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Chapter 18
Electric Currents
Ch 18
1
Simple Electric Cell
Carbon
Electrode
(+)
+
+
+
_
_
_
Zn
Electrode
(-)
Zn+ Zn+
Zn+ Zn+
Sulfuric acid
•Two dissimilar metals or carbon rods in acid
•Zn+ ions enter acid leaving terminal negative
•Electrons leave carbon making it positive
•Terminals connected to external circuit
•‘Battery’ referred to several cells originally
Ch 18
2
Electric Current
•If we connect a wire between the two terminals
electrons will flow out of the negative terminal and
toward the positive terminal we have an electric
current.
•Electric current I is defined as the net amount of charge
that flows past a given point per unit time.
Q
I 
t
1 C/s = 1A (ampere)
• An ampere is a large current and often currents are
mA (10-3 A) or A (10-6 A).
Ch 18
3
Electric Circuit
• It is necessary to have a complete circuit in order for
current to flow.
• The symbol for a battery in a circuit diagram is:
+
_
Device
+
Ch 18
4
“Conventional” current direction is opposite to actual
electron flow direction which is – to +.
Ch 18
5
Ohm’s Law
• For wires and other circuit devices, the current is
proportional to the voltage applied to its ends:
IV
• The current also depends on the amount of resistance that
the wire offers to the electrons for a given voltage V. We
define a quantity called resistance R such that
V = I R (Ohm’s Law)
• The unit of resistance is the ohm which is represented by
the Greek capital omega ().
• Thus
1 
Ch 18
V
A
6
Resistors
• A resistor is a circuit device that has a fixed resistance.
Resistor
Circuit symbol
Resistors obey Ohm’s law but not all circuit devices do (semiconductor diode, LED)
I
I
0
V
Resistor
Ch 18
0
V
non-ohmic device
7
Example: A person experiences a mild shock if a current of 80 A flows along a path
between the thumb and the index finger. The resistance of this path is 4.0x105 
when the skin is dry and 2000  when the skin is wet. Calculate the minimum
voltage difference between these two points that will produce a mild shock.
Ch 18
8
Example: A person experiences a mild shock if a current of 80 A flows along a path
between the thumb and the index finger. The resistance of this path is 4.0x105 
when the skin is dry and 2000  when the skin is wet. Calculate the minimum
voltage difference between these two points that will produce a mild shock.
V  IR
V
DRY

80 10
6
A4.0 10  
5
 32 V
V
W ET

80 10
6
A2.0 10  
3
 0.16 V
Ch 18
9
Example: Calculate the number of electrons per second that flow past
a point on the skin in the previous example.
Ch 18
10
Example: Calculate the number of electrons per second that flow past
a point on the skin in the previous example.
Q
I 
t
# electrons
 Q 


sec
 t 
 charge 


 electron 
 Q   1 electron 
 




 t   1.6 10 C 
19

 80 10
6
C

 1 electron 


s  1.6 1019 C 


electrons
 5.0 10
sec
14
Ch 18
11
Power in Electric Circuits
• Electrical circuits can transmit and consume energy.
• When a charge Q moves through a potential difference V,
the energy transferred is QV.
• Power is energy/time and thus:
energy
QV
Q   IV

P  power 

  V
time
t
 t 
and thus:
P  IV
Ch 18
12
Notes on Power
•The formula for power applies to devices that provide
power such as a battery as well as to devices that consume
or dissipate power such as resistors, light bulbs and electric
motors.
J
C  J 
P  IV      
 W  watt
s
 s C 
•For ohmic devices, the formula for power can be combined
with Ohm’s Law to give other versions:
P  IV  I ( I R)  I2 R
Ch 18
V2
V 
P  IV    V 
R
 R
13
Household Power
•Electric companies usually bill by the kilowatt-hour
(kWh.) which is the energy consumed by using 1.0 kW for
one hour.
•Thus a 100 W light bulb could burn for 10 hours and
consume 1.0 kWh.
•Electric circuits in a building are protected by a fuse or
circuit breaker which shuts down the electricity in the
circuit if the current exceeds a certain value. This prevents
the wires from heating up when carrying too much current.
Ch 18
14
Connection of
Household
Appliances
Ch 18
15
Example: A person turns on a 1500 W electric heater, a 100 W hair dryer and
then a 300 W stereo. All of these devices are connected to a single 120 V
household circuit that is connected to a 20 A circuit breaker. At what point
will the circuit breaker trip off?
Ch 18
16
Example: A person turns on a 1500 W electric heater, a 100 W hair dryer and
then a 300 W stereo. All of these devices are connected to a single 120 V
household circuit that is connected to a 20 A circuit breaker. At what point
will the circuit breaker trip off?
P  IV
P
I 
V
I
I
I
I
heater
bulb
dryer
stereo
Ch 18
1500 W

 12.5 A
120 V
100 W

 0.83 A
120 V
1000 W

 8.3 A
120 V
300 W

 2.5 A
120 V
 Circuit breaker
trips with dryer
17
Example: If electricity costs $0.1379 per kWh in Nova Scotia, calculate
the cost of operating all the appliances in the previous problem for 2.0
hours.
Ch 18
18
Example: If electricity costs $0.1379 per kWh in Nova Scotia, calculate
the cost of operating all the appliances in the previous problem for 2.0
hours.
Heater cos t  1.5 kW  2 h $0.1379 kWh 
 $ 0.41
Bulb
 $ 0.028
Dryer  $ 0.28
Stereo  $ 0.083
$ 0.80
Ch 18
19
Microscopic View of Current
•Read Example 18-13 on page 545. It studies a 5.0A current in a copper
wire that is 3.2 mm in diameter. It finds that the average “free” electron
moves with a velocity of 4.7 x 10-5 m/s in the direction of the current.
This is called the drift velocity.
•It also assumes the “free” electrons behave like an ideal gas and
calculates that the thermal velocity of the average electron is 1.2 x 105
m/s.
•Thus in a wire carrying a current, the electron motion is largely random
with a slight tendency to move in the direction of the current. Thus if
you could see electrons in a wire carrying current they would appear to
be moving randomly.
Ch 18
20
Summary of Units
Ch 18
21
Chapter 19
DC Circuits
Ch 19
22
EMF
• Devices that supply energy to an electric circuit
are referred to as a source of electromotive force.
Since this name is misleading, we just refer to
them as source of emf (symbolized by  and a
slightly different symbol in the book.)
• Sources of emf such as batteries often have
resistance which is referred to as internal
resistance.
Ch 19
23
Terminal Voltage

r
a
b
Vab
•We can treat a battery as a source of  in series with an internal
resistor r.
•When there is no current then the terminal voltage is Vab= 
•But with current I we have:
V   r
ab
•The internal resistance is small but increases with age.
Ch 19
24
Circuit Symbols
Ch 19
25
Resistors in Series - Derivation
•We want to find the single resistance Req that has the same effect as
the three resistors R1, R2, and R3.
•Note that the current I is the same throughout the circuit since
charge can’t accumulate anywhere.
•V is the voltage across the battery and also
V = V1 + V2 + V3
•Since V1 = I R1 etc., we can say
V  V1  V2  V3  IR1  IR2  IR3
V  I ( R1  R2  R3 )
The equivalent equation is V=IReq and thus
Ch 19
Req  R1  R2  R3
26
Summary - Resistors in Series
The current I is the same throughout the circuit since
charge can’t accumulate anywhere.
V  V1  V2  V3
Req  R1  R2  R3
Ch 19
27
Resistors in Parallel - Derivation
This is called a parallel circuit
•Notice V1 = V2 = V3 = V
•Since charge can’t disappear, we can say
I  I1  I 2  I 3  V1 R1  V2 R2  V3 R3
•We can combine these equations with
V = IReq to give
1
1
1
1



Req
R1
R2
R3
Ch 19
28
Summary - Resistors in Parallel
•The electric potential (voltage) is the same across each
resistor
V1 = V2 = V3
•The current through the battery splits several ways
I = I 1 + I2 + I3
•Can be 2, 3 or more resistors in parallel.
Ch 19
1
1
1
1



Req
R1
R2
R3
29
Example: A 3.0 V battery is connected to three resistors as shown. Calculate
the resistance of the equivalent circuit and the power dissipated in the
equivalent circuit. R1 = 500 Ω, R2 = 1000 Ω and R3 = 2000 Ω.
Ch 19
30
Example: A 3.0 V battery is connected to three resistors as shown. Calculate
the resistance of the equivalent circuit and the power dissipated in the
equivalent circuit. R1 = 500 Ω, R2 = 1000 Ω and R3 = 2000 Ω.
1
1
1
1



REQ R1 R2 R3
1
1
1



500  1000  2000 
REQ  286 
V  I R EQ
V
3.00 V
I

 10.5 mA
REQ
286 
P  I V  (10.5 mA) (3.00 V )
Ch 19
P  31.5 mW
31
Example: From the previous example, calculate the current and the power
dissipated in each resistor and the total power dissipated in the circuit.
Ch 19
32
Example: From the previous example, calculate the current and the power
dissipated in each resistor and the total power dissipated in the circuit.
V IR
I1 
3.00 V
V

500 
R1
 6.0 mA
P1  I1 V1  (6.0 mA) (3.00 V )  18.0 mW
V
3.00 V
I2 

 3.0 mA
R2
1000 
P2  I 2 V2  (3.0 mA) (3.00 V )  9.0 mW
3.00 V
V

 1.5 mA
I3 
2000 
R3
P3  I 3 V3  (1.5 mA) (3.00 V )  4.5 mW
I  I1  I 2  I 3  10.5 mA
P  P1  P2  P3  31.5 mW
Ch 19
33
Example: A 3.0 V battery is connected to 4 resistors as shown. Calculate the
resistance of the equivalent circuit and the current in the equivalent circuit. R1 =
500 Ω, R2 = 1000 Ω, R3 = 1000 Ω, and R4 = 2000 Ω.
Ch 19
34
Example: A 3.0 V battery is connected to 4 resistors as shown. Calculate the
resistance of the equivalent circuit and the current in the equivalent circuit. R1 =
500 Ω, R2 = 1000 Ω, R3 = 1000 Ω, and R4 = 2000 Ω.
R2 and R3 are in series
R23  R2  R3  1000   1000 
R23  2000 
R23 and R4 are in parallel
1
1
1
1
1




2000  2000 
R324 R23 R4
R234  1000 
R1 and R234 are in series
REQ  R1  R234  500   1000   1500 
Ch 19
V IR
V
3 .0 V
I

 2.0 mA
REQ
1500 
35
Ammeter
•To measure current ammeter must be connected in
series.
•Must have small internal resistance or it will
reduce current and give a faulty measurement.
Ch 19
36
Voltmeters
•To measure voltage difference, it must be connected in
parallel.
• Must have high internal resistance or it will draw too much
current which reduces voltage difference and gives a faulty
measurement.
Ch 19
37