Transcript Chapter 20 Electric Circuits
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 terminals
potential difference
between positive and negative • 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 = W ch /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.
I
q
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 or
V
IR I
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.
R
V I
3 .
0 V 0.40
A 7 .
5
Eoc #4
Suppose that the resistance between the walls of a biological cell is 1.00 x 10 10 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
many such ions flow in 0.2 s?
= +
e
), how
Eoc #4
Suppose that the resistance between the walls of a biological cell is 1.00 x 10 10 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
many such ions flow in 0.2 s? 1.19 x 10 7 = +
e
), how
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 R 1 is________ the resistance R 2 A. twice D. four times B. one half C. equal to E. 1/4
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
I
2
R P
V R
V
V
2
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 10 2 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 : |∆V R |= IR • The
equivalent
resistor (R s ) is made by adding up all resistors in series in the circuit.
∆V bat = |∆V circuit |= IR s
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
V 1 (voltage drop across the resistor1) = IR 1 V 2 (voltage drop across the resistor2) = IR 2 V bat = V cir = IR eq
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)
R S
6 .
00 3 .
00 9 .
00
I
V R S
12 .
0 V 9 .
00 1 .
33 A
P
I
2
R
1 .
33 A
P
I
2
R
1 .
33 A 6 .
00 10 .
6 W 3 .
00 5 .
31 W (c)
P
10 .
6 W 5 .
31 W 15 .
9 W
Series Resistors
What is the value of R (3 rd resistor in circuit)?
a. 50 Ω b. 25 Ω 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
I
I
1
I
2
V R
1
V R
2
V
1
R
1 1
R
2
V
1
R P
parallel resistors
1
R P
1
R
1 1
R
2 1
R
3
note that for 2 parallel resistors this equation is equal to : R p = (R 1 R 2 ) / (R 1 + R 2 )
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)
1
R P
1 8 .
00 1 4 .
00 3 8 .
00
(b)
I
rms
V
rms
R P
6 .
00 V 2 .
67 2 .
25 A
R P
2 .
67
20.7 Parallel Wiring (c)
I
rms
V
rms
R
6 .
00 V 8 .
00 0 .
750 A
I
rms
V
rms
R
6 .
00 V 4 .
00 1 .
50 A
(d)
P
I
rms
V
rms 0 .
750 A 6 .
00 V 4 .
50 W
P
I
rms
V
rms 1 .
50 A 6 .
00 V 9 .
00 W
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) I tot = V/R p = 24V/240 Ω = .10A
b) now go back to the 1st circuit in part c to calculate the voltage drop across A-B: V AB = IR AB = (.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 d. (ABC)
http://bcs.wiley.com/he bcs/Books?action=resource&bcsId=4678&ite 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 d. (ABC)
http://bcs.wiley.com/he bcs/Books?action=resource&bcsId=4678&ite 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/he bcs/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.