Resistance

Ohm’s law

Combining resistances

Battery: Emf and internal resistance

Power

Combining Capacitances

ELECTRICITY

Lecture 3.9

Simple Circuits

bulb

Simple circuit

I Battery: Potential difference

+ -

Simple Circuit

•Voltage (V) supplied by battery •causes a current (I)(charge per second Q/t) 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

Flow rate F ~

Narrow pipe provides resistance R to fluid flow.

Flow rate F ~ (P1-P2)/R

pressure difference resistance

Narrow pipe

Water pump

P1 P2

Simple Circuit, Ohm’s Law

+ - battery

resistance

I I

I I

Anything in the circuit that impedes the current is called the resistance R.

Resistance: •Electrons collide with atoms of material •Therefore depends on type of material

Resistance also depends on •geometrical factors •temperature

Simple Circuit, Ohm’s Law

+ - battery

resistance

I I

I I

I V∝V IR=VI

R=

Ohm’s Law: •current is proportional to potential difference •current inversely proportional to resistance

This image cannot currently be displayed.

VIR

=

The larger the resistance the smaller the current.

The Ohm is the unit of resistance and is denoted by the Greek letter Ω.

A voltage of 1 volt applied to a resistance of 1 Ohm results in a current of 1 amp.

+ - battery

resistance

I I

I I

SI unit of resistance named after German physicist Georg Ohm, 1789- 1854

Simple Circuit, Ohm’s Law

V IR=

Simple circuit: ----- a complete path or loop, Simple example: contains one source of potential difference and one resistance.

The current I is the same everywhere along the path and is equal to: I = V/R

Simple Circuit

-

R

+

V

Electron flow

In general a circuit can 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 flowing in the circuit goes through each resistance serially.

The total potential difference V is the sum of the potential differences 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

+ -

21V

8Ω 4Ω 2Ω

R1 R2 R3

Voltage Drop

As current is passed through a resistance the voltage is reduced.

Current is the same entering the resistance as exiting

There must be a voltage drop or potential difference across the resistance to cause charges to flow (current)

Resistances in series Example

Consider circuit as shown. Calculate (a) total series resistance (b) current flowing in the circuit (c) voltage drop across each resistance (d) total voltage drop across the 3 resistances.

+ -

21V

8Ω 4Ω 2Ω

(a) RS =Rtot = R1 + R2 + R3 = (8+4+2)Ω= 14Ω

(b)

V1 = IR1 = 1.5A* 8Ω =12V V2 = IR2 = 1.5A* 4Ω =6V V3 = IR3 = 1.5A* 2Ω =3V

(c)

Vtot = V1+V2+V3 =12v + 6V + 3V = 21V

R1 R2 R3

21 1.514total

total

V VI ampsR

= = =Ω

Each additional path results in an increase in the total current flowing:

All resistances are subject to the same voltage V therefore:

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 = I1 + I2 +I3 = 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

60Ω

60Ω

60Ω I2

I1

I3

Itot

9V

+ -

1 2 3

1 1 1 1

PR R R R= + +

1 1 1 1 160 60 60 20PR

= + + =

p

VIR

=

Example

Three resistances each of value 60Ω 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 = 20Ω

9 0.4520

I A= =

R3

R2

R1

R4 A B

Resistance Rp between A and B 1/Rp = 1/R1 +1/R2 + 1/R3

= 1/R1 +1/R2 + 1/R3 1

+R4 Rtotal

Calculate the total resistance between A and C when R1 = 2Ω, R2 = 3Ω, R3 = 6Ω and R4 = 12Ω.

Example

= 1/2Ω +1/3Ω + 1/6Ω 1

+12Ω Rtotal

Rtotal = (1+12)Ω = 13Ω

. .

= 1/R1 +1/R2 + 1/R3 1

Rp

. C

Total resistance Rtotal between A and C

Example There are four possible ways in which three resistances can be combined in a simple circuit. Calculate the total resistance in each of the four ways if individual resistances are of value 6 Ω.

Rs = 18 Ω

1 1 1 1

PR R R R= + +

1 1 1 16 6 6PR

= + +

RP = 2 Ω

1 1 1

PR R R= + RP = 3 Ω

RT = 3 Ω + 6 Ω = 9 Ω

This image cannot currently be displayed.

6 Ω 6 . 6 Ω

Series

Parallel

1 1 112 6PR

= + RT = 4 Ω

The potential difference V available to the external circuit will be lower than the emf:

A real voltage source (battery) is never perfect. It is defined by: Electromotive force (emf): ε (in Volts) internal resistance: r [in ohms (Ω)]

Part of the energy delivered by the source is ‘used up’ inside the source (which heats slightly).

V depends on the current and hence on the external resistance R: V = ε – Ir and V = IR

V = ε - Ir

Note: V = ε in an open circuit (I=0).

Battery: Emf and internal resistance

R

V

+ _ ε r

I

Battery I = ε /(R+r).

Example

Determine the potential difference between the ends of a 20 Ω resistance when it is connected to a battery of emf 10 V and internal resistance of 5 Ω?

V = ε - Ir I = ε /(R+r). V = IR

V = 10V – 0.4A x 5Ω

V = 8volts.

8V

20Ω

+ _

10V

I

Battery

5Ω

( )10 10 0.4

20 5 25V VI amps= = =+ Ω

0.4 20 8V IR amps volts= = × Ω =

Example

The potential difference between the ends of a 40 Ω resistance is 16V when it is connected to a battery of emf 18V. Find the internal resistance of the battery

V = ε - Ir

I = ε /(R+r).

V = IR

16V = 18V – 0.4amps x r

r = 5Ω

16V

40Ω

+ _

18V

I

Battery

5Ω

16 0.440

VI amps= =Ω

Electrical Power

+ -

V

R Most of the energy supplied by the battery 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 UPowertime t

= =

U qV=qVPt

=Therefore qIt

=

P IVPower current voltage

== ×

Electrical power consumed by any component in a circuit is P =IV

•power produced by the source, •consumed by resistive components

A resistance transforms the electrical energy into heat (due to ‘friction’ exerted on the electrons).

Electrical Power

Units of power are Joules per second or Watts

P=IV from Ohm’s law P = I(IR) = I2R

Also P =IV = (V/R)V = V2/R

P = IV P =I2R P= V2/R

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.

The power consumed in a resistance connected across a potential difference of 20 V is 4 Watts. What is (a) the value of the current in the resistance ? and (b) the value of the resistance

Example

P =IV I =4w/ 20V = 0.2 amps

R =V/I = 20V/ 0.2A = 100Ω

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 glow brightly

use less energy and are cool to the touch.

The capacitance (denoted C) is the ability to store charge, expressed as ratio of charge to potential difference:

where Q is the charge on either plate

Electrical Capacitance

-

- - - -

-

+

+ + + +

+

V

Voltage removed,charge remains on the plates

Q V C =

SI unit of capacitance Farad (F)

Capacitance

In series:

Capacitances in series act as a single capacitance Cs:

Combining Capacitances

q+ q- q- q+

Total voltage V = Vac + Vcb V = Vac + Vcb =q/C1 + q/C2 = q(1/C1+1/C2)=q(1/Cs)

Left hand plate of C1 is at the potential of the positive terminal of the battery Right hand plate of C2 is at the potential of the negative terminal of the battery

These two plates have charges of q+ and q- respectively

Hence charges of q+ and q- are induced on remaining plates

Each capacitor -- same quantity of charge q

a b c Vac Vcb

C1 C2

V

1 2

1 1 1

sC C C= +

Q = C1V + C2V =V(C1+C2)= CPV

Capacitances in parallel act as a single capacitance Cp, sum of all capacitances:

Combining Capacitances C1

C2

V

+q1 -q1

-q2 +q2

In parallel:

Potential difference V across each capacitance is the same

q2 = C2V q1 = C1V

Total charge stored Q = q1 + q2

Divide across by V 1 2pC C C= +

1 01

ACdε

=1

' 1.4 2 2.8C F Fµ µ= × =

1/Cs = 1/C1 + 1/C2

Two parallel plate capacitors C1 = 2µF andC2 = 3µF 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 =1.2µF

43%

'2

'2

1 1 11.2 2.8

2.1

CC Fµ

= +

=

C2 must decrease from 3µF to 2.1µF

So distance must increase by factor 3/2.1 = 1.43

Find new value of C2 such that Cs = 1.2µF

Example