ANSWER KEY
QUESTIONS AND CALCULATIONS
DIRECTIONS FOR ALL ACTIVITY CALCULATIONS–SHOW
YOUR WORK AS FOLLOWS:
1. Start by writing the short formula (for example, V = I x R ).
2. Next, rewrite it substituting any values given in the problems for their abbreviated
letters.
3. Remember to include the correct unit after your number in your boxed answer.
SERIES CIRCUIT
SCHEMATIC DIAGRAM OF A
SERIES CIRCUIT
ACTIVITY 1 – OHM’S LAW
Calculate the resistance of your circuit in ohms (Ω) given that the current running through this
circuit is 0.5 amperes (0.5 A) and that you are using a 6-volt battery.
Ohm's law states that resistance (R) equals voltage (V) divided by current (I).
R=V/I
R = 6 V / 0.5 A
R = 12 ohms or 12 Ω
ACTIVITY 2 – OPEN AND CLOSED CIRCUITS
Given a 1.5-volt battery and 0.5-ohm lightbulb, calculate the current of your circuit in amps.
Ohm's law states that current (I) equals voltage (V) divided by resistance (R).
I=V/R
I = 1.5 V / 0.5 Ω
I = 3 amperes or I = 3 amps or I = 3 A
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Activities 1 and 2 - TEACHER Page 1/1
ANSWER KEY
ACTIVITY 3 – SERIES AND PARALLEL CIRCUITS
ACTIVITY 3 – SERIES CIRCUIT QUESTIONS
SERIES
CIRCUIT
b 1. What happened to the brightness of the bulbs each time you added a new bulb wired in
series?
a. They got brighter. b. They got dimmer. c. They stayed the same.
In series circuits if voltage stays the same, as the resistance increases (from adding lights), the
current must decrease. Think of the increased resistance as slowing down the electron flow
(current).
a 2. How did adding a second battery in series affect the brightness of the bulbs?
a. They got brighter. b. They got dimmer. c. They stayed the same.
In series circuits if resistance stays the same, as the voltage increases (from adding batteries), the
current must increase. Think of the increased voltage as hurrying the electron flow (current).
ACTIVITY 3 – PARALLEL CIRCUIT QUESTIONS
PARALLEL
CIRCUIT
c 1.
What happened to the brightness of the bulbs each time you added a new, similar bulb
wired in parallel?
a. They got brighter. b. They got dimmer. c. They stayed the same.
Each circuit path gets the full voltage of the battery and has the same current (rate of electron
flow) as long as each bulb has the same resistance.
c 2.
What happened to the brightness of the bulbs when you added a new, similar battery in
parallel with the first one?
a. They got brighter. b. They got dimmer. c. They stayed the same.
Adding a second equal battery also does not affect the brightness of the bulbs, although the
circuit will last longer!
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Activity 3 - TEACHER Page 1/1
ANSWER KEY
ACTIVITY 5 – POWER AND ENERGY CALCULATIONS
Calculate Your Power Use and Electric Bill for a Single Light
1. Remember that Power = Current x Voltage or P = I V , so start by finding out what these
values are. First, use a form of Ohm's law (I = V / R ) to calculate the current, using a
voltage of 120 V and a lightbulb resistance of 240 ohms.
What is the current in amps?
I=V/R
(Current = Voltage ÷ Resistance)
I = 120 V ÷ 240 Ω
I = 0.5 A
2. Power is calculated in watts, but we're charged for kilowatt·hours (energy used), so you'll have to
convert the watts to kilowatts by multiplying by this fraction, which equals 1 (since 1 kW = 1,000 W): 1
kW / 1,000 W.
a. How many kilowatts of power does this circuit use?
P=I/V
(Power = Current x Voltage)
P = 0.5 A x 120 V
P = 60 watts (60 W)
P = 60 W x 1 kW / 1,000 W
P = 0.060 kW
b. If you run the circuit for 30 minutes, how many kilowatt·hours (kWh) of energy will the circuit use?
Remember to convert the time from minutes to hours by multiplying by this fraction, which equals
1 (since 1 hour = 60 min): 1 h / 60 min.
E=Pt
(Energy = Power x time)
E = 0.060 kW x 30 min x 1 h / 60 min
E = 0.030 kWh
c.
If electricity costs $0.08 per kWh, what will the power company charge to run this circuit for 30
minutes, seven days a week, for a 30-day month?
Cost = kWh / day x days x $ / kWh
Cost = 0.030 kWh / day x 30 days x $0.08 / kWh
Cost = $0.072
If this was all the energy you used for the month, it wouldn't amount to much on your
electric bill (a little more than 7 cents a month). But let's face it, we all use more electricity
than one 60-watt light for one half hour a day!
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Activity 5 - TEACHER Page 1/1
ANSWER KEY
ACTIVITY 6 – SERIES CIRCUIT CALCULATIONS
SERIES CIRCUIT
Observation: How is current (the rate of electron flow) affected by adding
lightbulbs?
a. The amount of current stays the same when bulbs are added.
b. The amount of current gets larger as each bulb is added.
c. The amount of current gets smaller as each bulb is added.
In series circuits as loads are increased, the total resistance to the flow of electrons increases.
Thus, current slows down. All of the lights appear dimmer.
Calculations
For the series circuit pictured, the three lightbulb loads are identical. The battery source is 6
volts ( VT = 6 volts). The total resistance ( RT ) is 1.5 ohms. Do the math.
1. Calculate the total current ( IT ), which is the same as the individual currents.
IT = VT / RT
(Current = Voltage / Resistance)
IT = 6 volts / 1.5 ohms
IT = 4 amps
2. Calculate the resistance of the individual bulbs ( R1, R2, and R3 ) for this series circuit.
In this case you apply a little logic and basic math. The total resistance is the sum of the three
individual resistances, which are all the same.
RT = R1 + R2 + R3
Since the resistances are the same, you can substitute R1 for R2 and R3:
RT = R1 + R1 + R1
or
R T = 3 x R1
Dividing both sides by 3 results in:
R1 = 1.5 ohms / 3
R1 = 0.5 ohms = R2 = R3
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Activity 6 - TEACHER Page 1/1
ANSWER KEY
ACTIVITY 7 – PARALLEL CIRCUIT CALCULATIONS
Observation: How is current (the rate of electron flow) through each circuit path
affected by adding lightbulbs in parallel?
a. The amount of current stays the same when bulbs are added.
b. The amount of current gets larger as each bulb is added.
c. The amount of current gets smaller as each bulb is added.
The amount of current through each path stays the same when bulbs are added in
parallel. As long as no bulbs are added to the same path, there is no increased
resistance to the flow of electrons in that circuit path.
PARALLEL CIRCUIT
Sample Calculation
You have a dry cell with a voltage of 6 volts (VT) connected to three lightbulbs wired in parallel
to each other. The three bulbs have resistances of 400, 500, and 600 ohms (R1, R2, and R3),
respectively. This is how to complete the calculations:
1. Total resistance (RT), calculated using the reciprocal formula
1
-----------------1
1
1
--- + --- + --R1 R2 R3
RT
=
RT
=
RT
=
1
----------------------------------------------0.0025 Ω + 0.0020 Ω + 0.0017 Ω
RT
=
1
------------0.0062 Ω
RT
1
-------------------------1
1
1
------ + ------- + ------400 Ω 500 Ω 600 Ω
= 161 Ω
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Activity 7 - TEACHER Page 1/4
ANSWER KEY
Activity 7 – Parallel Circuit Calculations
Sample Calculation (continued)
You have a dry cell with a voltage of 6 volts (VT) connected to three lightbulbs wired in parallel
to each other. The three bulbs have resistances of 400, 500, and 600 ohms (R1, R2, and R3),
respectively. This is how to complete the calculations:
2. Individual currents (I1, I2, and I3)
Remember, the total voltage (VT), which is the voltage of the source dry cell, equals the voltage
for each path. In other words, VT = V1 = V2 = V3.
I1 = V1 / R1
I1 = 6 V / 400 Ω
I1 = 0.015 A (or 15 mA)
I2 = V2 / R2
I2 = 6 V / 500 Ω
I2 = 0.012 A (or 12 mA)
I3 = V3 / R3
I3 = 6 V / 600 Ω
I3 = 0.010 A (or 10 mA)
3. Total current, IT (using the sum of the individual currents)
IT = I1 + I2 + I3
IT = 0.015 A + 0.012 A + 0.010 A
IT = 0.037 A
4. Total current (using Ohm's law)
IT = VT / RT
IT = 6 V / 161 Ω
IT = 0.037 A (or 37 mA)
5. Is total current the same using both methods of calculation?
Yes, IT = 0.037 A using both methods of calculation!
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Activity 7 - TEACHER Page 2/4
ANSWER KEY
Parallel Circuit Calculations
Given a battery voltage of 3 V and lightbulbs with individual resistances of 2, 3, and 4 ohms,
solve for the following values. First, write the short formulas. Next, substitute known values for
the letter abbreviations. Remember to show the units after all the numbers. Do the math.
1. The voltages across each lightbulb, V1, V2, and V3.
Did you remember that for parallel circuits, the
voltage of each circuit branch is equal to the
source voltage (the battery), or VT = V1 = V2 =
V3?
Thus, each branch has the same voltage across it as the battery given at the beginning of
the problem, 3 volts.
V1 = V2 = V3 = 3 V
2. The total resistance, RT.
This answer requires applying the reciprocal formula to calculate RT.
RT
=
RT
=
1
---------------1 1
1
--- + --- + --R1 R 2 R 3
1
---------------1
1
1
--- + --- + --2
3
4
The lowest common denominator for the numbers 2, 3, and 4 is 12. Thus, 1/2 becomes
6/12; 1/3 becomes 4/12; and 1/4 becomes 3/12. This will give new fractions whose
denominators are all 12.
RT
=
1
------------------6
4
3
--- + --- + --12 12 12
Adding the fractions in the denominator, you should get
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Activity 7 - TEACHER Page 3/4
ANSWER KEY
RT
1
= ----13
--12
The reciprocal of a fraction is the same as exchanging the numerator and denominator.
Thus,
12
RT = ----13
Expressing 12/13 (12 divided by 13) as a decimal gives us
RT = 0.923 ohms.
RT = 0.92 ohms
(rounded to two decimals)
3. The currents in each circuit path, I1, I2, and I3.
Each individual current can be found by applying Ohm's law, I = V / R (or I = V ÷ R) .
I1 = V1 / R1
I1 = 3 V / 2 Ω
I1 = 1.50 A
I2 = V2 / R2
I2 = 3 V / 3 Ω
I2 =1.00 A
I3 = V3 / R3
I3 = 3 V / 4 Ω
I3 = 0.75 A
4. The total current, IT.
For parallel circuits the total current equals the sum of the individual currents,
IT = I1 + I2 + I3.
Adding the individual currents from Question 3,
IT = 1.50 A + 1.00 A + 0.75 A
IT = 3.25 A
Do you know how smart you are for being able to solve this problem?
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