2. Topics
• Flow of Charge
• Electric Current
• Voltage Sources
• Electrical Resistance
• Ohm’s Law
• Direct Current and Alternating Current
• Speed and Source of Electrons in a Circuit
• Electric Power
• Electric Circuits
9. Electric Fields in Circuits
• Point away from positive terminal, towards negative
• Channeled by conductor (wire)
• Electrons flow opposite field lines (neg. charge)
E
E
E
electrons & direction of motion
E Electric field direction
10. Batteries produce a voltage
• Typical Alkaline cells produce 1.5 Volts
– AAA cells
– AA cells
– D cells
• Putting batteries in series, voltages add
• Putting batteries in parallel, same voltage
as a single cell, but can draw
more current, lasts longer (more water in reservoir)
13. Relationship between Voltage, Current and
Resistance: Ohm’s Law
• There is a simple relationship between voltage, current and
resistance:
V is in Volts (V)
I is in Amperes, or amps (A)
R is in Ohms ()
V = I R
Ohm’s Law
V I R
14. Examples
• What is the ratio of the currents that flow in these 2 circuits?
4 V 10 Ohms 8 V 20 Ohms
15. Class Problem - Ohm’s Law (V = I·R)
• How much voltage is being supplied to a circuit that contains a 1
Ohm resistance, if the current that flows is 1.5 Amperes?
• If a 12 Volt car battery is powering headlights that draw 0.5
Amps of current, what is the total resistance in the circuit?
16. Class Problem
(How much voltage is being supplied to a circuit that contains a 1
Ohm resistance, if the current that flows is 1.5 Amperes?)
• Use the relationship between Voltage, Current and Resistance,
V = IR.
• Total resistance is 1 Ohm
• Current is 1.5 Amps
So V = IR = (1.5 Amps)(1 Ohms) = 1.5 Volts
17. Class Problem
If a 12 Volt car battery is powering headlights that draw 0.5 Amps
of current, what is the total resistance in the circuit?
• Again need V = IR
• Know I, V, need R
• Rearrange equation: R = V divided by I
= (12 Volts)/(0.5 Amps)
= 24 Ohms
18. How about multiple resistances?
• Resistances in series simply add
• Voltage across each one is DV = IR
Total resistance is 10 + 20 = 30
So current that flows must be I = V/R = 3.0 V / 30 = 0.1 A
What are the Voltages across R1 and R2?
R1=10 R2=20
V = 3.0 Volts
19. Voltage is potential, bulb presents resistance
• Battery is like reservoir of elevated water
– The higher, the bigger the potential, or voltage
• Imagine bulbs as small tubes that let water drain
– Resistance is represented by length of tube
shorter: less resistance: more current flows
longer: more resistance: less current flows
Voltage
20. Parallel Resistances are a little trickier....
• Rule for resistances in parallel:
1/Rtot = 1/R1 + 1/R2
10 Ohms 10 Ohms 5 Ohms
Can arrive at this by applying Ohm’s Law to find equal current
in each leg. To get twice the current of a single10 , could use 5 .
22. Power Dissipation
• Physical model – electrons bumping into things!
• Kinetic Energy is turned into thermal energy (heat)
• Power = Voltage Current
• P = V I
A device with a voltage drop of 1 Volt that passes a current of 1 Amp
uses 1 Watt of power.
23. Multi-bulb circuits
Rank the expected brightness of the bulbs in the circuits shown,
e.g. A>B, C=B, etc. WHY?!
A
+
_
B
C
+
_
24. Answer:
• Bulbs B and C have the same brightness, since the same
current is flowing through them both.
• Bulb A is brighter than B and C are, since there is less total
resistance in the single-bulb loop, so
A > B=C.
25. Adding Bulbs
• Where should we add bulb C in order to get A to shine more
brightly?
C
A
B
+
_
26. Answer
• The only way to get bulb A to shine more brightly is to increase
the current flowing through A.
• The only way to increase the current flowing through A is to
decrease the total resistance in the circuit loop
• Since bulbs in parallel produce more paths for the current to
take, the best (and only) choice for C is to put it in parallel with
B, as illustrated on next page
27. How to get A to shine more brightly:
A
B
+
_
C
29. Exercises
If you double the voltage across a light bulb, while keeping the
current the same, by what factor does the power consumption
increase?
If you double the current through a resistor, by what factor does the
consumption change?
30. Answers
• If you double the voltage while keeping the current fixed, the
power consumption doubles
P = IV
• If you double the current though a resistor, the power used goes
up by a factor of 4! This is because both the current and the
voltage double
P = I V = I (IR) = I2 R
31. Flashlights
• “A holder for dead batteries”
• How does a flashlight work?
– Light source?
– Power source?
– Control device?
33. What limits bulb’s lifetime?
• Heated tungsten filament drives off tungsten atoms
• Tradeoff between filament temperature and lifetime
– Eventually the filament burns out, and current no longer flows – no
more light!
• How “efficient” do you think incandescent bulbs are?
34. Efficiency
• Ratio between energy doing what you want vs. energy supplied
Efficiency = (energy emitted as visible light)/(total supplied)
For incandescent bulbs, efficiency is at most 10% percent
– Where does the rest of the energy go?
35. Decorative Lights
• Strings of lights used to decorate contain many bulbs
• In some light sets, a single bulb going out can shut off the entire
set
– How do you think sets like this are wired up?
• How might you design a light set that still works, even if a bulb
goes out?
Series combo: one goesno light
37. Lights in your Car
• The car has a battery as part of its electrical system
– (as well as a generator, voltage regulator, etc..)
• Lights in a car include:
– Interior light, turn on when the door opens
– Turn signals
– Brake lights
– Headlights (high and low beam)
• The illustrations that follow are by no means the only way to
accomplish these tasks!
40. Class Problem
• The simple series circuit consists of three identical lamps powered by battery. When a wire is
connected between points a and b,
a) What happens to the brightness of lamp 3?
b) Does current in the circuit increase, decrease or remain the same?
c) What happens to the brightness of lamps 1 and 2?
d) Does the voltage drop across lamps 1 and 2 increase, decrease, or remain the same?
e) Is the power dissipated by the circuit increased, decreased, or does it remain the same?
41. Class Problem
• a) Lamp 3 is short-circuited. It no longer glows because no current passes through it.
b) The current in the circuit increases. Why? Because the circuit resistance is reduced.
Whereas charge was made to flow through three lamps before, now it flows through only two
lamps. So more energy is now given to each lamp.
c) Lamps 1 and 2 glow brighter because of the increased current through them.
d) The voltage drop across lamps 1 and 2 is greater. Whereas voltage supplied by the battery
was previously divided between three lamps, it is now divided only between two lamps. So
more energy is now given to each lamp.
e) The power output of the two-lamp circuit is greater because of the greater current. This
means more light will be emitted by the two lamps in series than from the three lamps in
series. Three lamps connected in parallel, however, put out more light. Lamps are most often
connected in parallel.
