Transcript 1. Quantity & Unit

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 electron
Units 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 mA
125 A = 125 x 10-3 mA = 0.125 mA
125 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 is
Q = q1 + q2 = 10 C + (-4) C = 6 C.
Time t = 2 ms
I = Q/t = (6 x 10-6)/(2 x 10-3) = 3 mA
(ii)The charge net flow is
Q = q1 – q2 = 10 C - (-4) C = 14 C.
I = Q/t = (14 x 10-6)/(2 x 10-3) = 7 mA
A
q1
Y
q2
X
 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
q2 = 1 C
B
q1
 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 as
R = l/A
or R = 1/(A)
 = 1/ = conductivity
 Resistance also depends on temperature and can be fomulated as
R 1 1  α 0 t1

R 2 1  α0 t 2
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.
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.
R
R1 1

R2 2
R 1 2 100  8000
R2 

 1143 
1
700
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.
1
R
A
R1 A2 d 22

 2
R 2 A1 d1
300 1.0 2

R 2 0.52
R2  75 
 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
R1  Ro 1   o1   2001  0.00428  80  268.5 
 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
V 10
At 20oC
R1  1 
 10 
I1
1
At 2oC
V2
10
R2 

 10.53 
I 2 0.95
R 1 1   o1 

R 2 1   o 2 
10.0 1  0.00428  20

10.53 1  0.00428   2 
 2  33.4o C
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
I
+
-
Vs
10V
R
R
1.2 k
(fixed resistor)
Rv
Rp
1
M
10 k
(potentiometer)
(
rheostat)
To generate a potential difference in the circuits
An 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 etc
Circuit’s symbols for DC voltage sources are
+
V
-
Ideal source
ri
+
V
-
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 (wt + )
where A is amplitude, w = 2p 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
+
ix
+
V
-
(a) Current
dependent source
vx
V
-
(b) Voltage
dependent source
 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(wt +)
I
+
i(t)
The current source which depends on other parameter
such as input current or input voltage of a device. The
symbols are as follows
ix
I
(a) Dependence on
current
vx
I
(b) Dependence on
voltage
For physical quantities
W= F(Newton) x d (m)
W= P t
1
W  mu 2
2
F=ma
For Electrical energy
W= I2 R t
W= VI t
Where F =force, d=distance, t = time , a= acceleration, P=power, m= mass
, u=velocity, V=voltage, I=current and R=resistance
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.
(a)
(b)
P  I 2 R  32 10  90W
W  Pt  90  5  60  2700 J
A heater takes a current of 8 A from a 230 V source for 12
h. Calculate the energy consumed in kilowatt hours
P  VI  230  8  1840W  1.84kW
W  Pt  1.84 12  22kWh
kWh is the unit used in determine amount of energy used in electricity
Physical quantities
W F .d
d
P

 F .  F .u
t
t
t
In case of rotating electrical machine
2 p Nr T
P T w 
 2 p nrT
60
T  F.r
Power efficiency

Po
Pin
T= torque , Nr = rotation speed (revolution per min), nr=revolution
per sec, Po= output power, Pin =input power
 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 = I2R
or 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
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.
Motor speed  1800 rpm
Torque  T  48 Nm
1800
 30 rps
60
Pout  wT  2p  30  48  9.048 kW
Pout
9048
Pin 

 10.282 kW

0.88
  88%
V  415 V
f 
and
P.F.  0.85
Pin  VI  P.F.  10.282 kW
Pin
10282
I

 29.15 A
V  P.F. 415  0.85
 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.
P  VI  230  0.26  60 W
Power for parallel lamps
P  60  60  120W
P 120
I 
 0.52 A
V 230
Calculate the dissipated power from a resistor of
2M when a current of 10 A flowing in it.
I = 10 A
P


R = 2 M
= I2R = (10 x 10-6)2 x 2 x 106
= 200 W
 For current and voltage which are not constant, the power must be
calculated as follow :
 Divide into time interval and the instantaneous power are
p1  i12 R
p 2  i22 R
p 3  i32 R
p 4  i42 R
p 5  i52 R
p 6  i62 R
p  ave  (p1  p 2  p3  p 4  p5  p6 ) / 6
 (i12 R  i22 R  i32 R  i42 R  i52 R  i62 R) / 6
2
 (i12  i22  i32  i42  i52  i62 ) R / 6  irms
R
 Rms =(root mean square )
i1  i 2  i 3  i 4  i 5  i 6

6
2
I rms
2
2
2
i  i 2  ............  i n
 1
n
2
I rms
2
2
Current (A)
2
2
masa (s)