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Chapter 20
Electric Circuits
The Battery
• A battery consists of
chemicals, called electrolytes,
sandwiched in between 2
electrodes, or terminals, made
of different metals.
• Chemical reactions do work to
separate charge, which
creates a potential difference
between positive and negative
terminals
• A physical separator keeps the
charge from going back
through the battery.
ε and ∆V
• For an ideal battery
potential difference
between the positive and
negative terminals, ∆V,
equals the chemical work
done per unit charge.
∆V = Wch /q = ε
• ε is the emf of the battery
• Due to the internal
resistance of a real
battery, ∆V is often
slightly less than the emf.
How do we know there is a
current?
20.1 Electromotive Force and Current
The electric current is the amount of charge per unit time that passes
through a surface that is perpendicular to the motion of the charges.
q
I
t
One coulomb per second equals one ampere (A).
20.1 Electromotive Force and Current
If the charges move around the circuit in the same direction at all times,
the current is said to be direct current (dc).
If the charges move first one way and then the opposite way, the current is
said to be alternating current (ac).
EOC #3
A fax machine uses 0.110 A of current in its normal mode
of operation, but only 0.067 A in the standby mode. The
machine uses a potential difference of 120 V.
(a) In one minute (60 s!), how much more charge passes
through the machine in the normal mode than in the
standby mode?
(b) How much more energy is used?
EOC #3
A fax machine uses 0.110 A of current in its normal mode
of operation, but only 0.067 A in the standby mode. The
machine uses a potential difference of 120 V.
(a) In one minute (60 s!), how much more charge passes
through the machine in the normal mode than in the
standby mode? 2.6 C
(b) How much more energy is used? 310 J
20.1 Electromotive Force and Current
Conventional current is the hypothetical flow of positive charges (electron
holes) that would have the same effect in the circuit as the movement of
negative charges that actually does occur.
20.2 Ohm’s Law
The resistance (R) is defined as the
ratio of the voltage V applied across
a piece of material to the current I through
the material.
20.2 Ohm’s Law
OHM’S LAW
The ratio V/I is a constant, where V is the
voltage applied across a piece of mateiral
and I is the current through the material:
V
 R  constant
I
or
V  IR
SI Unit of Resistance: volt/ampere (V/A) = ohm (Ω)
20.2 Ohm’s Law
To the extent that a wire or an electrical
device offers resistance to electrical flow,
it is called a resistor.
20.2 Ohm’s Law
Example 2 A Flashlight
The filament in a light bulb is a resistor in the form
of a thin piece of wire. The wire becomes hot enough
to emit light because of the current in it. The flashlight
uses two 1.5-V batteries to provide a current of
0.40 A in the filament. Determine the resistance of
the glowing filament.
V
3.0 V
R 
 7.5 
I 0.40 A
Eoc #4
Suppose that the resistance between the walls of a
biological cell is 1.00 x 1010 Ω.
(a) What is the current when the potential difference
between the walls is 95 mV?
(b) If the current is composed of Na+ ions (q = +e), how
many such ions flow in 0.2 s?
Eoc #4
Suppose that the resistance between the walls of a
biological cell is 1.00 x 1010 Ω.
(a) What is the current when the potential difference
between the walls is 95 mV? 9.5 x 10-12 A
(b) If the current is composed of Na+ ions (q = +e), how
many such ions flow in 0.2 s? 1.19 x 107
Ratio Reasoning using Ohm’s Law
In circuit #1, the battery has twice as much potential
difference as the battery in circuit #2. However, the
current in circuit # 1 is only ½ the current in circuit #2.
Therefore R1 is________ the resistance R2
:
A. twice
D. four times
B. one half
E. 1/4
C. equal to
20.4 Electric Power
ELECTRIC POWER
•Charges acquire EPE due to the potential difference across the battery. In
the circuit, this energy is transformed into KE, as the charges move.
•The rate at which energy is delivered is Power.
P = Energy/∆t
•When there is current in a circuit as a result of a potential, the electric
power delivered to the circuit is:
P  IV
SI Unit of Power: watt (W), where one watt equals one joule per second
Many electrical devices are essentially resistors, so we use 2 formulas for power
based on resistance. From Ohm’s Law:
P  I IR   I 2 R
V2
V 
P   V 
R
R
20.4 Electric Power
Example 5 The Power and Energy Used in a
Flashlight
In the flashlight, the current is 0.40A and the voltage
is 3.0 V. Find (a) the power delivered to the bulb and
(b) the energy dissipated in the bulb in 5.5 minutes
of operation.
In this case, the resistance of the bulb is not given,
so this determines the formula for power.
20.4 Electric Power
(a)
P  IV  0.40 A3.0 V  1.2 W
(b)
E  Pt  1.2 W 330 s   4.0 102 J
Power in a circuit
Through which resistor is the most power
dissipated?
A. b
B. bd C. d
D. a
Power rating of a household applicance
• Commercial and residential electrical systems are set up
so that each individual appliance operates at a potential
difference of 120 V.
• Power Rating or Wattage is the power that the appliance
will dissipate at a potential difference of 120 V (e.g. 100
W bulb, 1000 W space heater).
• Power consumption will differ if operated at any other
voltage.
• Energy is often expressed as kilowatt-hours, for metering
purposes.
kW-h
How much energy in a kilowatt-hour?
A) 1000 Joules
B) 0.28 Joules
C) 3.6 million Joules
D) 60,000 Joules
Resistors in Series
• Series wiring means that the resistors are connected so that there is
the same current through each resistor.
• The potential difference through each resistor is :
|∆VR |= IR
• The equivalent resistor (Rs ) is made by adding up all resistors in
series in the circuit.
∆Vbat = |∆Vcircuit |= IRs
20.6 Series Wiring
•By understanding that the
potential difference across any
circuit resistor will be a voltage
DROP, while the potential
difference back through the battery
(charge escalator) will be a GAIN,
one can replace the ΔV with “V” ,
and dispense with the absolute
value signs
V1 (voltage drop across the
resistor1) = IR1
V2 (voltage drop across the
resistor2) = IR2
Vbat = Vcir = IReq
20.6 Series Wiring
Resistors in a Series Circuit
A 6.00 Ω resistor and a 3.00 Ω resistor are connected in series with
a 12.0 V battery. Assuming the battery contributes no resistance to
the circuit, find (a) the current, (b) the power dissipated in each resistor,
and (c) the total power delivered to the resistors by the battery.
20.6 Series Wiring
(a)
(b)
RS  6.00   3.00   9.00 
V 12.0 V
I

 1.33 A
RS 9.00 
P  I 2 R  1.33 A  6.00    10.6 W
2
P  I 2 R  1.33 A  3.00    5.31 W
2
(c)
P  10.6 W  5.31 W  15.9 W
Series Resistors
What is the value of R (3rd resistor in circuit)?
a. 50 Ω
b. 25Ω c. 10 Ω d. 0
20.7 Parallel Wiring
Parallel wiring means that the devices are
connected in such a way that the same
voltage is applied across each device.
When two resistors are connected in
parallel, each receives current from the
battery as if the other was not present.
Therefore the two resistors connected in
parallel draw more current than does either
resistor alone.
20.7 Parallel Wiring
20.7 Parallel Wiring
The two parallel pipe sections are equivalent to a single pipe of the
same length and same total cross sectional area.
20.7 Parallel Wiring
1
 1 
V V
1 


I  I1  I 2  
 V     V  
R1 R2
 R1 R2 
 RP 
parallel resistors
1
1
1
1
 


RP R1 R2 R3
note that for 2 parallel resistors this
equation is equal to :
Rp = (R1R2 ) / (R1 + R2)
20.7 Parallel Wiring
Example 10 Main and Remote Stereo Speakers
Most receivers allow the user to connect to “remote” speakers in addition
to the main speakers. At the instant represented in the picture, the voltage
across the speakers is 6.00 V. Determine (a) the equivalent resistance
of the two speakers, (b) the total current supplied by the receiver, (c) the
current in each speaker, and (d) the power dissipated in each speaker.
20.7 Parallel Wiring
(a)
(b)
1
1
1
3



RP 8.00  4.00  8.00 
I rms
Vrms 6.00 V


 2.25 A
RP 2.67 
RP  2.67 
20.7 Parallel Wiring
(c)
(d)
I rms 
Vrms 6.00 V

 0.750 A
R
8.00 
I rms 
P  I rms Vrms  0.750 A 6.00 V   4.50 W
P  I rms Vrms  1.50 A 6.00 V   9.00 W
Vrms 6.00 V

 1.50 A
R
4.00 
Circuits Wired Partially in Series and Partially in Parallel
Example 12
Find a) the
total current
supplied by
the the
battery and
b) the
voltage
drop
between
points A
and B.
Circuits Wired Partially in Series and Partially in Parallel
Example 12
Find a) the total
current supplied
by
the the battery
and b) the
voltage drop
between points A
and B.
Circuits Wired Partially in Series and Partially in Parallel
Example 12
a) Itot = V/Rp = 24V/240Ω = .10A
b) now go back to the 1st circuit in part c to
calculate the voltage drop across A-B:
VAB = IRAB = (.10A)(130 Ω) = 13V
Your Understanding
• What is the ratio of
the power supplied
by the battery in
parallel circuit A to
the power supplied
by the battery in
series circuit B?
a. ¼ c. 2 e. 1
b. 4
d. 2
A.
B.
Rank the bulbs
Rank the brightness
of bulbs. “Out” is
the least bright.
Parentheses show
ties
a. A, B, C
b. A , (B, C)
c. (A, B), C
http://bcs.wiley.com/hebcs/Books?action=resource&bcsId=4678&ite
d. (ABC)
mId=0470223553&resourceId=15300
Rank the bulbs
Rank the brightness
of bulbs. “Out” is
the least bright.
Parentheses show
ties
a. A, B, C
b. A , (B, C)
c. (A, B), C
http://bcs.wiley.com/hebcs/Books?action=resource&bcsId=4678&ite
d. (ABC)
mId=0470223553&resourceId=15300
Switch closed
How does A’s
brightness compare
with when the
switch was open?
a. A gets brighter
b. A gets dimmer
c. A stays the same
Use Ohm’s Law and
then look at sim
33, Ch 20, link at
right.
closed
http://bcs.wiley.com/hebcs/Books?action=resource&bcsId=4678&ite
mId=0470223553&resourceId=15300
20.11 The Measurement of Current and Voltage
An ammeter must be inserted into
a circuit so that the current passes
directly through it.
The resistance of the ammeter
changes the current through the
circuit.
The ideal ammeter has a very low
resistance
20.11 The Measurement of Current and Voltage
To measure the voltage between two
points, in a circuit, a voltmeter is connected
in parallel, between the points.
A voltmeter takes some current away from
the circuit it measures.
The ideal voltmeter has a very large
resistance so it diverts a negligible current.