physics

2. Current flow versus Electron
flow
Conventional
current flows
this way.
Electrons
flow this
way.

3. What formula relates
Charge, 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

4. 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

5. 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!!

6. 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?

7. Worked example on Ohm’s Law
2 A 8 Ω
V = ?
V IxR=
= 2A x 8
= 16 V
Ω

8. Ammeters and Voltmeters
Ammeters measure current and are placed in series
in a circuit.
Voltmeters measure voltage and are placedVoltmeters measure voltage and are placed
in parallel in a circuit.in parallel in a circuit.
A
V

9. Rules for
Resistors in SERIES
RTotal = + +R R R1 2 3

10. Examples on Resistors in Series
6 Ω 9 Ω
Ans. = 15 Ω
4Ω 6 Ω 3 Ω
Ans. = 13 Ω
No. 1
No. 2

11. Rules for
Resistors in PARALLEL
1 1 1
R
1
R R R
This formula is shortened to
R
R R
R R
oduct
Sum
Total 1 2 3
Total
1 2
1 2
= + +
=
+
=
Pr

12. Examples on Resistors in Parallel
6 Ω
12 Ω
Ans. = 3 Ω
6 Ω
Ans. = 6 Ω
12 Ω
No. 1
No. 2

13. For you to do!!!!
16 Ω
6 Ω
16 Ω
Ans. = 14 Ω
No. 3

14. 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

15. Rules for SERIES CIRCUITS
• Same current but ……
• split voltage between them.

16. 18 V
6 V 6 V 6 V
?
Equal resistors share the
voltage between them!!

17. Rules for PARALLEL
CIRCUITS
• Same voltage but ……
• split current between them.

18. ? A
? A
4 A
? A
Equal
resistors
What will be the currents flowing
through each ammeter?

19. 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 W
Current = ----------- = 0.25 A
240 V
ElectricalElectrical Power = Potential difference * currentPower = Potential difference * current
Watts Volts AmpsWatts Volts Amps

20. A transformer is a device for increasing or decreasing
an a.c. voltage.

21. Structure of Transformer

22. Circuit Symbol for Transformer

23. How Transformer works
Laminated soft
iron core
Primary coil Secondary coil
Input voltage
(a.c.)
Output voltage
(a.c.)

24. 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.

25. 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.

26. 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
N
p
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.

27. Worked example No. 1
The diagram shows a transformer. Calculate the
voltage across the secondary coil of this transformer.
Step-up transformer!

28. Solution
V
V
N
N
Substituting
12
V
180
540
Crossmultiplying
180.V 12 x 540
V
12 x 540
180
V 36 V
P
S
P
S
S
S
S
S
=
=
=
∴ =
∴ =

29. Worked example No. 2
A 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 = 1380
VS = 6 V
NS = ?
Obviously, a Step-down transformer!!

30. Solution
V
V
N
N
Substituting
230
6
1380
N
Crossmultiplying
2300.N 6 x 13800
N
6 x 1380
230
N 36 turns
P
S
P
S
S
S
S
S
=
=
=
∴ =
∴ =