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Electricity

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Page 1: Electricity
Page 2: Electricity

Current flow versus Electron flow

Conventional current flows this way.

Electrons flow this way.

Page 3: Electricity

What formula relatesCharge, Current and Time?

A current of 1 Ampere is flowing when 1 Coulomb of charge flows past a point in a circuit in 1 second.

Charge = current x time (C) (A) (s)If a current of 5 A is flowing then 5 C of charge pass a point in

1 second. In general, if a steady current I (amperes) flows for time t

(seconds) the charge Q (coulombs) passing any point is given by

Q = I x t

Page 4: Electricity

Worked example

A current of 150 mA flows around a circuit for 1minute. How much electrical charge flows past a point in the circuit in this time?

Solution

Substituting into Q = It

gives Q = 0.15 A x 60 s

= 12 C

Page 5: Electricity

1. Convert the following currents into amperes:

a) 400 mA b) 1500 mA.

Ans. = a) 400 mA = 0.4 A b) 1500 mA = 1.5 A

2. What charge is delivered if a current of 6A flows for 10 seconds?

Ans. = 60 C

3. What charge is delivered if a current of 300 mA flows for 1 minute(60 seconds)?

Ans. = 18 C

For you to do!!

Page 6: Electricity

What is Ohm’s Law?The voltage dropped across a resistor is directly proportional to the

current flowing through it, provided the temperature remains constant.

Voltage (V) = Current (A) x resistance (Ω)

V = I x R

What is the formula for Ohm’s law?

Page 7: Electricity

Worked example on Ohm’s Law2 A 8 Ω

V = ?

V IxR = 2A x 8

= 16 V

Page 8: Electricity

Ammeters and VoltmetersAmmeters measure current and are placed in series

in a circuit.

Voltmeters measure voltage and are placed Voltmeters measure voltage and are placed in parallel in a circuit.in parallel in a circuit.

A

V

Page 9: Electricity

Rules forResistors in SERIES

RTotal R R R1 2 3

Page 10: Electricity

Examples on Resistors in Series

6 Ω 9 Ω

Ans. = 15 Ω

4Ω 6 Ω 3 Ω

Ans. = 13 Ω

No. 1

No. 2

Page 11: Electricity

Rules forResistors in PARALLEL

1 1 1

R

1

R R R

This formula is shortened to

RR R

R R

oduct

Sum

Total 1 2 3

Total1 2

1 2

Pr

Page 12: Electricity

Examples on Resistors in Parallel6 Ω

12 Ω

Ans. = 3 Ω

6 Ω

Ans. = 6 Ω

12 Ω

No. 1

No. 2

Page 13: Electricity

For you to do!!!!

16 Ω

6 Ω

16 Ω

Ans. = 14 Ω

No. 3

Page 14: Electricity

Ans. = 9 Ω

Ans. = 6 Ω

6 Ω6 Ω

12 Ω

10 Ω

3 Ω

10 Ω 2 Ω

2 Ω

No. 4

No. 5

Page 15: Electricity

Rules for SERIES CIRCUITS

• Same current but ……

• split voltage between them.

Page 16: Electricity

18 V

6 V 6 V 6 V

?

Equal resistors share the voltage between them!!

Page 17: Electricity

Rules for PARALLEL CIRCUITS

• Same voltage but ……

• split current between them.

Page 18: Electricity

? A

? A

4 A? A

Equal resistors

What will be the currents flowing through each ammeter?

Page 19: Electricity

Electrical Power

E.g. A study lamp is rated at 60 W, 240 V. How much current is the bulb carrying?Solution

60 W = 240 V * Current

60 WCurrent = ----------- = 0.25 A 240 V

ElectricalElectrical Power = Potential difference * current Power = Potential difference * current

Watts Volts AmpsWatts Volts Amps

Page 20: Electricity

A transformer is a device for increasing or decreasing

an a.c. voltage.

Page 21: Electricity

Structure of Transformer

Page 22: Electricity

Circuit Symbol for Transformer

Page 23: Electricity

How Transformer works

Laminated soft iron core

Primary coil Secondary coil

Input voltage

(a.c.)

Output voltage (a.c.)

Page 24: Electricity

All transformers have three parts:

1. Primary coil – the incoming voltage Vp

(voltage across primary coil) is connected across this coil.

2. Secondary coil – this provides the output voltage Vs (voltage across the secondary coil)

to the external circuit.

3. Laminated iron core – this links the two coils magnetically.

Notice that there is no electrical connection between the two coils, which are constructed using insulated wire.

Page 25: Electricity

Two Types of Transformer

A step-up transformer increases the voltage - there are more turns on the secondary than on the primary.

A step-down transformer decreases the voltage - there are fewer turns on the secondary than on the primary.

To step up the voltage by a factor of 10, there must be 10 times as many turns on the secondary coil as on the primary. The turns ratio tells us the factor by which the voltage will be changed.

Page 26: Electricity

Formula for Transformer

voltage across the primary coil

voltage across the secondary coil

number of turns on primary

number of turns on secondary

V

V

N

Np

s

p

s

Where Vp = primary voltage

Vs = secondary voltage

Np= Number of turns in primary coil

Ns = Number of turns in a secondary coil.

Page 27: Electricity

Worked example No. 1The diagram shows a transformer. Calculate the voltage across the secondary coil of this transformer.

Step-up transformer!

Page 28: Electricity

Solution

VV

NN

Substituting12V

180540

Crossmultiplying180.V 12 x 540

V 12 x 540180

V 36 V

P

S

P

S

S

S

S

S

Page 29: Electricity

Worked example No. 2A transformer which has 1380 turns in its primary coil is to be used to convert the mains voltage of 230 V to operate a 6 V bulb. How many turns should the secondary coil of this transformer have?

VP = 230 V

NP = 1380VS = 6 V

NS = ?

Obviously, a Step-down transformer!!

Page 30: Electricity

Solution

VV

NN

Substituting2306

1380N

Crossmultiplying2300.N 6 x 13800

N 6 x 1380230

N 36 turns

P

S

P

S

S

S

S

S


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