Section 15.5 Electrical Power

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© Manhattan Press (H.K.) Ltd. 1

Section 15.5Section 15.5

Electrical power Electrical power

Section 15.5Section 15.5

Electrical power Electrical power

• Heating effect of currentHeating effect of current

• Electrical power Electrical power




Heating effect of current

15.5 Electrical power (SB p. 67)

current passing through →

temperature ↑

Go to

Discussion 5Discussion 5

charges lose energy to atoms

by collision

Reason

→ atoms vibrate morevigorously

→ heats up

electrical energy→

heat




Electrical power


Go to

Discussion 6Discussion 6

takenTime

dtransferreenergyElectrical =

t

E P =

unit: watt (W)

Electrical

power




220 V 220 V

each secondconsumes 100 J

each secondconsumes 60 J

Electrical power





(1) Useful equations of electrical power

Electrical power


t

E P =

( )QV E t

QV ==

=t

QV

VI P =∴





Electrical power


VI P =

( )IR V I IR =×= R I P

2 =∴





Electrical power


VI P =

=

=

R

V I

R

V V

R V P

2

=∴



© Manhattan Press (H.K.) Ltd. 8Thinking 5Thinking 5

VI P =

R I P 2 =

R

V P

2

=

or

or

Electrical power





To section 15.6

http://e-ch15_06.ppt/

http://e-ch15_06.ppt/




Discussion 5:Discussion 5:

In each of the following electrical appliances,what kind of energy will the electrical energy be

converted to?

Electric cooker, fluorescent tube, electric iron,

vacuum cleaner, radio, washing machine, toaster

Electric cooker – heat, fluorescent tube – light,

electric iron – heat, vacuum cleaner – mechanical

energy, radio – sound, washing machine – mechanical

energy and toaster – heat.

Ans

wer


Return to

TextText




Discussion 6:Discussion 6:

In Chapter 6, we have learnt that power isdefined as

P =

Discuss with your classmates how to derive an

equation in terms of the electrical power (P ),

voltage (V ) and current ( I ) from their definitions.

Return to

TextText

Answer


TakenTimedtransferreEnergy

VI

t

Q

Q

E

t

E P

t Q I

Q E V

=×==∴

== ,




Thinking 5Thinking 5

In the circuit, L1, L2 and L3 are three identical lightbulbs.

1. Is the power dissipated in L3 the same as

the sum of power dissipated in L1 and L2?


L1

L3

L2

Ans

wer




Thinking 5 (Cont)Thinking 5 (Cont)


The current passing through L1 and L2 is only

half of the current passing through L3. Power

dissipated in L3 = I 2 R. The sum of power

dissipated in L1 and L2 =

power dissipated in L3.

R I

R I

22

22

+

≠=2

2 R I




Thinking 5 (Cont)Thinking 5 (Cont)

2. Arrange the light bulbs in descending order of brightness. Ans

wer

L1 and L2 have the same resistance and current

passing through. By P = I 2 R, their brightness

is the same. By P = I 2 R, L3 is brighter than L1

and L2. Therefore the order of the brightness is

L3 >

L1 =

L2.


Return to

TextText

L1

L3

L2

Section 15.5 Electrical Power

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