- New Quantities
- Formulas
- General Problem Solving Strategy
- Example Problem #1
- Example Problem #2
- Sample Problems
- Solutions to Sample Problems
- Go back to Tutorial Page

Here's the quantities you can know:

- V Voltage
- I Current
- R Resistance
- P Power
- Q Charge
- t Time

And here are the formulas that we have so far:

- V = IR

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

So now we have all the formulas we need for solving electric power problems:

P=IV

P=I^2R

P=(V^2)/R

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- Read the problem.
- Go through the problem and figure out what is given or implied
- Find any formula that will allow you to calculate
- 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

Make a list, and identify the quantities you know.

anything that you don't know, and apply it.

previous two steps until you are done.

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

Here is what you start with

- P=1000 W
- V=27V
- R=?

putting in numbers:

R=(V^2)/P=27^2/1000=.729 Ohms

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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

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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.

(29.2 W)

(.6173 ohms)

(4.32 * 10^6 joules)

($756)

($4.54 per week) Go back to: Table of Contents

(9.3 A less)

(1.69 *10 ^23 electrons)

(2.2 W)

(.33 W)

Solution:

P=QV/t

P= 350 C * 20-V / (4 min * 60 secs/min) = 29.2 W

Go to: Problem Formulas Table
of Contents

Solution:

P=I^2 R

50 W= (9 A)^2 R

R= .6173 ohms

Go to: Problem Formulas Table
of Contents

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

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

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

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

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

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

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