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

Here's the quantities you can know:

- V Voltage
- I Current
- R Resistance
- Q Charge

I = (delta Q)/(delta t).

A = C/s

For instance, if a wire is hooked up to a battery that creates a current of 6.7 A, and has been operating for 10 minutes, what is the total amount of charge that has passed through the wire?

The answer can be found by finding the quantity of current and total time. t = 10 * 60 = 600, and current was given as I = 6.7. 600 * 6.7 = 4020 C.

Georg Simon Ohm did further research in this subject to discover that in a given curcuit, current is proportional to potential difference in voltage. In other terms, I is proportional to V. This is known as Ohm's law. This can more easily be understood if you relate the situation to a river. If the river is steeper, creating a larger difference in height that the water has to travel, the water flows faster. The same can be applied to current. If there is a greater difference in voltage, the current will be greater as well.

In an electric curcuit, there often exists resistors, which basically serve the purpose of reducing current, like a kink in a hose would reduce water flow. Therefore the total current which flows through a curcuit depends upon the voltage and the total resistance. As votlage varies proportionally with current, resistance varies inversely with current.

I=V/R

This equation can be changed around to yield useful variations as well:

V=IR

R=V/I

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

- I=V/R
- V=IR
- R=V/I

- 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.

If a 9V battery is hooked up to a curcuit that has a current of 3A, what is the total resistance? How would the current change if the resistance was doubled?

Here is what you start with:

- V = 9V
- I(1) = 3A
- R =?
- I(2) = ?

Part 1:

9/3=? ?=3=R

Part 2:

9/(2*3)=? ?=1.5

Go back to: Table of Contents

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.

( 5 ohms)

(640-V)

(90 C)

(4.5 A)

(2.8125 * 10^23 electrons) Go back to: Table of Contents

6.4.0 * 10^12 electrons flow through a wire that has a resistance of
14 ohms and is attached to a 35-V source. Hwo long has the
wire bveen attached to the electrical source?

(256 s)

Solution:

V=IR

R= V/I= 10-V/ 2 A= 5 ohms

Go to: ProblemFormulas Table
of Contents

Solution:

V=IR

V= 20 A * 30 ohms = 640-V Go to: ProblemFormulasTable
of Contents

Solution:

I=V/R

I= 3-V / 4 ohms = .75 A

I= dQ/dt

.75 A = dQ / (2 minutes * 60 sec/min)

dQ = 90 C

Go to: Problem FormulasTable
of Contents

Solution:

R=V/I

R= 90 / 6 = 15 ohms

75% * 90-V = 67.5-V

I= V/R

I= 67.5-V/ 15 ohms = 4.5 A

Go to: ProblemFormulas Table
of Contents

Solution:

I=V/R

I= 200-V / 16 ohms = 12.5 A

I= dQ/dt

12.5 A = dQ / (60 min/hr * 60 sec/min)

dQ = 45000 C

charge on one electron is 1.6 * 10^-19 C

45000 C / (1.6 * 10^-19 C) = 2.8125 * 10^23 electrons Go
to: ProblemFormulas Table
of Contents

Solution:

I=V/R

I= 35-V / 14 ohms = 2.5 A

I = dQ/dt

dQ= 4.0 * 10^21 electrons * 1.6 * 10^-19 C/electron = 640 C

2.5 A = 640 C / dt

dt = 256 s

Go to: Problem FormulasTable
of Contents