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.