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8. Power in electric circuits
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Example: Two resistors, R1 = 5 , R2 = 10 , are connected in series. The battery has voltage of V = 12 V. a) Find the electric power delivered by the batteryb) Find the electric power dissipated in each resistor
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211 Power in the resistor R1:
WARIP RR 4.6)10()8.0( 22
222 Power in the resistor R2:
The total power: WWWPPP RRtot 6.94.62.321
VV
R
R
12
10
5
2
1
Example: Two resistors, R1 = 5 , R2 = 10 , are connected in parallel. The battery has voltage of V = 12 V. a) Find the electric power delivered by the batteryb) Find the electric power dissipated in each resistor
Power in the resistor R1:
Power in the resistor R2:
The total power:
VV
R
R
12
10
5
2
1
+1R
V
2R
I
I
1I 2I
10
3
10
1
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WWWPPP RRtotal 2.434.148.2821
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Example:
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WP
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50
100
110
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~
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WWW
PP
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WW
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150W
10050W
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22
21
122
22
21
211
A V
W
V
P I total 8.20
120
2500
Fuses and Circuit breakersTo prevent some damage in the electric circuit we use electric fuses. It will blow up due to a large heat if the current flowing through it will be larger than a certain critical value (10 A, 20 A, 100 A, etc.). Fuses are one-use items – if they blow, the fuse is destroyed and must be replaced.Circuit breakers, which are now much more common, are switches that will open if the current is too high; they can then be reset.
Example: Consider an electric hair dryer and electric iron which have 1000 W and 1500 W power when running on 120 V
Total power:
Total current (when used simultaneously):
WWWPtotal 250015001000
The fuse must to keep a current larger than 20.8 A
Example: How much energy does a typical appliance use? Let’s look at 1000 W hair dryer. We use it for 10 minutes, electricity costs ~10 cents per kWh.
How much did running the hair dryer cost?
cents 66.1min/60
min 10
hkW
cents 10 kW)(1 cost
h
Example: Three wires, of the same diameter, are connected in turn between two points maintained at a constant potential difference. Their resistivities and length are: 1) ρ and L; 2) 2ρ and 2L; 3) 0.9ρ and L. Rank the wires according to the rate at which energy is transferred to thermal energy within them.
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AV
R
VP
A
LR
22
L
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2
1 L
AVP
22
2
2
L
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9.0
2
3
312 PPP
1 kWh = 1000 J/s x 3 600 s = 3 600 000 J
What you pay for on your electric bill is not power, but energy – the power consumption multiplied by the time (E = Pt). We have been measuring energy in joules, but the electric company measures it in kilowatt-hours, kWh. Electric meter monitors power consumption.
9. EMF and terminal voltage
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WextDefinition:
ER
IAn ideal emf device: r internal =0
E
An real emf device: r internal =0
Disconnected battery: R=∞
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rRI
r
a
b
IrVab Terminal voltage:
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IVIP
rIIIVP ab2
Potential in Closed Circuit
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IrVab
ab
Example: An electric bulb with resistance of 22 Ω is connected to the battery with emf of 12 V and internal resistance 2 Ω. Find current, terminal voltage, and potential difference across the bulb.
V11225.0
V11)2)(A5.0(V12
R
AIRV
IrVab
A5.0222
V12
rRI
-+
A
V
I I
abV
2r
22R
V12
ba