Download Power in the resistor R 1

8. Power in electric circuits
I
R
W  QV
V
W QV
P

 IV
t
t
Q
I
t
V  IR
2
V
P  IV  I R 
R
P  1W  1V  A  1J / C 1C / s  1J / s
2
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 battery
b) Find the electric power dissipated in each resistor
R1
R1  5
I
V
R2  10
+
V  12V
R2
I
Req  R1  R2  5   10   15 
V 2 12 V 
P

 9.6 W
Req
15
V 12 V
I

 0.8 A
Req 15 
2
Power in the resistor R1:
PR1  I R1 R1  (0.8 A) 2 (5 )  3.2 W
Power in the resistor R2:
PR 2  I R 2 R2  (0.8 A) 2 (10 )  6.4 W
The total power:
Ptot  PR1  PR 2  3.2 W  6.4 W  9.6 W
2
2
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 battery
b) Find the electric power dissipated in each resistor
I
R1  5
R2  10
V
V  12V
+
I
1
1
1
1
1
2
1
3







Rtot R1 R2 5  10  10  10  10 
V 2 12 V 
PB 

 43.2 W
R 10 / 3
R1
I1
R2
10 
Rtotal 
 3.33 
3
2
Power in the resistor R1:
I2
V 2 (12 V ) 2
PR1 

 28.8 W
R1
5
Power in the resistor R2:
V 2 (12 V ) 2
PR 2 

 14.4 W
R2
10 
The total power:
Ptotal  PR1  PR 2  28.8 W  14.4 W  43.2 W
V2
P
 I2R
R
Example:
V  110V
P1  100W
2
V 2 110V 
R2 

 242
P2
50W
2
R1
V
P2  50W
V 2 110V 
R1 

 121
P1
100W
R2
R  R1  R2  363
V
R2
R1
2
V 110V
I 
 0.30 A
R 363
~
2
P1  I 2 R 1  0.30 A 121  11W
~
2
P2  I 2 R 2  0.30 A 242  22W
 P2 
~
 50W  100W
  100W 
P1  P1 
 11W
 
9
 150W 
 P1  P2 
2
2
 P1 
4
~
 100W 
  50W 
P2  P2 

50
W
 22W

9
 150W 
 P1  P2 
2
Fuses and Circuit breakers
To 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:
Ptotal  1000W  1500W  2500W
Total current (when used simultaneously):
I 
Ptotal 2500 W

 20.8 A
V
120 V
The fuse must to keep a current larger than 20.8 A
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.
L
R
A
V 2 V 2A
P

R
L
2
V A
P1 
L
V 2A
P2 
2 2L
V 2A
P3 
0.9 L
P2  P1  P3
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.
1 kWh = 1000 J/s x 3 600 s = 3 600 000 J
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 kWh.
How much did running the hair dryer cost?
 10 cents
cost  (1 kW)  
 kW  h
  10 min 

  1.66 cents
  60 min/ h 
9. EMF and terminal voltage
Wext

Q
Definition:
I
An ideal emf device: r internal =0
Disconnected battery: R=∞
Vab    Ir
P  Vab I  I  I 2 r
a
P  VI  I
 V
E
An real emf device: r internal =0
I 0
  VR  Vr  IR  Ir
I
R
E
I
r
b
[ ]  1V
  V  IR
R
E
Terminal voltage:
Units:

Rr
Potential in Closed Circuit
-
b
+
r 2
I 2A
a
R4
V 0
V
12 V
Ir  4 V
8V
IR  8 V
0
I

Rr

12V
 2A
4  2
Vab    Ir  12V  (2A)( 2)  8V
VR  IR  2 A  4  8V
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.
Vab

12V
I

 0.5A
R  r 22  2
Vab    Ir  12V  (0.5A)( 2 )  11V
VR  IR  0.5 A  22  11V
V
+
a
r  2
I
A
-
b
  12V
I
R  22