Section 15.5 Electrical Power
Transcript of Section 15.5 Electrical Power
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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
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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
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Electrical power
15.5 Electrical power (SB p. 67)
Go to
Discussion 6Discussion 6
takenTime
dtransferreenergyElectrical =
t
E P =
unit: watt (W)
Electrical
power
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220 V 220 V
each secondconsumes 100 J
each secondconsumes 60 J
Electrical power
15.5 Electrical power (SB p. 67)
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(1) Useful equations of electrical power
Electrical power
15.5 Electrical power (SB p. 68)
t
E P =
( )QV E t
QV ==
=t
QV
VI P =∴
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(2) Useful equations of electrical power
Electrical power
15.5 Electrical power (SB p. 68)
VI P =
( )IR V I IR =×= R I P
2 =∴
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(3) Useful equations of electrical power
Electrical power
15.5 Electrical power (SB p. 68)
VI P =
=
=
R
V I
R
V V
R V P
2
=∴
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VI P =
R I P 2 =
R
V P
2
=
or
or
Electrical power
15.5 Electrical power (SB p. 68)
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To section 15.6
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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
15.5 Electrical power (SB p. 66)
Return to
TextText
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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.
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TextText
Answer
15.5 Electrical power (SB p. 67)
TakenTimedtransferreEnergy
VI
t
Q
Q
E
t
E P
t Q I
Q E V
=×==∴
== ,
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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?
15.5 Electrical power (SB p. 68)
L1
L3
L2
Ans
wer
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Thinking 5 (Cont)Thinking 5 (Cont)
15.5 Electrical power (SB p. 68)
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
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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.
15.5 Electrical power (SB p. 68)
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TextText
L1
L3
L2