

At 2oC

Potential different (V) across the resistor is proportional to current (I) Thus we can write as

V I (A graph V-I is linear and then V/I is a constant)

This constant is called a resistance (R):

Hence V = RI

E.g.A simple dc circuit consist of voltage source V=10V and resistor producing a current of 4 mA. What is the value of resistance R? From Ohm’s law:

R = V/I = 10/(4 x 10-3) = 2.5 k

+

-

Vs10V

R

I

R

1.2 k

(fixed resistor)

Rp

10 k

(potentiometer)

Rv

1 M

(rheostat)

+

-

V

+

-

ri

V

To generate a potential difference in the circuitsAn ideal voltage source is that can supply a constant voltage between the two terminals independent of current withdraws from the circuit. A practical voltage source usually experience a voltage drop at the terminals due to internal resistance, ri.E.g of DC voltage sources are battery , dry cell , solar panel etcCircuit’s symbols for DC voltage sources are

Ideal source

Practical source

Generate a variable potential difference with time The variation of potential different follows a sine

waveform, so the voltage varied in amplitude and phase which can represented by

V (t) = A sin (t + )

where A is amplitude, = 2 f , f is frequency , t time and is phase displacement

~ v(t)

The voltage source depends on other parameter such as input current or input voltage of a device. The symbols are as follows

+

-

V

+

-

Vix vx

(a) Current dependent source

(b) Voltage dependent source

I

i(t)+

DC current source supplies a constant current to the circuit connected to it.

For an ideal current source supplies a constant current independence of any value of voltage across its terminals.

AC current source supplies a current amplitude varies with time. The variation of current amplitude normally in sinusoidal waveform. This can be represented by

i(t) = Io sin(t +)

I Iix vx

(a) Dependence on current

(b) Dependence on voltage

The current source which depends on other parameter such as input current or input voltage of a device. The symbols are as follows

W= F(Newton) x d (m)

Where F =force, d=distance, t = time , a= acceleration, P=power, m= mass , u=velocity, V=voltage, I=current and R=resistance

W= VI t

W= I2 R t

W= P t 2

2

1muW

For Electrical energy

For physical quantities

F=ma

A current of 3 A flows through a 10 W resistor. Find:(a)The power developed by the resistor(b)(b) the energy dissipated in 5 min.

WRIP 9010322

JPtW 270060590

(a)

(b)

A heater takes a current of 8 A from a 230 V source for 12 h. Calculate the energy consumed in kilowatt hours

kWWVIP 84.118408230

kWhPtW 221284.1

kWh is the unit used in determine amount of energy used in electricity

in

o

P

P

rFT .

uFt

dF

t

dF

t

WP ..

.

TnTN

TP rr 2

60

2

In case of rotating electrical machine

Physical quantities

T= torque , Nr = rotation speed (revolution per min), nr=revolution per sec, Po= output power, Pin =input power

Power efficiency

Power is the rate of consuming the energy (The rate of work done)

For (ac) current the power in rms :P = VI

From Ohm’s law V = IR P = I2Ror from Ohm‘s law;I = V/R P = V2/R

Unit for power is in watt (W) A power of 1 W is a rate of energy consuming for 1 J per

sec. Thus 1 W = 1 J/s

rpm 1800speedMotor rps 3060

1800f

Nm 48Torque T kW 048.948302 TPout

%88 kW 10.28288.0

9048

out

in

PP

V 415V 85.0P.F.

kW 10.282P.F. VIPinA 29.15

85.0415

10282

P.F.

V

PI in

and

An electric motor has a torque of 48 Nm at rotation speed of 1800 revolution per minute (r.p.m). The efficiency of the motor is 88%. If the power factor is 0.85, calculate the current drawn by the motor when it is connected to the main supply of 415 V.

A 230 V lamp is rated to pass a current of 0.26 A. Calculate its power output. If a second similar lamp is connected in parallel to the lamp, calculate the supply current required to give the same power output in each lamp.

AV

PI 52.0

230

120

WP 1206060

WVIP 6026.0230

Power for parallel lamps

Calculate the dissipated power from a resistor of 2M when a current of 10 A flowing in it.

P = I2R = (10 x 10-6)2 x 2 x 106 = 200 W

R = 2 MI = 10 A

For current and voltage which are not constant, the power must be calculated as follow :

Divide into time interval and the instantaneous power are

Rms =(root mean square )

Ri266p

masa (s)

Current (A)

Ri211p Ri2

22p Ri255p Ri2

33p Ri244p

6/)( 26

25

24

23

22

21 RiRiRiRiRiRi

6/)pppppp(p 654321 ave

RiRiiiiii rms22

625

24

23

22

21 6/)(

n

i............iiI

2n

22

21

rms

6

iiiiiiI

26

25

24

23

22

21

rms