Current elctricity student packet key

Assignment #1: Circuit Building
1.) Which of the following circuits will light the light bulb? Explain your choice:
answers vary
1 2 3 4 5
(two wires)
6 7 8 9 10
2.) Try hooking up the light bulb, wires, and battery in each of the scenarios above.
Which one/s light? ____4, 6, 9_____________________
3.) Was your prediction correct? ________
4.) Based on your results, what feature(s) must a successful circuit have? Explain.
A closed loop and a power supply (battery)

Assignment #2: Basic Circuits Internet Activity
Open you web browser and go to:
www.explorelearning.com/index.cfm?method=cResource.dspView&ResourceID=29
Click on the Intro to E & M activity.
Click on the Moving Charge button. Go through the activity and answer the following
questions.
1) What happens inside a wire when an electromotive force is applied? (2 nd screen)
Electrons will flow through the wire (electrons jump from one nucleus to the next).
2) What is current? (3 rd screen)
The charge that passes over a certain amount of time
q charge
Current I
t time
3) How is a car driving down a flat open road like an electron moving through a wire with low
resistance? (4 th screen)
Easily being able to move down the road is similar to electron flow not being hindered
when going through a wire.
4) What happens when electrons get “joggled” around trying to move through a substance with
moderate resistance? (4 th screen)
They heat up the substance and prohibit a good electron flow.
5) What is needed to make an electron move away from its nucleus? (5 th screen)
ENERGY!
6) What kind of energy is stored in a battery? (7 th screen)
Chemical energy (voltaic cell – battery!)

Click on the button for Circuits.
7) What four things make up a common circuit? (2 nd screen)
1 – battery (power supply)
2 – wire (to connect 2 ends of battery)
3 – switch (on/off)
4 – device (uses electricity to perform a function)
8) Before closing the switch of the circuit, move your mouse onto the battery, wire and light bulb.
What is happening in each when the switch is open? (4 th screen)
- Battery: no reaction
- Wire: electrons move about the closest nuclei
- Light Bulb: no electrons move through the bulb, therefore there is no light
9) Now close the switch of the circuit. Move your mouse onto the battery, wire and light bulb. What
is happening in each when the switch is closed? (5 th screen)
- Battery: reaction occurs – allows a positive and negative end of the battery
- Wire: force supplied by battery causes electron to move towards positive side of
battery
- Light Bulb: electrons pass through the bulb. The filament has high resistance, causing
material to get hot and give off light.
10) In our labs we use resistors that cannot change. But in the next activity, set the battery to a
certain voltage and use the slider to change the resistance. What happens to the current as
resistance decreases? Explain.
If voltage is kept constant and resistance is decreased, the current increases.
(less resistance = less hindrance = greater flow of electrons)

Assignment #3: What is Current?
We have looked at charges being transferred through friction, polarization, conduction, grounding,
and induction. Now we are going to look at not only how much but how fast charge moves.
1. What is the unit for Charge? __Coulomb (C)__
2. What is a rate? a measured quantity per unit of time
3. If I want to know the rate at which charges are moving, what mathematical operation should I use?
__division __
4. If Current is defined as the rate at which charges flow, what is the equation for current?
Current
charge
time
q
I
t
5. Check your reference table for an equation involving current and charge. Was your prediction
correct? If not, fix the equations above!
***The units for Current are Amperes (Amps), abbreviated with a capital letter, A. The SI unit for
Current is Coulombs per seconds or C/s
6. Current is often compared to the amount of water flowing in a faucet or stream. If you have two
streams flowing the other how do you know which one is moving faster or slower?
More water flowing through a section in a faster amount of time
7. Using the previous analogy, how can you tell if the current is faster or slower?
More charge in a smaller amount of time
8. 20 Coulombs of charge pass a given point in a conductor in 4.0 seconds. What is the current in
the conductor?
q
20C
I = 5 A (or 5 C/s)
t 4s
9. In a circuit with a battery supplying 12 volts, the current is 10 amperes.
(a) How much charge flows through the circuit in 2.0 minutes?
I
q
t
q
10 A ∆q = 1200 C
120 s
(b) How many electrons were transferred during these 2.0 minutes?
1e
1.610
1 e = 1.6 x 10 -19 21
C 1200C 7.510
e
19
C

Assignment #4: What is Ohm’s Law?
Question: What are voltage, current and resistance? How do they work?
An Analogy:
Current
Resistance
Voltage
Putting It All Together…
a.) The gumballs in this demonstration represent the flow of _electrons_ through the pipe.
b.) The height the pipe is raised above the desk represents potential difference, or
_voltage_.
c.) The nails placed in the pipe represent _resistance_ and slow down the flow of the
gumballs.
d.) The higher we raise the pipe, the _faster_ the gumballs flow; the more nails we place in
the pipe, the __slower_ the gumballs flow.
In summary, the current (rate of gumball flow per second) through the pipe depends on
the voltage (height the end of the pipe is raised) and the resistance (number of nails in
the pipe).
Ohm’s Law is the mathematical formula used to explain this relationship.

Ohm’s law states that voltage is equal to current times resistance.
Find a version of this equation on p. 4 of your Reference Tables & write it below!
Questions: Show all work!
R = resistance and has the unit of ohms (Ω)
V = potential difference, or voltage, and has units of volts (V)
I = current and has the unit of amperes (or amps, A)
1. We can rearrange this equation to solve for V or I.
a.) Solve for potential difference (V).
V = IR
b.) Solve for current (I) in Ohm’s Law.
I = V
R
R
V
I
2. What voltage is used by an electrical appliance that draws 0.4 amps of current and has a
resistance of 3 ohms?
I = 0.4 A
R = 3 Ω
V = IR = (0.4 A)(3 Ω) = 1.2 V
3. A light bulb uses 240 volts of electricity and draws a current of 2 A. What is its resistance?
V = 240 V
I = 2 A
R = V / I = (240 V) / (2 A) = 120 Ω
4. Calculate the current used by a television that runs on 240 volts and has a resistance of 600 ohms.
V = 240 V
R = 600 Ω
I = V / R = (240 V) / (600 Ω) = 0.4 A
5. A 20 ohm resistor has 40 Coulombs of charge passing through it in 5 s. What is the potential
difference across this resistor?
R = 20 Ω I = ∆q / t = (40 C) / (5 s) = 8 A V = IR = (8 A) (20 Ω) = 160 V
q = 40 C
t = 5 s

6. A wire carries a current of 2.0-amperes. How many electrons pass a given point in this wire in 1.0
second?
I = 2 A I = ∆q / t ∆q = It = (2 A)(1 s) = 2 C
t = 1 s
1 e = 1.6 x 10 -19 C
1e
19
2C 1.2510
e
19
1.610
C
7. Complete the following table: V = IR
Current - I
(amps)
Resistance - R
(ohms)
Voltage - V
(volts)
(a) 0.1 600 60
(b) 0.25 1,000 250
(c) 0.4 600 240
(d) 0.5 480 240
(e) 0.2 1,000 200

Assignment #5: Ohm’s Law Calculations
1. What current flows through a circuit in which the resistance is 20 ohms and the potential
difference is 90 volts?
R = 20 Ω
V = 90 V
V
R
I
90V
20 I = 4.5 A
I
2. An ammeter and voltmeter in a circuit respectively read: 2.48 A and 88.4 V. What is the
resistance of the circuit?
I = 2.48 A
V = 88.4 V
R
V
I
88.4V
= 35.6 Ω
2.48A
3. A battery supplies a potential difference of 120 volts to a circuit in which the resistance is 15
ohms. What current flows in this circuit?
R = 15 Ω
V = 120 V
V
R
I
120V
15 I = 8 A
I
4. What potential difference is needed to produce a current of 345 milliamps in a circuit having a
resistance of 8.74 ohms?
I = 345 mA = 0.345 A
R = 8.74 Ω
V = IR = (0.345 A)(8.74 Ω) = 3.0 V
5. When a wire is connected to a 9.0 V battery, the current is 0.020 A. What is the resistance of the
wire?
I = 0.02 A
V = 9.0 V
R
V
I
9V
= 450 Ω
0.02A

6. A potential difference of 12 Volts is applied across a circuit having a 4 ohm resistance. What is
the current in the circuit?
R = 4 Ω
V = 12 V
V
R
I
12V
4 I = 3 A
I
For each of the following circuits use the rules and equations for electric circuits to determine the
voltage (V), current (I), and resistance (R).
7.
V = 50 volts
R
V
I
R1 = ?
Circuit V I R
Total (T) 50 V 2 A 25 Ω
8.
V = 200 volts
R1 = 20 ohms
Circuit V I R
Total (T) 50 V 10 A 20 Ω
9.
R1 = 35
Circuit V I R
Total (T) 140 V 4 A 35 Ω

Assignment #6: Ohm’s Law Problems
V
__1___1. A graph of voltage (y-axis) versus current for a given resistor would produce a
1) straight line 2) hyperbola 3) parabola 4) slope of zero
__3___2. The rate of flow of electrons in a circuit is called
R = V / I
I
1) resistance 2) voltage 3) current 4) power
__3___3. A volt is a
V = W / q = J / C
1) joule/sec 2) coulomb/sec 3) joule/coulomb 4) coulomb/joule
__2___4. An ampere is a
I = Δq / t = C / s
1) joule/sec 2) coulomb/sec 3) joule/coulomb 4) coulomb/joule
__2___5. A 10 ohm resistor is connected to a power supply producing 5 amperes of current
in that resistor. The voltage drop across the resistor must be
R = V / I
1) 2 volts 2) 50 volts 3) 0.5 volts 4) 15 volts V = IR
V = (5A)(10Ω)
__1___6. If the voltage provided to a resistor increases, the current through that resistor
1) increases 2) decreases 3) remains the same I = V / R
I & V have direct relationship!
__2___7. As the resistance of a circuit increases, the current through that circuit
1) increases 2) decreases 3) remains the same R = V / I
R & I have inverse relationship!
__3___8. Ten coulombs of charge pass a given point within a circuit in 2 seconds. The
current in the circuit is
I = Δq / t = 10C / 2s
1) 20 amps 2) 12 amps 3) 5 amps 4) 0.2 amps
__3___9. The opposition to the flow of electrons in a circuit is defined as electrical
1) current 2) power 3) resistance 4) voltage drop
__1___10. How much work is done in moving 100 coulombs of charge through a potential
difference of 10 volts?
V = W / q 10 V = W / 100 C
1) 1000 joules 2) 100 joule 3) 10 joules 4) 1 joule

Assignment #7: Conductors vs. Insulators
In order to understand resistance, we first have to look at the types of materials with which
we’re trying to allow a current to flow.
Directions: Obtain a conductivity tester from your teacher. Test the conductivity of 15 objects found
throughout your classroom. You may want to consult p. 508 in your textbook to help in answering the
questions that follow.
Data:
Object
Conductivity
1.) Define electrical conductivity.
The ability of charge to flow through a material.
2.) Out of the materials you tested, which would you rank as being conductors? Why?
Varies – because it had the greatest amount of flashes with the conductivity
tester.
3.) What is a conductor?
A material that easily allows a flow of electrons through it.
4.) Out of the materials you tested, which would you rank as being non-conductors? Why?
Varies – because it had the lowest amount of flashes with the conductivity tester.
5.) Materials that are poor conductors are often called “insulators.” What is an insulator?
A material that does NOT easily allow a flow of electrons through it.
6.) Why are conductors important? Answers vary

7.) Why are insulators important? Answers vary
8.) Most cords and electrical appliances have some sort of plastic surrounding the metal
components. What is the purpose of the metal components? What is the purpose of the
plastic?
Metal components allow the flow of electrons (electricity), while the plastic
prevents people from being shocked.
View your teacher’s computer simulations for the following questions:
9.) According to the atomic model, what do you observe as the difference between conductors
and insulators? What about similarities?
Difference – electrons move much more with conductors
Similarities – both are positive ions with electrons on the outside. Only the e − can move.
10.) A simple circuit is created using both an insulator and a conductor. Which material allows
the bulb to be lit? Explain why.
The conductor allows the bulb to be lit because it enables electrons to flow through it,
while the insulator does not.
NOTE: What do you notice about the direction of the flow of electrons in the circuit?
Electrons flow away from the negative pole of the battery, and toward the positive pole
of the battery.

Assignment #8: Resistance and Resistivity
L
A
1.) A 0.686-meter-long wire has a cross-sectional area of 8.23 × 10 –6 meter 2 and a resistance of
0.125 ohm at 20° Celsius. This wire could be made of
R
R = ρL / A
a) aluminum b) nichrome c) copper d) tungsten
ρ = RA / L = (0.125Ω)(8.23x10 -6 m 2 ) / (0.686 m)
2.) A copper wire at 20°C has a length of 10 m and a cross-sectional area of 1.00 × 10 –3 m 2 . The wire
is stretched, becomes longer and thinner, and returns to 20°C.
a.) What effect does this stretching have on the wire’s resistance?
R = ρL / A Stretching causes an increase in L and a decrease in A.
Therefore, resistance will increase.
b.) What effect does this stretching have on the wire’s resistivity?
NO EFFECT! ρ depends only on the material. As long as the copper is still copper,
resistivity isn’t affected.
3.) A 6.0 m long copper wire has a cross-sectional area of 3.0 mm 2 . What is the resistance of the
wire?
A = 3 mm 2 x (1 m) 2 = 3 x 10 -6 m 2
(1000mm) 2
R = ρL / A = (1.72 x 10 -8 Ω·m)(6 m) / (3 x 10 -6 m 2 ) = 0.0344 Ω
4.) What is the resistance of aluminum wire that is 4.0 m long and has a radius of 0.25 mm?
r = 0.25 mm = 0.00025 m
A = πr 2 = π((0.00025m) 2 ) = 1.96 x 10 -7 m 2
R = ρL / A = (2.82 x 10 -8 Ω·m)(4 m) / (1.96 x 10 -7 m 2 ) = 0.58 Ω
5.) A 10.0-meter length of copper wire is at 20°C. The radius of the wire is 1.0 × 10 –3 meter.
a.) Determine the cross-sectional area of the wire.
A = πr 2 = π((1.0 x 10 -3 m) 2 ) = 3.14 x 10 -6 m 2
b.) Calculate the resistance of the wire.
R = ρL / A = (1.72 x 10 -8 Ω·m)(10 m) / (3.14 x 10 -6 m 2 ) = 0.055 Ω

Assignment #9A: Series and Parallel Circuits Internet Activity
Open your web browser to www.explorelearning.com
Sign in using the Username: rh student 4 and Password: 4
Search “Circuits” Run the gizmo Circuits
Series Circuits
1. Build the series circuit with three light bulbs and one battery shown at
the right. To do this, drag the necessary parts from the COMPONENTS
pane to the CIRCUIT BOARD.
2. Click on the battery in your circuit, and then vary the voltage across the circuit by dragging the
Selected battery voltage (volts) slider. What happens to the brightness of the lights when the
voltage is increased?
They get brighter
3. When it is decreased?
They get dimmer
4. Click Show current. Set the Selected battery voltage to 1 volt and click Show current. The
current is represented by moving arrows. Is there more or less current leaving the battery than
flowing through ONE bulb?
NO – the current stays the same!
5. How does the flow of current change when the voltage is increased?
Current increases
6. When it is decreased?
Current decreases
7. Does the current ever reach an intersection in the wires?
No
8. Increase the voltage to 30 Volts and observe the brightness of the bulbs. Add three more bulbs to
your circuit in series. How does the brightness of the bulbs change as more are added?
They get dimmer

Parallel Circuits
1. Build a parallel circuit as shown with three light bulbs and one
battery. To do this, drag the necessary parts from the
COMPONENTS pane to the CIRCUIT BOARD.
2. Click on the battery in your circuit, and then vary the voltage
across the circuit by dragging the Selected battery voltage (volts)
slider. What happens to the brightness of the lights when the voltage
is increased?
They get brighter
3. When it is decreased?
They get dimmer
4. Set the Selected battery voltage to 1 volt and click Show current. The current is represented by
moving arrows. Is there more or less current leaving the battery than flowing through ONE bulb?
More current leaves battery than flowing through the bulb
5. What happens to the current when it reaches an intersection in the wires?
It increases when joining back to “main line” from a parallel line (decreases when going
from parallel line back to “main line”)
6. Does an equal amount of current flow through each bulb?
Yes
7. Add one more bulb to the circuit in parallel. What happens to the brightness of the bulbs as more
are added?
No noticeable difference
8. Keeping the voltage set to 1 volt, what happens to the current leaving the battery as more bulbs
are added?
Current increases
9. Explain three differences between series and parallel circuits:
For series circuits, adding resistors (lights) doesn’t change the current. For parallel,
total current differs from the current going through each resistor.
Adding bulbs in parallel doesn’t affect brightness, whereas in series the bulbs get
dimmer.
Placement: Series have resistors within same line, whereas parallel has them “in
parallel” to one another

Assignment #9B: Series and Parallel Snap Circuits Activity
Part 1: Series Circuits
Use your supplies to build a circuit with single bulb connected to the battery as shown in the
diagram.
2. Observe the brightness of the one bulb.
3. Draw a circuit showing TWO bulbs connected in a series circuit.
4. Use your supplies to build this circuit. What happens to the brightness of the two bulbs
when compared to one bulb alone?
5. Gently unscrew one of the bulbs. What happens to the other bulb? Why?
6. Add another battery pack to your circuit to increase the voltage. What happens to the
brightness of the bulbs?
7. What do you think is happening to the current flowing in the circuit as voltage increases?

Part 2: Parallel Circuits
1. Connect the same two bulbs in a parallel circuit as shown.
Explain how is this circuit different from a series circuit.
Add another bulb to the circuit in parallel. Draw a picture of this parallel circuit in the space below.
What happens to the brightness of the two bulbs in a parallel circuit when compared to three
bulbs?
Gently unscrew one of the bulbs. What happens to the two other bulbs? Why?
Conclusion:
Three light bulbs are connected in series and the filament in one of the bulbs burns out or breaks!
How will this affect the other two bulbs? Explain your answer.
Three light bulbs are connected in parallel and the filament in one of the bulbs burns out or
breaks! How will this affect the other two bulbs? Explain your answer.
Do you think the electrical circuits in your house are wired in series or in parallel? Explain.

Series Circuit Example:
Assignment #10: Series Circuits Problems
2.0-ohm (R 1 ), 3.0-ohm (R 2 ), and 1.0-ohm (R 3 ) resistors
are connected in series to a voltage source of 12. volts
(V B ) with a current of 2.0 amperes (I), as seen in the
diagram to the right.
Find the voltage drop at each resistor.
Equations: I = I 1 = I 2 = I 3
V = V 1 + V 2 + V 3
R eq = R 1 + R 2 + R 3
R = V / I (V = IR)
Circuit V I R
Total (T) 12. V 2 A 6.0 Ω
1 4.0 V 2 A 2.0 Ω
2 6.0 V 2 A 3.0 Ω
3 2.0 V 2 A 1.0 Ω
___A___ 1.) As resistors are added to a circuit in series, the total resistance of the circuit
a) increases b) decreases c) remains the same R eq = R 1 + R 2 + …
___A___ 2.) As resistors are added to a circuit in series, the voltage across each resistor
a) increases b) decreases c) remains the same V = V 1 + V 2 + …
___C___ 3.) As resistors are added to a circuit in series, the current in the circuit
a) increases b) decreases c) remains the same
I = I 1 = I 2 = …
Base your answers to questions 4-8 on a circuit consisting of two 10 ohm light bulbs in series
with a 40 volt power supply.
___B___ 4.) The total resistance in the circuit is
R eq = R 1 + R 2 = 10Ω + 10Ω
a) 10 ohms b) 20 ohms c) 5 ohms d) 100 ohms
___D___ 5.) The total current in the circuit is I = V / R = (40 V) / (20 Ω)
a) 800 Amps b) 20 Amps c) 4 Amps d) 2 Amps
___B___ 6.) The voltage drop across each light bulb is
V = IR = (2A)(10Ω)
a) 40 Volts b) 20 Volts c) 10 Volts d) 5 Volts

___B___ 7.) If a third bulb is added to a circuit, the brightness of the original two bulbs will
a) increase b) decrease c) remain the same
___C___ 8.) If one of the light bulbs should burn out, the other two bulbs will
a) dim b) become brighter c) go out
*there is no longer
a complete circuit!
Base your answers to questions 9 – 11 on the diagram below and the following information.
A 4.0 ohm resistor (R 1 ), a 6.0 ohm resistor (R 2 ), and an unmarked resistor (R 3 ) are connected to a
24-volt source as shown. The ammeter reads a current of 2.0 amps.
9.) Determine the voltage drop across the 4.0 ohm resistor. Show all work.
V 1 = IR 1 = (2 A)(4 ohm) = 8 V
10.) Determine the value of R 3 . Show all work.
R eq = V T = 24 V = 12 ohm
I T 2 A
R eq = R 1 + R 2 + R 3
12 ohm = 4 ohm + 6 ohm + R 3 R 3 = 2 ohm
11.) If the 4.0 ohm resistor (R 1 ) were removed from the circuit and the remaining circuit elements
were reattached in series, what would be the reading on the ammeter? Show all work.
R eq = R 2 + R 3 = 6 ohm + 2 ohm = 8 ohm
I T = V T
R eq
= 24 V = 3 A
8 ohm

Assignment # 11: OUTSMARTING THE REGENTS EXAM
Although the NYS Regents Physics Exam is a difficult test, many of the questions you will face
require NO knowledge of physics whatsoever!! All you need to answer the questions correctly is a
Reference Table, eyeballs and just a little bit of thinking.
Here is a typical problem taken from the January 2008 Regents Exam:
A circuit consists of a 10.0-ohm resistor, a 15.0-ohm resistor, and a 20.0-ohm resistor connected
in parallel across a 9.00-volt battery. What is the equivalent resistance of this circuit?
STEP 1: Read the problem twice. Pick out the key words: circuit, parallel, and equivalent resistance.
STEP 2: Look at your Reference Tables. Again, using only
your eyeballs, look on the Electricity page in the Circuits
section and find the equations for Parallel Circuits.
It doesn’t even matter if you know what a parallel circuit is!
Just find the equation you need and solve it. Ka-ching!
1/Req = 1/R1 + 1/R2 + 1/R3
1/Req = 1/10 + 1/15 + 1/20
1/Req = 0.1 + 0.067 + 0.05
1/Req = 0.2167
Req = 4.62 ohms
STEP 3: Think carefully. Predict what the other choices will be! The Regents WILL try to trick you
and offer choices that may seem right if you don’t think.
So if the correct answer is choice (a) 4.62 ohms, choice (b) will be 0.2167 ohms for the
students who forget to cross multiply in the last step.
Choice (c) is going to be 45 ohms, for the students who did not check the Reference Tables
and simply added the resistors to find the equivalent resistance. (This would be correct for a
series circuit!)
Choice (d) will probably be 9.00 volts, for students who don’t realize voltage has nothing to
do with the question.

Now you try one. Here’s another problem from a Regents Exam.
A circuit consists of 5.0-ohm resistor, a 10.0-ohm resistor, and a 15.0-ohm
resistor connected in series across a 9.00 volt battery. What is the current
flowing through the 10.0-ohm resistor?
What is the correct answer? Show all work.
(a) __0.3 A______
R eq = R 1 + R 2 + R 3 = 5 Ω + 10 Ω + 15 Ω = 30 Ω
R = V / I 30 Ω = 9 V / I I = 9 V / 30 Ω = 0.3 A
I = I 1 = I 2 = I 3 = 0.3 A
What will choice (b) be? Explain.
(b) __3.33 A_____
3.33 A for those who do 30 Ω / 9 V when solving for I.
What will choice (c) be? Explain.
(c) __0.9 A______
0.9 A for those who thought V = V 1 = V 2 = V 3 and do I = V / R = 9 V / 10 Ω
What will choice (d) be? Explain.
(d) __30 A______
30 A for those who just add up all of the resistors.
Here’s another:
A circuit consists of three 10.0-ohm resistors connected in parallel across
an 18.0 volt battery. What is the equivalent resistance of this circuit?
What is the correct answer? Show all work.
(a) __3.33 Ω____
1/Req = 1/R1 + 1/R2 + 1/R3
1/Req = 1/10 + 1/10 + 1/10
1/R eq = 3/10
R eq = 10/3 = 3.33 Ω
What will choice (b) be? Explain.
(b) __0.3 Ω_____
0.3 Ω for those that forget to cross multiply the last step
What will choice (c) be? Explain.
(c) __ 30 Ω_____
30 Ω for those that use the series circuit rules
What will choice (d) be? Explain.
(d) ___18 Ω____
18 Ω for those that think it has something to do with the total voltage

Assignment #12A: Parallel Circuits Problems
Practice filling in the following VIR tables.
10 Ω
Circuit V I R
1 1.5 V 0.3 A 5 Ω
2 1.5 V 0.15 A 10 Ω
Total (T) 1.5 V 0.45 A 3.333 Ω
4 A
2 Ω
4 Ω
Circuit V I R
1 3.2 V 1.6 A 2 Ω
2 3.2 V 1.6 A 2 Ω
3 3.2 V 0.8 A 4 Ω
Total (T) 3.2 V 4 A 0.8 Ω

Assignment #12B: Parallel Circuits Problems
For each of the following circuits use the rules and equations for circuits to determine the voltage (V),
current ( I ) and resistance (R) for each resistor and the total circuit by making a VIR chart.
1.)
V = 200 volts
R1 = 20
Circuit V I R
1 200 V 10 A 20 Ω
2 200 V 5 A 40 Ω
T 200 V 15 A 13.33 Ω
R2 = 40
V = ?
2.)
i = 5 a
R1 = 30
Circuit V I R
T 100 V 5 A 20 Ω
1 100 V 3.33 A 30 Ω
2 100 V 1.67 A 60 Ω
R2 = 60
3.)
V = 60 volts
i = 2 a
R1
R2
Circuit V I R
1 60 V 0.5 A 120 Ω
2 60 V 1.5 A 40 Ω
T 60 V 2 A 30 Ω
i = 1.5 a

Base your answers to questions 4 – 6 on the diagram below and the following information.
A 20.0 ohm resistor (R 1 ), a 30.0 ohm resistor (R 3 ), and an unmarked resistor (R 2 ) are connected to a
24-volt source as shown. The ammeter reads a current of 3.0 amps.
4.) Determine the voltage drop across the 20.0 ohm resistor. Show all work.
V = V 1 = V 2 = V 3 = 24 V
5.) Determine the value of R 2 . Show all work.
R eq = V / I = 24 V / 3 A = 8 Ω
1 / R eq = 1/R 1 + 1/R 2 + 1/R 3
(1 / 8 Ω) = (1 / 20 Ω) + (1 / R 2 ) + (1 / 30 Ω)
0.125 = 0.0833 + (1 / R 2 )
1 / R 2 = 0.0417
R 2 = 24 Ω
6.) If the 20.0 ohm resistor were removed from the circuit and the remaining circuit elements were
reattached in parallel, what would be the reading on the ammeter? Show all work.
I 2 = V 2 / R 2 = 24 V / 24 Ω = 1 A
I 3 = V 3 / R 3 = 24 V / 30 Ω = 0.8 A
I = I 2 + I 3 = 1 A + 0.8 A = 1.8 A

Answer the following questions:
__B____ 1.) As resistors are added to a circuit in parallel, the total resistance of the circuit
a) increases b) decreases c) remains the same
__C____ 2.) As resistors are added to a circuit in parallel, the voltage across each resistor
a) increases b) decreases c) remains the same
__A____ 3.) As resistors are added to a circuit in parallel, the current in the circuit R T decreases,
a) increases b) decreases c) remains the same therefore
I T decreases
Base your answers to questions 4-9 on a circuit consisting of two 10 ohm light bulbs in
parallel with a 40 volt power supply.
4.) Using the circuit symbols found in the Reference Tables for Physical Setting/Physics,
draw a diagram of these components in a parallel circuit.
A
Ammeter goes in series
with the rest of the circuit,
before any junctions
V
Voltmeter goes in parallel
with the rest of the circuit
to measure how much
voltage (potential) is added
by the battery
5.) In the diagram above, draw in a voltmeter and ammeter in the correct positions to
measure the total amount of voltage and current, respectively, that are being delivered
to the circuit.
1/R eq = 1/10Ω + 1/10Ω = 2/10Ω R eq = 5Ω
__A____ 6.) The total current in the circuit is
I = V / R = (40V) / (5Ω)
a) 8 Amps b) 20 Amps c) 4 Amps d) 2 Amps
__A____ 7.) The voltage drop across each light bulb is V = V 1 = V 2
a) 40 Volts b) 20 Volts c) 10 Volts d) 5 Volts
__C____ 8.) If a third bulb is added to a circuit in parallel, the brightness of the original two
bulbs will
a) increase b) decrease c) remain the same
__D____ 9.) If one of the light bulbs should burn out, the other two bulbs will
a) dim b) become brighter c) go out d) remain the same

Assignment #12C: Parallel Circuits Problems
1. A battery of 24 volts is
connected to a circuit as
shown at the right.
6
4
Determine each of the following:
a) The equivalent resistance of the 4 ohm and 8 ohm
resistors.
8
R s = R 1 + R 2 = 4 Ω + 8 Ω = 12 Ω
b) The equivalent resistance of the 4 ohm and 8 ohm resistors plus the 12 ohm resistor.
1/R p = 1/R 12 + 1/R s = 1/12 Ω + 1/12 Ω = 2/12 Ω = 6 Ω
c) The total resistance of the circuit
R s = R 6 + R p = 6 Ω + 6 Ω = 12 Ω
d) The current in the 6 ohm resistor.
I = V/R eq = 24 V / 12 Ω = 2 A
e) The rate at which energy is dissipated (POWER) in the 12 ohm resistor.
Power (Topic #5) = I 2 R = (1 A) 2 (12 Ω) = 12 Watts

2. Three resistors are arranged in a circuit as shown. The battery has an unknown but constant
voltage. Determine the potential difference across resistor R1.
.
Circuit V I R
T 4.8 V 0.4 A 12 Ω
1 1.2 V 0.3 A 4 Ω
2 1.2 V 0.1 A 12 Ω
3 3.6 V 0.4 A 9 Ω
*R 1 & R 2 are in parallel!
Find R T for these, then R eq !
**PARALLEL RULES FOR R 1 & R 2 !**
3. Determine the resistance of R 1
i = 4 a
V = 100 volts
R1 = 10
R 2 = ?
R1 = ?
Circuit V I R
T 100 V 4 A 25 Ω
1 15 V 1.5 A 10 Ω
2 85 V 1.5 A 56.67 Ω
3 100 V 2.5 A 40 Ω
*R 1 & R 2 are in series!
Find R T for these, then R eq !
R3 = 40
**SERIES RULES FOR R 1 & R 2 !**

Assignment #13: Electric Power Problems
1.) How much time is required for an operating 100 watt light bulb to dissipate 10 Joules of electrical
energy?
W = Pt
a) 1 sec b) 0.1 sec c) 10 sec d) 1000 sec
2.) While operating at 120 Volts, an electric toaster has a resistance of 15 ohms. The power used by
the toaster is P = V 2 / R
a) 8 W b) 120 W c) 960 W d) 1800 W
3.) An electric clothes dryer consumes 6 x 10 6 J of energy while operating at 220 volts for 30
minutes. During operation, the dryer draws a current of W = VIt
a) 10 Amps b) 15 Amps c) 20 Amps d) 25 Amps
4.) What is the approximate amount of electrical energy needed to operate a 1600 Watt toaster for
60 seconds? W = Pt
a) 27 J b) 1500 J c) 1700 J d) 96,000 J
5.) Electric power is measured in
a) Volts b) Joules/sec c) Coulombs/sec d) Joules/ Coulomb
6.) Electric power can be calculated by
a) IV b) V 2 /R c) I 2 R d) all of the above
7.) Electric energy is measured in
a) Joules b) Watts c) Coulombs d) Ohms
8.) Electric energy can be calculated by
a) Pt b) VIt c) I 2 Rt d) all of the above
9.) A potential difference of 60 V is applied across a 15 ohm resistor. What is the power dissipated
in the resistor? P = V 2 / R
a) 240 J b) 240 W c) 2.4 W d) 12 W
10.) A current of 0.4 amps is measured in a 150 ohm resistor. How much energy is expended by the
resistor in 30 seconds? W = I 2 Rt
a) 72 J b) 240 J c) 600 J d) 720 J

Base your answers to the following questions on the diagram below. Show all work!
1.) What is the current reading in ammeter A1?
I = I 1 + I 2
10 A = I 1 + 4 A
I 1 = 6 A
2.) Determine the power developed in resistor R2 alone.
P 2 = I 2 2 R 2 = (4 A) 2 (30 Ω) = 480 W
3.) Determine the power developed in resistor R1 alone.
P 1 = I 1 2 R 1 = (6 A) 2 (20 Ω) = 720 W
4.) How much electric energy is used by the 30 ohm resistor in 10 minutes?
t = 10 min = 600 s
W 2 = P 2 t = (480 W)(600 s) = 288,000 J
5.) What would happen to the current in ammeter A 2 if resistor R 1 was removed? Explain.
It wouldn’t change! Since A 2 depends solely on the voltage and R 2 (Ohm’s Law), its
value is unaffected by any changes to R 1 .
6.) Given the following information for a radio:
P = 12 W
V = 120 V
Used for 30 min
Cost of energy is 7 cents ( 0.07 $) per kilowatt-hour
Calculate:
a) the current through the radio:
P = VI I = P / V = (12 W) / (120 V) = 0.1 A
b) the amount of energy used (in kilowatt-hours)
P = 12 W = 0.012 kW
W = Pt = (0.012 kW)(0.5 hr) = 0.006 kW-hr
c) the cost of your listening pleasure
0.006 kW-hr x $0.07 = $0.00042
1 kW-hr

Assignment #14: Circuit Simulation – Review
1. Open up Internet Explorer and navigate to:
http://phet.colorado.edu/en/simulation/circuit-construction-kit-dc
2. Click on “Run Now!”
3. Using four light bulbs and four batteries as well as a switch and as many wires as you need, build a
circuit similar to used for Holiday lights, where one bulb burning out causes all bulbs to fail. Draw
the schematic below (show resistors for light bulbs)
4. What type of circuit is this? ______series________ (series/parallel)
5. Add the Ammeter to the circuit. How much current is flowing through each of the light bulbs (L)?
L 1 : ___0.9 A___ L 2 : __0.9 A ____ L 3 : __0.9 A ____ L 4 : __0.9 A ___
6. What do you notice about the currents across each light bulb? ___the same__________
7. Apply the Voltmeter across all four batteries. How much voltage is being applied to the entire
circuit? _____36 V________
8. Apply the Voltmeter across each of the light bulbs. How much Voltage is dropped across each?
L 1 : __9 V____ L 2 : ___9 V____ L 3 : ___9 V___ L 4 : __9 V___
9. How does your answer for question #7 compare to the sum of your answers for question #8?
The same! (36 V = 9 V + 9 V + 9 V + 9 V)
10. How much power is being dissipated by each of the light bulbs? Show all work:
L 1 : __8.1 W__
P = VI = (9 V)(0.9 A) = 8.1 W
L 2 : __8.1 W__
L 3 : __8.1 W__
L 4 : __8.1 W__

11. Using four light bulbs and four batteries, and as many wires and resistors as required, design a
circuit such that when one bulb goes out, the other three bulbs continue to function. Draw the
schematic below (show resistors for light bulbs)
12. What type of circuit is this? ____parallel______ (series/parallel)
13. Add the Ammeter to the beginning of the circuit (next to the batteries). How much current is being
delivered into the entire circuit? _____14.4 A_______
14. Add the Ammeter to the circuit. How much current is flowing through each of the light bulbs (L)?
L 1 : __3.6 A___ L 2 : __3.6 A___ L 3 : __3.6 A___ L 4 : __3.6 A___
15. How does your answer for question #13 compare to the sum of your answers for question #14?
The same! (14.4 A = 3.6 A + 3.6 A + 3.6 A + 3.6 A)
16. Apply the Voltmeter across all four batteries. How much voltage is being applied to the entire
circuit? _____36 V________
17. Apply the Voltmeter across each of the light bulbs. How much Voltage is dropped across each?
L 1 : ___36 V_____ L 2 : ___36 V_____ L 3 : ___36 V_____ L 4 : ___36 V_____
18. What do you notice about the voltage drop across each light bulb? _____the same!________
19. How does your answer to question #16 compare to you answer to question #17? ___the same!_
20. How much power is being dissipated by each of the light bulbs? Show all work:
L 1 : __130 W____
P = VI = (36 V)(3.6 A) = 130 W
L 2 : __130 W____
L 3 : __130 W____
L 4 : __130 W____

21. Using four light bulbs and four batteries, design a circuit such that each light bulb has a different
brightness. Draw the circuit schematic below. Indicate which bulb is the brightest, the dimmest, etc.
50 Ω 10 Ω 76 Ω 25 Ω
#3 brightest dimmest #2
(#1) (#4)
22. You have thirteen batteries in a single circuit and one light bulb. There are no resistors or
anything to calm the flow of energy (amps). What will the circuit ultimately do? Build it and see.
Explain below.
Combust! Too much electrical energy, therefore extra is converted to
heat/thermal energy
23. Now experiment with different resistors and/or light bulbs to see how many ohms it takes to calm
the thirteen batteries it takes. Record what it took and explain how certain variables effect flow of
energy.
It takes about 12 ohms of resistance. More resistance = less current = less energy
24. Build a smaller circuit (2 or 3 batteries with one light bulb is big enough). Click on the “Advanced”
Settings in the bottom right corner. Describe how the brightness of the bulb changes as the “Wire
Resistivity” is changed from “Almost none” to “lots”.
More resistivity = dimmer light

Assignment #15: Graphing & Current Electricity
1.) The graph to the right shows the relationship between the
potential difference across a metallic conductor and the electric
current through the conductor at constant temperature T 1 .
Which graph best represents the relationship between potential
difference and current for the same conductor maintained at a
higher constant temperature, T2?
R = V / I = slope of line. Higher T = higher R = higher slope
2.) A long copper wire was connected to a voltage source. The voltage was varied and the
current through the wire measured, while temperature was held constant. The collected data are
represented by the graph below.
Using the graph, the resistance of the copper wire is approximately
(1) 8.0 Ω (2) 25 Ω (3) 30 Ω (4) 2.5 Ω

3.) A variable resistor was connected to a battery. As the resistance was adjusted, the current
and power in the circuit were determined. The data are recorded in the table below.
Using the information in the data table, construct a line
graph on the grid provided, following the directions below.
a.) Plot the data points for power versus current.
Allow 1 credit for accurately plotting the data points (±0.3 grid space) for power vs. current.
b.) Draw the best-fit line.
Allow 1 credit for drawing a straight best-fit line. If one or more points are plotted incorrectly in
question 3a, but a best-fit line is drawn, allow this credit.
c.) Using your graph, determine the power delivered to the circuit at a current of 3.5 amperes.
10.5 W ± 0.3 W
Allow credit for an answer that is consistent with the student’s answer to question 3b.
d.) Calculate the slope of the graph. [Show all calculations, including the equation and substitution
with units.]
Note: The slope may be determined by direct substitution of
data points ONLY IF the data values are on the best-fit line.
e.) What is the physical significance of the slope of the graph?
Voltage and/or Potential Difference

4.) An experiment was performed using various lengths of a conductor of uniform crosssectional area.
The resistance of each length was measured and the data recorded in the table below.
Using the information in the data table, construct a graph on the grid provided following the
directions below.
a.) Mark an appropriate scale on the axis labeled “Length (m).”
b.) Plot the data points for resistance versus length.
c.) Draw the best-fit line.
d.) Calculate the slope of the best-fit line. [Show all work, including the equation and substitution
with units.]

1.) 2
2.) 2
3.) 1
4.) 1
5.) 4
6.) 2
7.) 4
8.) 4
9.) 1
10.) 1
11.) 3
12.) 2
13.) 3
14.) 4
15.)
Assignment #16: Regents Current Electricity Problems
16.)

17.)
18.)
19.)

20.)
21.)
22.)