Electrical Power
by Mark Landmine, Joey Trachea, and Tyler Meatloaf
Table of Contents
Here's the quantities you can know:
-
V Voltage
-
I Current
-
R Resistance
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P Power
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Q Charge
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t Time
These quantities are defined and explained on other pages, except for electric
power.
New Quantities
And here are the formulas that we have so far:
-
V = IR
Defining Electric Power
Electric power is the convesion of electric energy into light or thermal
energy. In a lightbulb, if the current is large enough the kenitic
energy of the electrons will be transfered to the atom that it collides
with. This causes an increase in temperature of the atoms of the
wire, and eventually the emission of light.
Power defined is the rate at which energy is transformed.
So we have two new formulas:
P=(QV)/t
Since we have previously found that Q/t is current, (I) the formula
can be simplifed into P=IV. Using Ohm's Law,(V=IR) this formula can
be written in two other ways:
P=I(IR)=I^2R
P=(V/R)V=(V^2)/R
Formulas
So now we have all the formulas we need for solving electric
power problems:
P=IV
P=I^2R
P=(V^2)/R
Go back to: Table of Contents
General Problem Solving Strategy:
-
Read the problem.
-
Go through the problem and figure out what is given or implied
Make a list, and identify the quantities you know.
-
Find any formula that will allow you to calculate
anything that you don't know, and apply it.
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Add what you just found in the last step to your list of knowns.
-
Check to see if you have found the answer. If not, repeat the
previous two steps until you are done.
Go back to: Table of Contents
Example problem 1
Calculate the resistance of a 1000 Watt microwave designed for 27 Volts.
Here is what you start with
So, if you look at the formulas, P=(V^2)/R seems most logical.
putting in numbers:
R=(V^2)/P=27^2/1000=.729 Ohms
Go back to: Table of Contents
Example problem 2
A pool heater running on 150V, and draws a current of 30A. How
much energy does it use per week if it is used for 9 hours a day.
Here's what we know:
-
If you look at the formulas: P=IV is easiest to use sence we arn't
given R.
Plugging numbers into the formula:
P=150*30=4500W
4500W=4500J/s, *60*60*9*7=1.0206 *10^9
Go back to: Table of Contents
Sample Problems
The answers to each problem follow it in parentheses. They also link to a solution to
the problem. Try the problem, check your answer, and go to the solution if you do not
understand.
1.
In a hairdryer, a charge of 350 C moves through a potential defference
of 20-V for 4 minutes. How much power is transformed?
(29.2 W)
2.
Calculate the resistance of an electric blanket that draws 9 A to transform
50 W of energy.
(.6173 ohms)
3.
How many joules of energy are necessary to run a telecision for 30 munutes
if it is hooked up to a 120-V line and has a resistance of 6 ohms?
(4.32 * 10^6 joules)
4.
How much does it cost to run a refridgerator all day, for 6 months (30
days/month) if it requires 3.5 kW and the charge is $0.05 per kWh?
($756)
5.
How much does it cost per week to operate a stereo for 3 hours per day
if it draws 16 A on a 150-V line and the charge is $0.09 per kWh?
($4.54 per week) Go back to: Table
of Contents
6.
A generator produces 140 kW of power and delivers the electricity at 6,000-V.
How much less current will be received if the voltage is upped to 10,000-V?
(9.3 A less)
7.
A microwave transforms 50 W of energy when it is linked to a 20-V source.
How many electrons are leaving it in 3 hours?
(1.69 *10 ^23 electrons)
8.
5.0 * 10^22 electrons flow through a wire with a resistance of 16 ohms
in 6 hours. How much power was used?
(2.2 W)
9.
900 C flow through a flashlight with 3 ohms of resistance and is on for
45 minutes. How much power was used?
(.33 W)
Solutions to Sample Problems
1.
In a hairdryer, a charge of 350 C moves through a potential defference
of 20-V for 4 minutes. How much power is transformed?
Solution:
P=QV/t
P= 350 C * 20-V / (4 min * 60 secs/min) = 29.2 W
Go to: Problem Formulas Table
of Contents
2.
Calculate the resistance of an electric blanket that draws 9 A to transform
50 W of energy.
Solution:
P=I^2 R
50 W= (9 A)^2 R
R= .6173 ohms
Go to: Problem Formulas Table
of Contents
3.
How many joules of energy are necessary to run a telecision for 30 munutes
if it is hooked up to a 120-V line and has a resistance of 6 ohms?
Solution:
P=V^2/R
P= (120-V)^2 / 6 ohms
P= 2400 W = 2400 J/s
2400 J/s * (30 min * 60 secs/min) = 4320000 = 4.32 * 10^6 joules
Go to: Problem Formulas Table
of Contents
4.
How much does it cost to run a refridgerator all day, for 6 months (30
days/month) if it requires 3.5 kW and the charge is $0.05 per kWh?
Solution:
(3.5 kW) * (6 months * 30 days/month * 24 hours/day) = 15120
hours
15120 hours * $0.05 = $756
Go to: Problem Formulas Table
of Contents
5.
How much does it cost per week to operate a stereo for 3 hours per day
if it draws 16 A on a 150-V line and the charge is $0.09 per kWh?
Solution:
P=IV
P= 16 A * 150-V = 2400 W = 2.4 kW
3 hours/day * 7 days/week = 21 hours/week
(2.4 kW) * (21 hours/week) * ($0.09 /kWh) = $4.54 per week
Go to: Problem Formulas Table
of Contents
6.
A generator produces 140 kW of power and delivers the electricity at 6,000-V.
How much less current will be received if the voltage is upped to 10,000-V?
Solution:
P=IV
I= 140,000 W / 6000-V = 23.3 A
I= 140,000 W / 10,000-V = 14 A
23.3 A - 14 A = 9.3 A less
Go to: Problem Formulas Table
of Contents
7.
A microwave transforms 50 W of energy when it is linked to a 20-V source.
How many electrons are leaving it in 3 hours?
Solution:
P=IV
50 W = I * 20-V
I = 2.5 A
I=dQ/dt
2.5 A=dQ/ (3 hours * 3600 secs/hour)
dQ = 27,000 C
27,000 C / (1.6 * 10^-19 C/electron) = 1.69 *10 ^23 electrons
Go to: Problem Formulas Table
of Contents
8.
5.0 * 10^22 electrons flow through a wire with a resistance of 16 ohms
in 6 hours. How much power was used?
Solution:
dQ = (5.0 * 10^22 electrons) * (1.6 * 10^-19 C/electron) = 8000
C
I=dQ/dt
I= 8000 C / ( 6 hours * 3600 secs/hour) = .37 A
P=I^2 R
P= (.37 A)^2 * 16 ohms =2.2 W Go to: ProblemFormulas
Table of Contents
9.
900 C flow through a flashlight with 3 ohms of resistance and is on for
45 minutes. How much power was used?
Solution:
I=dQ/dt
I = 900 C / (45 min * 60 sec/min) = .33 A
P= I^2 R
P= (.33 A)^2 * 3 ohms = .33 W
Go to: Problem Formulas Table
of Contents