Ohm's Law

by Joe Trachta, Tyler Motlagh, and Mark Lander 

Table of Contents


Here's the quantities you can know:     Voltage and charge have been explained on previous pages.


New Quantities

Defining Current

A curcuit can be defined as a continuous conducting path for an electric charge.  The flow of that charge is current.  Current is more specifically the amount of charge that passes through a particular point along that curcuit, and is measured in amperes (amps or A), which is coulombs per second.

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
 
 
 
 
 

Formulas

So now we have all the formulas we need for solving current problems:
  1. I=V/R
  2. V=IR
  3. R=V/I
Go back to: Table of Contents

General Problem Solving Strategy:

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

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

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

    6. previous two steps until you are done.
Go back to: Table of Contents

Example Problem


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:

putting numbers into equations:
Part 1:
9/3=?  ?=3=R

Part 2:
9/(2*3)=?  ?=1.5
 
 
 

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.

A radio is designed to take in 2 A when connected to a 10-V source.  What is its net  resistance?
( 5 ohms)

2.

A spotlight has a resistance of 20 ohms and requires 32 A of current to function.   What amount of voltage does the light need to be plugged into?
(640-V)

3.

What charge has passed through a wire with 4 ohms of resistance if it is attached to  a 3-V battery for 2 minutes?
(90 C)

4.

A blender draws 6 A at 90-V.  What current would it receive if the voltage suddenly  dropped 25%?
(4.5 A)

5.

An electrical device is attached to a 200-V source and has a resistance of 16 ohms.   How many electrons are leaving the electrical device per hour?
(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)
 
 

Solutions to Sample Problems


1.

A radio is designed to take in 2 A when connected to a 10-V source.  What is its net  resistance?

 Solution:
 V=IR
 R= V/I= 10-V/ 2 A= 5 ohms
Go to: ProblemFormulas Table of Contents

2.

A spotlight has a resistance of 20 ohms and requires 32 A of current to function.   What amount of voltage does the light need to be plugged into?

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

3.

What charge has passed through a wire with 4 ohms of resistance if it is attached to  a 3-V battery for 2 minutes?

 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

4.

A blender draws 6 A at 90-V.  What current would it receive if the voltage suddenly  dropped 25%?

 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

5.

An electrical device is attached to a 200-V source and has a resistance of 16 ohms.   How many electrons are leaving the electrical device per hour?

 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

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?

 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