Ohm’s law and resistance
Electric power
Alternating current
Combining resistances
Combining capacitances
ELECTRICITY
Lecture 14
Simple Circuit
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.
+ -
battery bulb
Ohm’s Law
Flow rate I =
Fluid flow analogy:
helps visualise flow of electricity through
a resistive component
Narrow pipe provides resistance R
to fluid flow.
Flow rate
pressure difference
resistance
Narrow pipe
Water pump
P1 P2
1 2P PI
R
Simple Circuit, Ohm’s Law
+ -
battery
resistance
I I
I I
+ -
battery
resistance
I I
I I
I VV IR
Anything in the circuit that impeds the
current is called the resistance R.
VI
R
Ohm’s Law:
current is proportional to potential difference
and inversely proportional to resistance:
1I
R
VI
R
The larger the resistance the smaller the current.
The Ohm is the unit of resistance and is
denoted by the Greek letter W.
A voltage of 1 volt
applied across a resistance of 1 W
results in a current of 1 A.
+ -
battery
resistance
I I
I I
named after German physicist, Georg Ohm,
1789- 1854
Simple Circuit, Ohm’s Law
V IR
Electrical Power
+ -
V
R Most of the energy
is lost as heat
in the resistance
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
In an electrical circuit the energy
and since
energy EPower
time t
E qV
qVP
tTherefore
qI
t
Power currrent voltage
P IV
Electrical power consumed by any
component in a circuit is
Electrical energy supplied by the battery,
or generator,
is transformed into thermal energy by
resistive components in the circuit.
Electrical Power
Units of power are Joules per second or Watts
Ohm’s law
Also since
P = IV P =I2R P= V2/R
V IR
VI
R
2P IV I IR I R
2V VP IV V
R R
P IV
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 (direct) voltage
Current in a circuit powered by a battery
is also steady and is called direct current (dc)
t
V
V0
Direct voltage
Nearly all the electricity we use is in the form of
alternating voltage (& current ) termed ac
t
V
+V0
-V0
Voltage changes
in magnitude
and direction
periodically and
Alternating Current (AC):
is generally sinusoidal.
The current also alternates
t
I
I0
Direct current dc
constant
voltage
& current
ac voltage and current are always
characterized by their
effective (or root mean square ) values:
where Vo, and Io are the amplitudes or maximum
values of the voltage and the current.
Alternating Current (Voltage)
T is the periodic time
t
V
+V0
-V0
Veff
T
V0 is the maximum voltage value or
voltage amplitude
Average voltage value is zero.
0
and 2
Vo Veff = Ieff =
2
Io
T
Alternating Current
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
t
periodic time (time for one cycle)
T= (1/50)s = 20x10-3 s = 20 ms
SI unit of frequency named after German
Physicist Heinrich Hertz 1857-1894.
ac mains---frequency 50 Hertz ----50Hz
t
+V0
-V0
Veff
T
0
T
2
Vo Veff =
Ieff = 2
Io
AC power is:
P = IeffVeff
Power in an AC circuit
since V0=I0R
since I0 = V0/R
( IoVo)
2 P =
(Io)2R
2 P =
(Vo)2
2R P =
P = (I0/√2) (V0/√2)
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 = Veff * 2 = 311V
The effective current going through a
100W light bulb is:
Ieff = P/Veff = 100/220 = 0.45 amps
Alternating Current
Mains voltage swings from +311V to -311V 50
times every second giving an effective voltage
of 220V.
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?
Example
Veff = Vo/2
V0 = Veff 2 = 2202 = 311V
t
V
+V0
-V0
0
Veff
(a)
(b) Average value is zero volts
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
b) In parallel: each
resistance is on a parallel
path for the current
Combining Resistances
R3
R2
R1
I2
I1
I3
Itot
V
+ -
+ -
V
R1 R2 R3
I
The current I is the same in each resistance
because there is only one path.
Charge goes through each resistance
serially.
total potential difference V is the sum of the
potential difference across each resistance:
Resistances in series act as a single
resistance Rs, sum of all the resistances:
Resistances in series
V = V1+V2+V3 = IR1 + IR2 + IR3
RS = R1 + R2 + R3
+ -
V
R1 R2 R3
I
V1=IR1 V2=IR2 V3=IR3
V = I(R1 + R2 + R3) = IRS
+ -
V
R1 R2 R3
Voltage Drop
Current is the same entering the resistance
as exiting
There must be a potential difference or
voltage drop across the resistance to
sustain the flow of electrons (current)
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.
+ -
21V
8W 4W 2W
(a) RS =Rtot = R1 + R2 + R3 = (8+4+2)W= 14W
(b) Using Ohms law Itot = V/Rtot= 21V/14W
Itot = 1.5A
V1 = IR1 = 1.5A* 8W =12V
V2 = IR2 = 1.5A* 4W =6V
V3 = IR3 = 1.5A* 2W =3V
(c)
Vtot = V1+V2+V3 =12v + 6V + 3V = 21V
R1 R2 R3
Each additional path
results in an increase in
the total current flowing:
All resistances are subject
to the same voltage V :
I1 = V/R1 I3 = V/R3 I2 = V/R2
The total current is thus:
Resistances in parallel act as a single
resistance Rp such that:
Resistances in parallel
Itot = I1 + I2 + I3
Itot = V/R1 + V/R2 + V/ R3
R3
R2
R1
I2
I1
I3
Itot
V
+ -
1 2 3
1 1 1 1
PR R R R
Itot = V(1/R1 + 1/R2 + 1/ R3 )=V/RP
60W
60W
60W
I2
I1
I3
Itot
9V
+ -
1 2 3
1 1 1 1
PR R R R
1 1 1 1
60 60 60PR
p
VI
R
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?
RP = 20W
90.45
20I A
1 1
20PR
The potential difference V available to the
external circuit will be lower than the Emf:
A real voltage source in never perfect.
It is defined by:
Electromotive force (Emf): e (in V)
internal resistance: r [in ohms (W)]
Part of the energy delivered
by the source is ‘used up’
inside the source (which
slightly heats).
V = e – Ir
I = e /(R+r).
Emf and internal resistance of a battery
R
V I
+
_ e
r
Battery
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
V = 10V – 0.4A x 5W
V = 8volts.
8V
20W
+ _
10V
I
Battery
5W
10 10
0.420 5 25
V VI amps
W
Also
0.4 20 8V IR amps Volts W
In series:
Capacitances in series act
as a single capacitance Cs:
In parallel:
Capacitances in parallel act
as a single capacitance Cp
sum of all capacitances:
Combining Capacitances
Vac Vcb
C1
C2
V
+q1 -q1
-q2 +q2
Each carries the same amount of
charge
V = Vac + Vcb
=Q/C1 + Q/C2 = Q/CS
Q = q1 +q2 = C1V + C2V = CPV
Q+ Q- Q- Q+
a b c
C1 C2
V
1 2
1 1 1
sC C C
1 2pC C C
1 01
AC
d
e
1
' 1.4 2 2.8C F F
Two parallel plate capacitors C1 = 2F andC2 = 3F 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
= 1
1/2 + 1/3
=1.2F
43%
'
2
'
2
1 1 1
1.2 2.8
2.1
C
C F
C2 must decrease from 3F to 2.1F
So distance must increase by factor 3/2.1 = 1.43
Find new value of C2
such that Cs = 1.2F
Example
1 2
1 1 1
sC C C
R3
R2
R1
R4
A B
Rtotal = Rp + R4
Derive an expression for the total resistance
between A and B
1 2 3
1 1 1 1
pR R R R
4
1 2 3
1
1 1 1totalR R
R R R
Example
Resistance between A and C is RP
C
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
Electrical Power
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.
Example
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.
100W 21W
Energy Saving bulbs:
•work in the same way as fluorescent lights
•electric current passes through gas in a tube
Traditional
filament light bulb
consumes
Energy Saving
light bulb consumes
Equal amounts of visible light output
•Atoms of gas are excited
•UV radiation emitted
•UV radiation incident on fluorescent material
coated on inside of tube
•coating glows brightly
use less energy and are cool to the touch.