Lecture 14
ELECTRICITY
Ohm’s law and resistance
Electric power
Alternating current
Combining resistances
Combining capacitances
Simple Circuit
+ -
bulb
battery
Voltage (V) (potential difference)
•supplied by the battery
•causes a current (I) (charge per second)
in the wires and the filament of the bulb.
Current will continue as long as voltage
is supplied by the battery.
Ohm’s Law
Fluid flow analogy:
helps visualise flow of electricity through
a resistive component
Narrow pipe
P1
P2
Water pump
Narrow pipe provides resistance R
to fluid flow.
Flow rate I =
pressure difference
resistance
P1  P2
Flow rate I 
R
Simple Circuit, Ohm’s Law
resistance
I
I
+
-
I battery I
Anything in the circuit that impeds the
current is called the resistance R.
resistance
I
I
+
-
I battery I
Ohm’s Law:
current is proportional to potential difference
and inversely proportional to resistance:
I V
1
I
R
V
I
R
V  IR
Simple Circuit, Ohm’s Law
resistance
I
I
+
-
I battery I
The Ohm is the unit of resistance and is
denoted by the Greek letter W.
named after German physicist, Georg Ohm,
1789- 1854
V
I
R
V  IR
The larger the resistance the smaller the current.
A voltage of 1 volt
applied across a resistance of 1 W
results in a current of 1 A.
Electrical Power
R
+
-
Most of the energy
is lost as heat
in the resistance
V
The flowing charges carry energy from the
battery to the resistance. This is converted to
thermal energy as the electrons collide with
the atoms of the resistive material.
Power is the rate of
doing work or
expending energy
energy E
Power 

time
t
In an electrical circuit the energy
qV
P
t
E  qV
q
and since I 
t
Power  currrent  voltage
Therefore
P  IV
Electrical Power
Electrical power consumed by any
component in a circuit is
P  IV
Units of power are Joules per second or Watts
Ohm’s law
V  IR
V
Also since I 
R
P = IV
P  IV  I  IR   I 2 R
V2
V 
P  IV   V 
R
R
P =I2R
P= V2/R
Electrical energy supplied by the battery,
or generator,
is transformed into thermal energy by
resistive components in the circuit.
Electrical Power
Example
A 100 W bulb operates at a voltage of 220V.
Determine the resistance of its filament?
P = V2/R
or
R =V2/P
R = (220)2/100 = 484W
Example
How much energy does a 100W light bulb
consume in 15 minutes?
Power is the rate of doing work or expending
energy
Power = E/t or
E = Power x time.
E = 100W x 15x60 Joules = 90x103Joules.
Alternating Current (ac)
Batteries are sources of steady or direct voltage.
Current in a circuit powered by a battery
is also steady and is called direct current (dc)
Direct voltage
Direct current dc
I
V
constant
Voltage
& current
V0
I0
t
t
Alternating Current (AC):
Nearly all the electricity we use is in the form of
alternating voltage (& current ) termed ac
V
Voltage changes
+V0
in magnitude
t and direction
periodically and
-V0
is generally sinusoidal.
The current also alternates
Alternating Current (Voltage)
V
0
+V0
Veff
T
T
t
-V0
V0 is the maximum voltage value or
voltage amplitude
T is the periodic time
Average voltage value is zero.
ac voltage and current are always
characterized by their
effective (or root mean square ) values:
Veff =
Vo
2
and
Ieff =
Io
2
where Vo, and Io are the amplitudes or maximum
values of the voltage and the current.
Alternating Current
+V0
Veff
0
T
T
Veff =
t
-V0
Ieff =
Vo
2
Io
2
Example
An alternating current with a maximum value of
3 amps will produce the same heating effect in a
resistance as a direct current of (3/√2)amps
ac mains varies at a rate of 50 times per
second– frequency of 50 cycles per second
ac mains---frequency 50 Hertz ----50Hz
SI unit of frequency named after German
Physicist Heinrich Hertz 1857-1894.
periodic time (time for one cycle)
T= (1/50)s = 20x10-3 s = 20 ms
Power in an AC circuit
AC power is:
P = IeffVeff
P = (I0/√2) (V0/√2)
( IoVo)
P=
2
since V0=I0R
since I0 = V0/R
2R
(I
)
P= o
2
2
(V
)
P= o
2R
Alternating Current
ESB provides electricity at a voltage of
220V, and a frequency of 50Hz.
The period of the oscillations is:
T = 1/50Hz = 0.02s = 20 ms.
220V is the effective value of the voltage:
Veff = 220V
amplitude V0 is:
V0 = Veff * 2 = 311V
Mains voltage swings from +311V to -311V 50
times every second giving an effective voltage
of 220V.
The effective current going through a
100W light bulb is:
Ieff = P/Veff = 100/220 A = 0.45A
Example
What is (a) the maximum value and (b) the
average value of potential difference of the AC
mains if the effective potential difference is
220 V?
V
+V0
Veff
0
t
-V0
Veff = Vo/2
V0 = Veff 2 = 2202 = 311V
Combining Resistances
In general a circuit may contain many resistances.
Resistances can be combined
in two main ways:
a) In series: all the resistances are on the
same path for the current
R1
R2
R3
+ I
V
b) In parallel: each
resistance is on a parallel
path for the current
R3
I3
I2
R2
R1
I1
Itot
+ V
Resistances in series
V1=IR1
R1
V2=IR2
R2
V3=IR3
R3
+ I
V
The current I is the same in each resistance
because there is only one path.
Charge goes through each resistance
serially.
The total potential difference V is the sum of
the potential difference across each
resistance:
V = V1+V2+V3 = IR1 + IR2 + IR3 = IRS
Resistances in series act as a single
resistance Rs, sum of all the resistances:
RS = R1 + R2 + R3
Voltage Drop
R1
R2
R3
+ V
Current is the same entering the resistance
as exiting
There must be a potential difference or
voltage drop across the resistance to
sustain the current
As current is passed through a resistance the
voltage across the resistance is reduced.
Resistances in series
Example
Consider circuit as shown below.
(a) Calculate total series resistance.
(b) Calculate the current flowing in the circuit.
(c) Calculate the voltage drop across each resistance and
the total voltage drop across the 3 resistances together.
8W
4W
2W
R1
R2
R3
+ 21V
(a)
RS =Rtot = R1 + R2 + R3 = (8+4+2)W= 14W
(b) Using Ohms law Itot = V/Rtot= 21V/14W
Itot = 1.5A
(c)
V1 = IR1 = 1.5A* 8W =12V
V2 = IR2 = 1.5A* 4W =6V
V3 = IR3 = 1.5A* 2W =3V
Vtot = V1+V2+V3 =12v + 6V + 3V = 21V
Resistances in parallel
Each additional path
results in an increase in
the total current flowing:
Itot = I1 + I2 + I3
R3
I3
I2
R2
R1
I1
Itot
All resistances are subject to the
same voltage V :
+ V
I1 = V/R1 I2 = V/R2 I3 = V/R3
The total current is thus:
Itot = V/R1 + V/R2 + V/ R3 =V/RP
Resistances in parallel act as a single
resistance Rp such that:
1
1 1
1
  
RP R1 R2 R3
Example
Three resistances each of value 60W are
connected in parallel. (a) What is the total
resistance of the combination? (b) What is the
total current in the circuit when the parallel
combination is connected to a 9V battery?
60W
I3
I2
60W
60W
I1
Itot
+ 9V
1
1 1
1
  
RP R1 R2 R3
1
1
1
1

 
RP 60 60 60
1
1

RP 20
RP = 20W
V
I
Rp
9
I
 0.45 A
20
Emf and internal resistance of a battery
A real voltage source in never perfect.
It is defined by:
Electromotive force (Emf): e (in V)
internal resistance: r [in ohms (W)]
The potential difference V available to the
external circuit will be lower than the Emf:
R
Part of the energy delivered
by the source is ‘used up’
inside the source (which
I
slightly heats).
V = e – Ir
V
r
+
I = e /(R+r).
Battery
e
_
Example
Determine the potential difference between
the ends of a 20 W resistance when it is
connected to a battery of emf 10 V and
internal resistance of 5 W?
V = e - Ir
I = e /(R+r).
V = IR
20W
10V
10V
I

 0.4amps
 20  5 25W
8V
I
10V
V = 10V – 0.4A x 5W
5W
+
V = 8volts.
Battery
_
Combining Capacitances
In series: Each carries the same amount of
charge
C2
C1
Q+ Q- Q+ QV = Vac + Vcb
a
c
Vac
b
=Q/C1 + Q/C2 = Q/CS
Vcb
V
Capacitances in series act
as a single capacitance Cs:
1
1
1
 
Cs C1 C2
In parallel: Q = q1 +q2 = C1V + C2V = CPV
Capacitances in parallel act
as a single capacitance Cp
sum of all capacitances:
C p  C1  C2
C1
+q1 -q1
+q2
-q2
C2
V
Example
Two parallel plate capacitors C1 = 2F andC2 = 3F are
connected in series. Calculate the total capacitance of
the combination. If the area of the plates of C1 is
increased by 40%, by what % must the separation of the
plates of C2 change such that the total capacitance of
the combination remains unchanged.
Cs =
1
1/C1 + 1/C2
A1e 0
C1 
d
=
1
1/2 + 1/3
1
1
1
 
Cs C1 C2
=1.2F
C1  1.4  2 F  2.8 F
'
1
1
1

 '
1.2 2.8 C2
Find new value of C2
such that Cs = 1.2F
C2'  2.1 F
C2 must decrease from 3F to 2.1F
So distance must increase by factor 3/2.1 = 1.43
43%
Derive an expression for the total resistance between
A and B
R3
R4
R2
A
R1
1
1 1
1
  
Rp R1 R2 R3
Rtotal =
Rtotal
Rp + R4
1

 R4
1 1
1
 
R1 R2 R3
B
Electrical Power
Example
Will a 220V dental-surgery circuit protected by
a 15A circuit breaker be able to operate a
200W dental drill, a 1200W x-ray machine and
eight 100W lights simultaneously?
Veff = 220V
Ieff = 15A
The maximum power consumption allowed is:
P = IeffVeff = 3300W
All the apparatus will consume:
P = 200W + 1200W + 8*100W = 2200W
So all apparatus will be able to operate
simultaneously.
Electrical Power
Example
How much energy does a 800W microwave
oven consume in 3 minutes?
Power is the rate of doing work or expending
energy
Power = E/t or
E = Power x time.
E = (800W x 3 x 60) Joules = 144x103Joules.
Energy saving light bulbs
Traditional (filament) bulbs waste a lot of energy
by turning it into heat rather than light.
Energy Saving bulbs:
•work in the same way as fluorescent lights
•electric current passes through gas in a tube
•Atoms of gas are excited
•UV radiation emitted
•UV radiation incident on fluorescent material
coated on inside of tube
•coating glows brightly
Equal amounts of visible light output
Traditional
filament light bulb
consumes
100W
Energy Saving
light bulb consumes
21W
use less energy and are cool to the touch.