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4. In the space below, summarize the explanation of Ohm’s Law provided on this screen.
The voltmeter indicates that the potential difference across the light bulb is 10.0 volts.
The ammeter indicates that the current intensity through the light bulb is 1.0 amperes.
R = V / I 10.0 V / 1.0 A = 10 Ohm
7. Do Practice Problems 1 to 5 on page 282.
I did the practice problems in the Science Focus textbook.
1. 12 V / 2.4 V = 5 Ohm
2. 120 / A = 145 Ohm 120 / 145 = A = .827 amp
3. 1500 Ohm= x / 0.075 1500 x 0.075 = 112.5 V
4. 120 / 15 = 8 Ohm
8. Check your answers by visiting the Ohm’s Calculator web page
(http://webhome.idirect.com/~jadams/electronics/ohms_calc.htm)
This site did not work.
Voltage, Current, Resistance – pages 284 - 285
3. Draw a circuit diagram of this set up in the space below.
6. Repeat steps 4 and 5 above for the other two resistors.
Ohm’s Law Data Chart
Colour Code
of Resistor
Potential
Difference
(volts)
Blue
Light purple
Lime (green)
Orange
9.996 volts
9.9999
4.9998
3.3332
Current
(amps)
(use
ammeter)
0.10
0.0500
0.5
0.3333
Calculate
Resistance
(ohms)
(R=V/I)
99.96
199.998
9.9996
10.0006
Use the
Visualize
button to see
the resistance
of the resistor
100 ohms
200
10
20
Types of Circuits – pages 286 to 291
Circuit Review
3. In the space below, identify the main components of any electrical circuit and provide
an example.
(a) power supply - battery
(b) conductor - wire
(c) electrical device – light bulb
(d) switch – on and off
4. Click on the “But How Does This Work Button” at the bottom of the web page. You
should be taken to the page called – “What Is A Circuit?”
5. (a) Describe what happens to the electrons in the conductor when the switch is “open.”
There is no force directing electrons down the path, so they just keep orbiting the
nucleus, or they drift around randomly. Electricity can’t flow.
(b) Describe what happens to the electrons in the conductor when the switch is “closed.”
The electrons are pushed through the circuit by the power supply adding energy to the
electrons. Called “push” voltage – that’s why electrical power is measured in volts.
Series Circuits
Series circuits only have one current path.
All the moving charges travel through each component in the circuit.
When you put batteries into a flashlight from end to end, they are creating a series circuit.
When a light bulb burns out or is removed from the circuit, the circuit is broken and the
other light bulbs go out.
7. You should now be on the web page called “What is a series circuit?”
8. Follow the instructions that are printed in “gold” at the bottom of this web page.
(a) What happens when you click on the “on” button?
When the switch is closed and all parts of the circuit are connected, electrons jump from
atom to atom and electricity can flow. The bulbs light up.
(b) What happens when you click on the “light bulb” to break the circuit?
Electricity can only flow in one direction. It is interrupted when any part of a series
circuit is missing. Now power gets to the other light bulb.
(c) What happens when you click on the “battery” to break the circuit?
The flow of electricity stops and the bulbs don’t light up. It wouldn’t matter whether the
batteries were placed before or after the light bulbs in the circuit.
9. Draw a circuit diagram of a series circuit with one cell, two lamps, and a switch.
10. What happens when you remove a light bulb in a series circuit?
The circuit is broken.
11. If you add more light bulbs to a series circuit, what do you think will happen to the
brightness of each bulb?
It won’t be a bright.
12. Click on the “give a different path” button at the bottom of this web page. You should
be taken to the page called “What is a parallel circuit?”
Parallel Circuits
Parallel circuits have several current paths.
The total current is divided, with some of the moving charges travelling through each
branch, or part of the circuit.
You can think of a parallel circuit as a set of side branches connected side by side.
So parallel circuits have more than one path for electricity to follow! Even if one light
bulb burns out or is removed, the other light bulb is not affected.
13. Follow the instructions that are printed in “gold” at the bottom of this web page.
(a) What happens when you click on the “on” button?
Electrons flow through all available paths in a parallel circuit. Electricity has a choice of
which way to go.
(b) What happens when you click on the “left light bulb” in the circuit?
Electrons stop going down the path that’s been broken, but there is still another path for
the electrons to travel. This means the second bulb will stay lit.
(c) What happens when you click on the other “light bulb” in the circuit?
The electrons flow down the other path this time. You would have to remove both bulbs
to completely break the circuit and stop the flow.
14. Draw a circuit diagram of a parallel circuit with two cells, two lamps, and a switch.
15. What happens when you remove a light bulb in a parallel circuit?
Only the circuit of the removed light bulb stops – electricity still will flow in the other
circuit.
16. If you add more light bulbs to a parallel circuit, what do you think will happen to the
brightness of each bulb?
The brightness of each bulb would stay the same.
Congratulations on finishing this science944 notes/worksheet!
Topic 3 Review Answers
1. Electric resistance relates to something that stops
or slows the .ow of electric current through a circuit
or conductor.
2. The ohm is equal to R = V/I, so ohms would relate
to a volt/ampere.
3. Ohm’s Law: resistance =
Resistance (R) is measured in ohms (); Voltage,
measured in volts (V); current , measured in
amperes (A).
4. A series circuit has only one conducting pathway
for electric current, while a parallel circuit has several
alternate or branching pathways for electric
current.
5. A wire’s electric resistance is affected by the gauge
or thickness of the wire, the material from which
the wire is made, the length of the wire, and the
temperature of the wire.
6. The parallel circuit would produce a higher electric
current. Evidence for this is found in the observation
that adding additional bulbs to a parallel circuit
does not decrease their brightness.
7. Apply V = R _ I
V = 32 _ 6.8 A
V = 217.6 V
The water heater would have a voltage of 217.6 V.