Electric Revision (1)

7/31/2019 Electric Revision (1)

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Current flow versus Electron

flowConventional

current flows

this way.

Electrons

flow 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 in1 second.

In general, if a steady current I (amperes) flows for time t(seconds) the charge Q (coulombs) passing any point isgiven by

Q = I x t



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



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!!



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?



Worked example on Ohm’s Law2 A

8 Ω

V = ?

V IxR== 2A x 8

= 16 V

Ω



Ammeters and Voltmeters

Ammeters measure current and are placed in seriesin 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



Rules for

Resistors in SERIES

R Total = + + R R R1 2 3



Examples on Resistors in Series

6 Ω 9 Ω

Ans. = 15 Ω

4Ω 6 Ω 3 Ω

Ans. = 13 Ω

No. 1

No. 2



Rules for

Resistors in PARALLEL1 1 1

R

1

R R R

This formula is shortened to

R R R

R R

oduct

Sum

Total 1 2 3

Total1 2

1 2

= + +

=+

=Pr



Examples on Resistors in Parallel6 Ω

12 Ω

Ans. = 3 Ω

6 Ω

Ans. = 6 Ω

12 Ω

No. 1

No. 2



For you to do!!!!

16 Ω

6 Ω

16 Ω

Ans. = 14 Ω

No. 3



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



Rules for SERIES CIRCUITS

• Same current but ……

• split voltage between them.



18 V

6 V 6 V 6 V

?

Equal resistors share the

voltage between them!!



Rules for PARALLEL

CIRCUITS

• Same voltage but ……

• split current between them.



? A

? A

4 A? A

Equal

resistors

What will be the currents flowing

through each ammeter?



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 * currentWatts Volts AmpsWatts Volts 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

Primary coil Secondary coil

Input voltage

(a.c.)

Output voltage

(a.c.)



All transformers have three parts:

1. Primary coil – the incoming voltage V p

(voltage across primary coil) is connected

across this coil.

2. Secondary coil – this provides the outputvoltage V

s(voltage across the secondary coil)

to the external circuit.

3. Laminated iron core – this links the two coilsmagnetically.

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

voltage across the secondary coil

number of turns on primary

number of turns on secondary

V

V

N

N

p

s

p

s

=

=

Where V p

= primary voltage

Vs = secondary voltage

N p= 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

VV

N N

Substituti ng

12V

180540

Crossmulti plying

180. V 12 x 540

V12 x 540

180V 36 V

P

S

P

S

S

S

S

S

=

=

=

∴ =

∴ =



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!!



Solution

V

V

N

N

Substituti ng

230

6

1380

NCrossmulti plying

2300. N 6 x 13800

N

6 x 1380

230 N 36 turns

P

S

P

S

S

S

S

S

=

=

=

∴ =

∴ =

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Electric Revision (1)