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Lecture Comments (4)

3 answers

Last reply by: Professor Dan Fullerton
Wed Jan 29, 2014 6:24 AM

Post by Nawaphan Jedjomnongkit on May 8, 2013

Why in example 3 the relationship of I and R is not in linear?

Ohm's Law & Power

  • Ohm’s Law is an empirical relationship for conductors and resistors in circuits. V=IR
  • Ohmic materials exhibit linear plots of voltage vs. current.
  • The rate at which electrical energy is expended is electrical power. P+IV
  • The brightness of a lightbulb is related to its power output.

Ohm's Law & Power

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Objectives 0:06
  • Ohm's Law 0:21
    • Relates Resistance, Potential Difference, and Current Flow
  • Example 1: Resistance of a Wire 1:22
  • Example 2: Circuit Current 1:58
  • Example 3: Variable Resistor 2:30
  • Ohm's 'Law'? 3:22
    • Very Useful Empirical Relationship
    • Test if a Material is 'Ohmic'
  • Example 4: Ohmic Material 3:58
  • Electrical Power 4:24
    • Current Flowing Through a Circuit Causes a Transfer of Energy Into Different Types
    • Example: Light Bulb
    • Example: Television
  • Calculating Power 5:09
    • Electrical Energy
    • Charge Per Unit Time Is Current
    • Expand Using Ohm's Law
  • Example 5: Toaster 7:43
  • Example 6: Electric Iron 8:19
  • Example 7: Power of a Resistor 9:19
  • Example 8: Information Required to Determine Power in a Resistor 9:55

Transcription: Ohm's Law & Power

Hi folks and welcome back to

This lesson is going to be about Ohm's Law and Power.0003

Our objectives are going to be to define and calculate resistance, current, and voltage using Ohm's Law and to calculate the power dissipated and energy used in electric circuits.0006

This will be the basis of much more of our circuit analysis moving forward.0017

Ohm's Law -- Ohm's Law relates resistance, potential difference, and current flow.0021

It can be written in several ways, V = IR where (V) is the potential drop across a circuit element -- is equal to the current flow times the resistance.0027

You can look at it in a couple of different ways.0037

You use it so much that sometimes I have used these circles to help me understand or to quickly derive exactly the version of the formula that I need.0039

If I am looking for (V), what I would do is I would cover up on the circle, the (V).0049

So if I cover up (V), I find out that V = IR.0054

In the middle, if I want to find current, I cover up the current.0059

(I), therefore, must equal V/R.0064

And one more version -- if I want to find resistance, I cover up resistance and I find that (R) is equal to V/I -- just a quick little way to manipulate that formula that may save you some time.0068

Now let us start by talking about resistance of a wire.0082

The current in a wire is 24 amperes (A) -- so (I) = 24 A, when connected to a 1.5-volt battery -- (V) = 1.5 volts.0085

Find the resistance of the wire.0096

I will use Ohm's Law -- (R) = V/I, which is 1.5 volts/24 A or 0.625 ohms -- pretty straight forward.0098

Let us take a look here at circuit current.0119

In a simple electric circuit, a 24-ohm resistor -- so now (R) = 24 ohms -- is connected across a 6-volt battery, (V) = 6 volts. What is the current?0121

I will use another version of Ohm's Law -- (I) = V/R which is 6 volts/24 ohms, or 0.25 A or 250 milliamps (mA).0132

How about a graphical problem?0151

A constant potential difference is applied across a variable resistor held at constant temperature -- which graph best represents the relationship between the resistance of the variable resistor and the current through it?0153

Well, (R) = V/I and we are holding (V )constant. 0166

So we have (R) on the x and current on the y.0171

What is going to happen to (R) as I increase (I)?0175

What is going to happen to (I) -- I could write it also as V/R.0178

As (R) gets bigger, current must be getting smaller, right?0184

So right away, I can see that that must be answer 1.0188

We have an inverse relationship -- as (R) gets bigger, current gets smaller and as (R) gets smaller, current gets bigger.0193

So a little more on Ohm's Law.0203

Ohm's Law is not truly a law of physics.0205

Not all materials obey this relationship, but it is a very useful empirical relationship or experimental relationship that holds up in many different circumstances.0208

It is great for conductors and resistors and an easy way to test if a material is ohmic is to plot the voltage versus current through it.0217

If you get a straight line, the material is what we would call "ohmic".0225

The slope of that line tells you the resistance of that ohmic material -- the resistance of that resistor.0229

The graph below represents the relationship between the potential difference (V) across a resistor and the current through the resistor.0239

Through which interval does the resistor obey Ohm's Law or through which region -- which interval -- is it ohmic?0245

Well it is ohmic where this is a linear relationship.0252

So that is going to be from B to C in this section right there.0255

So BC is where it obeys Ohm's Law.0260

Electrical power -- when current is flowing through a circuit it causes a transfer of energy into different types.0264

The ray that which electrical energy is transformed into other types of energy is the electrical power dissipated.0270

In a light bulb this corresponds electrical energy being transferred to light and heat -- where typically the brightness of a bulb is related to its power output.0276

Let us say it is proportional to power output.0292

In television, for example, electrical energy is converted to light, sound and heat, and other types of energy, but the rate at which you convert it to these other types is the power dissipated.0298

Now to calculate power -- when you do work on an object, you change its energy.0309

So electrical energy is equal to the product of the electrical charge and potential difference -- we have done that before.0314

So then power is also the work done divided by the time, the rate at which work is done, which is QV/t.0319

And charge per unit time is current so if P = QV/t -- charge per unit time -- Q/t is current, therefore we could write that Power = current times voltage.0329

Well if Power is current times voltage, we can also apply Ohm's Law here.0348

We know that V = IR so I could write this as Power equals (I) times -- I am going to replace (V) with IR, therefore Power = I2 × R.0353

Or if Power = IV -- we also know that I = V/R, so Power = V/R × V, which implies that Power = V2/R.0369

So we have a couple equations to help us find power -- Power = current times voltage, Power = I2R, and Power = V2/R.0385

All of these are different versions of this power formula when we apply Ohm's Law to it.0397

Now as we talk about calculating power, I want to bring up briefly what you may see on an electrical bill.0402

You may see a unit known as a kilowatt-hour (kWh).0409

Well notice that watts (W) are power -- an hour is time, so that is a power times a time.0412

Well, that is a unit of work or energy -- a kilowatt-hour is a unit of energy.0420

So when you pay so much per kilowatt-hour, you are actually paying for energy.0426

It would be much more simple to talk about this in joules (J).0432

But if we think about it, a kWh is 1000 W × 3600 s which is going to give us 3.6 x 106J = 1 kWh.0435

So that is a unit conversion there for some day if you are trying to figure out what your electrical bill is corresponding to.0456

Let us take an example of a toaster here.0464

A 110-volt toaster -- V = 110 volts -- draws a current of 6 A -- I = 6 A -- on its highest setting as it converts electrical energy into thermal energy and hopefully gives you a nice, tasty delicious breakfast or a toast.0466

What is the toaster's maximum power rating?0482

Well Power = I × V, where 6 A × 110 V -- which is going to be 660 W.0484

Or how about an electric iron?0500

We have to go make our shirts nice and wrinkle-free.0501

An electric iron operating at 120 V -- V = 120 V -- draws 10 amperes of current -- how much heat energy is delivered by the iron in 30 s?0505

And if we are looking for heat energy, what we are looking for is either work or energy.0518

So let us say we are looking for W. 0523

Well if Power = work/time then Work = power × time, but power is V × I, so this is going to be VIt.0525

Therefore the heat energy is going to be the voltage (120 volts) times the current (10 A)times our 30 s or 3.6 × 104 J.0538

How about the power dissipated by a resistor?0560

A potential drop of 50 V is measured across a 250-ohm resistors, so our resistance is 250 ohms.0563

What is the power developed in the resistor?0571

So we are looking for P.0574

Well this time I am going to use Power = V2/R -- so that will be 50 V2/250 ohms or 10 W -- another version of our power formula.0576

Let us try one more.0595

What is the minimum information needed to determine the power dissipated in a resistor of unknown value.0597

Remember that Power = current × voltage.0604

If we have just the potential difference across the resistor -- just (V) -- we do not have enough information.0607

If we have just the current through the resistor, we are not going to do it.0613

Current and potential difference -- Yes! That will do it.0616

And current, potential difference and time -- No we do not need time, that cannot be the answer.0620

Correct answer? -- Current and potential difference only.0625

All right, hopefully that gets you a great start on Ohm's Law.0629

Thanks for watching and make it a great day.0632