8. Power in electric circuits
I
R
W  QV
V
W QV
P

 IV
t
t
Units:
V  IR
Q tI
2
V
P  IV  I R 
R
2
P  1W  1V  A  1J / C 1C / s  1J / s
1
R1
Example: Two resistors 5  and 10  are connected
in series. The battery has voltage of 12 V.
a) Find the electric power delivered by the battery.
b) Find the electric power dissipated in each resistor.
R1  5
V  12V
V
R2
Req  R1  R2  5   10   15 
R2  10
I
I
I
V 12 V

 0.8 A
Req 15 
V 2 12 V 
PB 

 9.6 W
Req
15
2
Power delivered by the battery:
Power in the resistor R1:
PR1  I R21R1  (0.8 A) 2 (5 )  3.2 W
Power in the resistor R2:
PR 2  I R2 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
Example: Two resistors 5  and 10  are connected
in parallel. The battery has voltage of 12 V.
a) Find the electric power delivered by the battery.
b) Find the electric power dissipated in each resistor.
I
I1
V
R1  5
R1
I2
R2
I
R2  10
V  12V
1
1 1
1
1
3
 




Req R1 R2 5  10  10 
Req 
10 
 3.33 
3
12 V   43.2 W
V2
PB 

Req 10 / 3
2
Power delivered by the battery:
Power in the resistor R1:
Power in the resistor R2:
The total power:
V 2 (12 V ) 2
PR1 

 28.8 W
R1
5
V 2 (12 V ) 2
PR 2 

 14.4 W
R2
10 
Ptotal  PR1  PR 2  28.8 W  14.4 W  43.2 W
3
Example (2 light bulbs labeled 100W and 50W):
V 2 120V 
R1 

 144
P1
100W
V2
P
 I2R
R
P1  100W
P2  50W
2
V 2 120V 
R2 

 288
P2
50W
2
V  120V
R1
V
What power is delivered if these
resistors are connected in series?
R2
R  R1  R2  432
V 120V

 0.278 A
R 432
~
2
P1  I 2 R 1  0.278 A 144  11W
~
2
P2  I 2 R 2  0.278 A 242  22W
I
V
R1
R2
4
Example (2 light bulbs labeled 100W and 50W – analytical solution):
V2
R1 
;
P1
V2
R2 
P2
P1  P2
R  R1  R2  V
P1 P2
2
I
V
P1 P2

R V P1  P2 
2
 P2 
~
 50W  100W
  100W 
P1  I 2 R 1  P1 
 11W
 
9
 150W 
 P1  P2 
2
2
 P1 
4
~
 100W 
2
  50W 
P2  I R 2  P2 
  50W  22W
9
 150W 
 P1  P2 
2
5
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.
R
L
A
V 2 V 2A

P
L
R
V 2A
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. 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   10 min 
  1.66 cents
cost  (1 kW)  
  
 kW  h   60 min h 
6
8a. 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
7
9. EMF and terminal voltage

Definition:
Wext

Q
An ideal emf device: r internal =0

Disconnected battery: R=∞
Units:
I
R
  V  IR
P  VI  I
 V

I 0
 V
An real emf device: r internal = 0
R
a
Terminal voltage:
Vab    Ir  IR
I
I

r
b
[ ]  1V
R
 Vr  IR  Ir

Rr
P  Vab I  I  I 2 r  I 2 R
8
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

12V
I

 2A
R  r 4  2
Vab    Ir  12V  (2A)( 2)  8V
VR  IR  2 A  4  8V
9
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.

12V
I

 0.5A
R  r 22  2
Vab    Ir  12V  (0.5A)( 2 )  11V
VR  IR  0.5 A  22  11V
Vab
V
+
a
r  2
I
A
-
  12V
b
I
R  22
10