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NASA-Threads
Electricity and Magnetism
Lesson 32: Electrical Power
Lesson 32: Electrical Power
Today’s topic will be electrical power. Before we discuss what electrical power is, let’s
review what we already know about electricity. What is the governing law of electricity?
OHM’S LAW
It is expressed in the following equation:
V = IR
The V in Ohm’s Law is the voltage of the circuit expressed in Volts (V). I represents the
current passing through the circuit; it is expressed in Ampere (A). Finally, the R is the
electrical resistance which is caused by frictional losses when the electrons move.
Resistance is expressed as Ohms (Ω).
Let’s take a closer look at the units for voltage and current. A Volt is a Joule (J) per Coulomb
(C), and an Ampere is a Joule (J) per second (s). The below relations show these statements
symbolically:
J
C
V=
A=
C
s
Keep these relations in mind as we move on to talk about power. Power (P) is work
performed per unit time. So let’s look at what the units might be for power:
Power =
Work
Joule
=
= Watt
Time second
Thus the units of electrical power are called Watts, and a Watt is a Joule (J) per second (s).
So knowing the units of power we can deduce the formula to find electrical power. Relying
on what we know about electricity, what can we do to get a J/s (Watt)? Well If a Volt is a J/C
and an Ampere is a C/s, then we can multiply them and get a Joule (J) per second (s) and in
turn the equation for finding electrical power.
P = IV
This equation expressed in terms of units is shown below. You can see how the Coulombs
cancel out and you get Watts.
C J
J
P= ∗ = =W
s C s
Using Ohm’s Law we can set up alternate forms of the power equation. If V = IR then we can
substitute IR for V and obtain the following relation for electrical power.
P = I2 R
Similarly, rearranging Ohm’s Law can show that I=V/R which gives us another expression
for electrical power.
V2
P=
R
NASA-Threads
Electricity and Magnetism
Lesson 32: Electrical Power
To summarize in one line, electrical power can be expressed in the following ways, each in
units of Watts:
V2
2
P = VI = I R =
R
You might be thinking that three different power equations is too much to remember, but if
you know Ohm’s Law and remember one of the power relations (P=IV possibly being the
easiest to remember), you can solve for any of the other forms of the electrical power
equation.
Hands On
It is important to note that these alternative forms of the electrical power equation do not
always apply to all circuits depending on the circuit configuration. For instance, if you have
Tip: When solving problems, if you know and two out of the four V,
P, I, & R, you can find the other two easily using these relationships.
a DC motor, there are no resistors so you cannot use the alternate forms to solve for power.
Taking that into account for this hands on activity, let’s only use the form P=IV to find the
electrical power of the circuits.
Using a circuit created on the breadboard of the Boe-Bot we are going to calculate the
overall electrical power associated with the circuit as well as the power through each
resistor.
1. Each student should obtain one 1kΩ, one 2kΩ, and one 4.7kΩ resistor.
2. Once materials are gathered, each student should create the following circuit on their
Boe-Bot:
1kΩ
+
Vd
d
2kΩ
4.7k
Ω
3. Once the circuit is created, turn the power on for the Boe-Bot.
4. Use the multimeter to figure out the different components in order to calculate the
electrical power using the equation P=VI. Use this table to help you organize the data.
NASA-Threads
Electricity and Magnetism
Voltage (V)
Current (A)
Lesson 32: Electrical Power
Power (W)
Overall
R1
R2
R3
For the Teachers only, the students should get the following (these are approximate):
Voltage (V)
Current (A)
Power (W)
Overall
4.97
0.0007
0.003479
R1
0.65
0.0007
0.000455
R2
1.28
0.0007
0.000896
R3
3.04
0.0007
0.002128