# Electrical Power

by Mark Landmine, Joey Trachea, and Tyler Meatloaf

Here's the quantities you can know:
• V Voltage
• I  Current
• R Resistance
• P Power
• Q Charge
• 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:
1. 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

## General Problem Solving Strategy:

2. Go through the problem and figure out what is given or implied

3. Make a list, and identify the quantities you know.
4. Find any formula that will allow you to calculate

5. anything that you don't know, and apply it.
6. Add what you just found in the last step to your list of knowns.
7. Check to see if you have found the answer. If not, repeat the

8. previous two steps until you are done.

## Example problem 1

Calculate the resistance of a 1000 Watt microwave designed for 27 Volts.

• P=1000 W
• V=27V
• R=?
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

## 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:

• V=150V
•  I=30A
• P=?
• 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

## 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?

## 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)

## 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

## 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

## 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

## 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

## 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

## 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