Download document

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).
20.1 Electromotive Force and Current
Example 1 A Pocket Calculator
The current in a 3.0 V battery of a pocket calculator is 0.17 mA. In one hour
of operation, (a) how much charge flows in the circuit and (b) how much energy
does the battery deliver to the calculator circuit?
q
I
t
20.1 Electromotive Force and Current
Example 1 A Pocket Calculator
The current in a 3.0 V battery of a pocket calculator is 0.17 mA. In one hour
of operation, (a) how much charge flows in the circuit and (b) how much energy
does the battery deliver to the calculator circuit?
(a)
(b)


q  I t   0.17 103 A 3600 s   0.61 C
Energy  Charge 
Energy
 0.61 C3.0 V   1.8 J
Charge
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 thatactually 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
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:
A. twice
D. four times
B. one half
E. 1/4
C. equal to
the resistance R2
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:
A. twice
D. four times
B. one half
E. 1/4
C. equal to
the resistance R2
Using ratio reasoning (R1/R2), the answer is D
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 voltage, 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  Pt  1.2 W 330 s   4.0 102 J
Power rating of a light bulb or
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.
• Power consumption will differ if operated at any
other potential difference (i.e. 220, such as is
standard in Europe and many other countries).
Resistors in Series
• The current, I, is the same through each of the resistor.
• The potential difference through each resistor is :
∆VR = (-)IR
• The equivalent resistor is made by adding up all resistors in series in
the circuit.
∆Vbat = (-) ∆Vcircuit = (-)IRs
• Why the negative sign?
20.6 Series Wiring
• Current is a vector. Positive
direction is from positive to
negative terminal.
• ΔVbat is positive; positive
terminal has more potential than
negative.
•ΔVR is negative; it is higher at the
initial value than at the final
•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 negative
sign in Ohm’s Law.
V1 (voltage drop across the
resistor1) = IR1
V2 (voltage drop across the
resistor2) = IR2
Vbat = Vcir = IReq
20.6 Series Wiring
Example 8 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
Power in series
A 60 W light bulb and a 100 W
bulb are placed one after the
other in a circuit. The battery’s
emf is 120V. Which one glows
more brightly and why?
a. 100 W because more current flows through it
b. 100 W because of the higher potential difference
across the bulb
c. 60 W because more current flows through it
d. 60 W because of the higher potential difference
across the bulb
Power in series
A 60 W light bulb and a 100 W
bulb are placed one after the
other in a circuit. The battery’s
emf is 120V. Which one glows
more brightly and why?
a. 100 W because more current flows through it
b. 100 W because of the higher potential difference
across the bulb
c. 60 W because more current flows through it
d. 60 W because of the higher potential difference
across the bulb
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 
Power in parallel
A 60 W light bulb and a 100 W
bulb are placed in parallel in a
circuit. The battery’s emf is
120V. Which one glows more
brightly and why?
60W
a. 100 W because more current flows through it
b. 100 W because of the higher potential difference
across the bulb
c. 60 W because more current flows through it
d. 60 W because of the higher potential difference
across the bulb
100W
Power in parallel
A 60 W light bulb and a 100 W
bulb are placed in parallel in a
circuit. The battery’s emf is
120V. Which one glows more
brightly and why?
a. 100 W because more current flows through it
b. 100 W because of the higher potential difference
across the bulb
c. 60 W because more current flows through it
d. 60 W because of the higher potential difference
across the bulb
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
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
What happens to A?
With switch closed, A
is____________:
closed
a. Brighter than
b. Less bright than
c. Equal to what
it was when switch
was open.
Use Ohm’s Law
http://bcs.wiley.com/hebcs/Books?action=resource&bcsId=4678&ite
mId=0470223553&resourceId=15300
What happens to A?
With switch closed, A
is____________:
closed
a. Brighter than
b. Less bright than
c. Equal to what
it was when switch
was open.
Use Ohm’s Law
http://bcs.wiley.com/hebcs/Books?action=resource&bcsId=4678&ite
mId=0470223553&resourceId=15300
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.
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
b. 4
c. ½
d. 2
e. 1
A.
B.
20.9 Internal Resistance
Batteries and generators add some resistance to a circuit. This resistance
is called internal resistance.
The actual voltage between the terminals of a battery is known as the
terminal voltage.
Recall the ideal voltage is the EMF
20.9 Internal Resistance
Example 12 The Terminal Voltage of a Battery
The car battery has an emf of 12.0 V and an internal
resistance of 0.0100 Ω. What is the terminal voltage
when the current drawn from the battery is (a) 10.0 A
and (b) 100.0 A?
(a)
V  Ir  10.0 A0.010   0.10 V
12.0 V  0.10 V  11.9V
(b)
V  Ir  100.0 A0.010   1.0 V
12.0 V 1.0 V  11.0V
20.11 The Measurement of Current and Voltage
A dc galvanometer. The coil of
wire and pointer rotate when there
is a current in the wire.
20.11 The Measurement of Current and Voltage
An ammeter must be inserted into
a circuit so that the current passes
directly through it.
20.11 The Measurement of Current and Voltage
If a galvanometer with a full-scale
limit of 0.100 mA is to be used to
measure the current of 60.0 mA, a
shunt resistance must be used so that
the excess current of 59.9 mA can
detour around the galvanometer coil.
20.11 The Measurement of Current and Voltage
To measure the voltage between two points
in a circuit, a voltmeter is connected between
the points.
Series Resistors
What is the value of R (3rd resistor in circuit)?
a. 50 Ω
b. 25Ω c. 10 Ω d. Other
Series Resistors
What is the value of R (3rd resistor in circuit)?
a. 50 Ω
b. 25Ω c. 10 Ω d. Other
Rs = R1 + R2 + R3 = ΔV/I.