Download Transformers - Port Hope High School

Current flow versus Electron
flow
Electrons
flow this
way.
Conventional
current flows
this way.
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=Ixt
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
For you to do!!
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
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.
What is the formula for Ohm’s law?
Voltage (V) = Current (A) x resistance (Ω)
V=IxR
Worked example on Ohm’s Law
2A
8Ω
V=?
V  IxR
= 2A x 8
= 16 V
Ammeters and Voltmeters
Ammeters measure current and are placed in series
in a circuit.
A
V
Voltmeters measure voltage and are placed
in parallel in a circuit.
Rules for
Resistors in SERIES
R Total  R1  R2  R3
Examples on Resistors in Series
No. 1
9Ω
6Ω
Ans. = 15 Ω
No. 2
4Ω
6Ω
3Ω
Ans. = 13 Ω
Rules for
Resistors in PARALLEL
1
1
1
1



R Total R1 R 2 R 3
This formula is shortened to
R1R 2
Pr oduct
R Total 

R1  R 2
Sum
Examples on Resistors in Parallel
6Ω
No. 1
Ans. = 3 Ω
6Ω
No. 2
12 Ω
Ans. = 6 Ω
12 Ω
For you to do!!!!
No. 3
16 Ω
6Ω
16 Ω
Ans. = 14 Ω
6Ω
6Ω
No. 4
12 Ω
Ans. = 6 Ω
No. 5
10 Ω
2Ω
3Ω
10 Ω
2Ω
Ans. = 9 Ω
Rules for SERIES CIRCUITS
• Same current but ……
• split voltage between them.
Equal resistors share the
voltage between them!!
18 V
?
6V
6V
6V
Rules for PARALLEL
CIRCUITS
• Same voltage but ……
• split current between them.
What will be the currents flowing
through each ammeter?
4A
?A
?A
Equal
resistors
?A
Electrical Power
Electrical Power = Potential difference * current
Watts
Volts
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
Amps
A transformer is a device for increasing or decreasing
an a.c. voltage.
Structure of Transformer
Circuit Symbol for Transformer
How Transformer works
Laminated soft
iron core
Input voltage
Output voltage
(a.c.)
(a.c.)
Primary coil
Secondary coil
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.
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.
Formula for Transformer
voltage across the primary coil
number of turns on primary

voltage across the secondary coil number of turns on secondary
Vp
Vs

Np
Ns
Where Vp = primary voltage
Vs = secondary voltage
Np= Number of turns in primary coil
Ns = Number of turns in a secondary coil.
Worked example No. 1
The diagram shows a transformer. Calculate the
voltage across the secondary coil of this transformer.
Step-up transformer!
Solution
VP N P

VS N S
Substituting
12 180

VS 540
Crossmultiplying
180.VS  12 x 540
12 x 540
 VS 
180
 VS  36 V
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
Obviously, a Step-down transformer!!
VS = 6 V
NS = ?
Solution
VP N P

VS N S
Substituting
230 1380

6
NS
Crossmultiplying
2300.N S  6 x 13800
6 x 1380
 NS 
230
 N S  36 turns