Power and Power Dissipated by a resistor .. in layman's terms


by lollol
Tags: dissipated, layman, power, resistor, terms
lollol
lollol is offline
#1
Dec31-07, 12:04 AM
P: 25
Can someone give me a reply Intro Physics style? :D

P = IV is known as the "power of the circuit" right?

But then, we have questions asking about the heat dissipated by a resistor. We can use P = IV, P=I^2 R , or P=V^2/R to answer such questions

Since the voltage drops all add up to the original voltage.. won't adding IV + IV for each resistor yield the original Power... implying that ALL energy is dissipated as heat?

I guess I don't understand the difference between the generic "Power" of the circuit and the power/head dissipated by resistors :(
Phys.Org News Partner Physics news on Phys.org
Information storage for the next generation of plastic computers
Scientists capture ultrafast snapshots of light-driven superconductivity
Progress in the fight against quantum dissipation
Cyrus
Cyrus is offline
#2
Dec31-07, 12:09 AM
Cyrus's Avatar
P: 4,780
Yes, all energy goes into heat for a resistor.
lollol
lollol is offline
#3
Dec31-07, 12:22 AM
P: 25
So ALL the energy is dissipated as heat?

Isn't energy needed to run the circuit though?

artemis52
artemis52 is offline
#4
Dec31-07, 12:26 AM
P: 6

Power and Power Dissipated by a resistor .. in layman's terms


Think of a resistor's purpose in a way like a light bulb: a light bulb takes current and out comes some heat, but some light too, and less voltage out the other side because you used some of that energy to make heat and light. Now take away the light part. A resistor takes down your voltage, duh - but the energy has to go somewhere, ex. conservation of energy applies to voltages as well. So it comes off as heat, and this is very, very dependent on the resistance and voltage, because these things are a measure of how much energy goes in and comes out of a resistor.

But also, not all of the current drops to zero; you can string multiple light bulbs and get varying amounts of light and heat to come out of them. So using a resistor effectively takes your voltage down a notch, by using some of that stored potential energy to heat up the resistor itself.
Gokul43201
Gokul43201 is offline
#5
Dec31-07, 12:28 AM
Emeritus
Sci Advisor
PF Gold
Gokul43201's Avatar
P: 11,154
Quote Quote by lollol
Isn't energy needed to run the circuit though?
No - at least nothing more than what is dissipated.

If there was more energy supplied than energy lost, then with each looping of the circuit, the electrons would have a higher speed. You end up making energy out of nothing. Not allowed.
gabee
gabee is offline
#6
Dec31-07, 12:35 AM
gabee's Avatar
P: 179
Quote Quote by artemis52 View Post
Think of a resistor's purpose in a way like a light bulb: a light bulb takes current and out comes some heat, but some light too, and less current out the other side because you used some of that energy to make heat and light. Now take away the light part. A resistor takes down your voltage, duh - but the energy has to go somewhere, ex. conservation of energy applies to voltages as well. So it comes off as heat, and this is very, very dependent on the resistance and voltage, because these things are a measure of how much energy goes in and comes out of a resistor.

But also, not all of the current disappears; you can string multiple light bulbs and get varying amounts of light and heat to come out of them. So using a resistor effectively takes your voltage down a notch, by using some of that stored potential energy to heat up the resistor itself.
A correction here: current isn't depleted after passing through a resistor or a light bulb. The same current enters the resistor as that which exits (by Kirchoff's laws). It is the voltage that is "diminished"; i.e. there is a potential drop from one side of the resistor to the other.
artemis52
artemis52 is offline
#7
Dec31-07, 12:42 AM
P: 6
Sorry for the confusion, I am a little rusty with the circuitry, and my edits = the slow. Further clarification:

The set of equations mentioned, P = IV, P=I^2 R , or P=V^2/R are all a result of the first, namely using P=IV and substituting V=IR. These hold for voltages across either the entire circuit or individual resistors then. If you add up IV for each voltage drop per resistor, then you should get the entire power dissipated, if the entire circuit is composed of resistors: if all of the voltage drops across all of the resistors, then the energy comes out entirely as heat.
lollol
lollol is offline
#8
Dec31-07, 12:42 AM
P: 25
Oh right... my bad

That's the purpose of the Battery... to provide new voltage everytime

So... as current travels through the circuit... energy has to be converted to another form.. in this case, heat .... at the end of the circuit, right before hitting the battery again, circuit energy is 0. However, it is replenished again with the voltage... am I getting it?
artemis52
artemis52 is offline
#9
Dec31-07, 12:59 AM
P: 6
Just remember that current and voltage are two separate but related parts of a circuit.

The voltage is stored electric potential energy. It's the electrical version of potential gravitational energy; you put a ball high on a hill, it rolls down fast. You put it higher, it rolls faster.

Current is a measure of how much electric charge you have moving. So to use the ball rolling, it is how many balls you have rolling. Current is the mechanism used to discharge the potential difference between the positive and negative sides of the battery.

So if you put one or many resistors in any complicated configuration, you have the potential difference coming down after it passes a resistor - the balls roll down the hill, all the way to zero if you used up all the voltage. But your current is still conserved - you still have the same number of balls, even though some of them may have went in different directions, at the end of the day your current is conserved.
DaleSpam
DaleSpam is offline
#10
Dec31-07, 08:56 AM
Mentor
P: 16,469
Quote Quote by lollol View Post
But then, we have questions asking about the heat dissipated by a resistor. We can use P = IV, P=I^2 R , or P=V^2/R to answer such questions
I have only one thing to add to the previous comments. P=IV is the general relationship and works for all components including capacitors, inductors, and transistors. The other two relationships are just for resistors. Also, you can apply P=IV for a whole circuit or for any individual component.
lollol
lollol is offline
#11
Dec31-07, 07:32 PM
P: 25
Sorry to bug you guys but:

Electric power for transmission wires is "stepped up" to a very high voltage in order to:

D) cut down heat loss in transmission wires

there were other options.. just didn't include them

If all the electric energy will be lost to heat in the resistors.. or wire... how can "heat loss" be reduced? Especially.. since the power of the circuit is fixed.

This question is on the MCAT and since they only expect a basic conceptual understanding of the formulas, P = IV and V = IR... can somebody explain this conceptually using these formulas? Thanks
gabee
gabee is offline
#12
Dec31-07, 08:43 PM
gabee's Avatar
P: 179
If you envision the transmission line as a big resistor, you have a circuit that looks like an AC power supply connected to that big resistor. You could send the same amount of power through the line by having either low voltage/high current or high voltage/low current, or somewhere in between.

Obviously, we would like to minimize the power lost to heat in our circuit (in our resistor). Since V=IR across our resistor circuit, and P=IV, our power lost to heat is P=I(IR)=(I^2)*R. By minimizing the current we have to pass through the line, we minimize the power lost to heat. This is why transmission lines step the voltage up to a high level.
DaleSpam
DaleSpam is offline
#13
Dec31-07, 08:44 PM
Mentor
P: 16,469
The point is that you want to deliver the power to the customer, not lose it in the transmission line. By using transformers you can basically deliver the power at any voltage that you like. Since the customer is in series with the transmission line the current is the same through both so to calculate the power lost you use P=I^2 R to determine the power lost in the transmission line. To minimize this for a fixed R you minimize I. To minimize I you maximize the delivered voltage.
BriK
BriK is offline
#14
Jan1-08, 05:23 PM
P: 14
If all the electric energy will be lost to heat in the resistors.. or wire... how can "heat loss" be reduced? Especially.. since the power of the circuit is fixed
P = I * E

E = V = voltage
P = power
I = current

I like the PIE formula better because its easy to remember the word pie...

If P stays the same then only I or E can change...so if you raise E then I must be reduced to keep P the same and vice versa...

As someone states above the higher voltage for transmission lines is used to raise the efficency of the transmission lines by reducing the amount lost as heat because the current is lower to keep P the same...

Using someone elses example above, the number of balls are reduced (lower current) and they are poured down a pipe on a taller hill (higher voltage) with fewer balls there is less congestion in the pipe and lower heat loss...

Heat lose is reduced by raising the voltage...making the wire (resistor) more efficiant (less heat lose)...

As a side note...most wiring today is copper, but what do you think transmission lines are made of?
Aluminum...to keep the wires lighter...
lollol
lollol is offline
#15
Jan2-08, 11:20 PM
P: 25
I guess I am getting confused between:

Ohm's Law V = IR

and Power

I'm still a lil' confused....
I understand if you increase V, I drops

So if the resistance of the wire is "fixed".... I^2R = less than it was before

But if the circuit has a fixed "power", I don't understand how I^R is changing.. if anyone can clear this up, I'd really appreciate it. According to a poster above, all the "power" of the circuit is transferred to heat...

I thought Power Dissipated in the resistors should add up to the overall Circuit Power
russ_watters
russ_watters is offline
#16
Jan3-08, 05:22 AM
Mentor
P: 21,997
Draw yourself a diagram. The power dissipated by the load is fixed, the power dissipated by the transmission lines is a function of voltage. So if you want to deliver a certain power with the least loss, you need a high voltage in the transmission lines.
Sojourner01
Sojourner01 is offline
#17
Jan3-08, 05:28 AM
P: 373
Depends what you're fixing. In a real circuit, the power of the implement creating the potential (battery, dynamo, black box, whatever) is generally fixed. Within this, you can set the voltage and current any way you like as long as combining them as I*V still equals the power rating of the 'box'. You could, of course, devise a black box power supply that always fixes the voltage - in which case the power dissipated by the circuit will vary instantaneously as you vary the current.
DaleSpam
DaleSpam is offline
#18
Jan3-08, 05:39 AM
Mentor
P: 16,469
Quote Quote by lollol View Post
But if the circuit has a fixed "power", I don't understand how I^R is changing.. if anyone can clear this up, I'd really appreciate it.
There are two parts of the circuit: the transmission lines, and the load (customers). Power will be dissapated in both parts, but the power company only gets paid for the power dissapated in the load. The power dissapated in the transmission lines is wasted. The goal is to minimize this fraction of wasted power. Only the power to the load is considered fixed in this analysis, the power wasted in the transmission lines is considered variable.


Register to reply

Related Discussions
max power delivered to variable resistor Engineering, Comp Sci, & Technology Homework 2
power dissipated in capacitor Introductory Physics Homework 7
Power dissipated without a load? Electrical Engineering 3
Energy Dissipated by Resistor Introductory Physics Homework 3
power dissipated in circuit Introductory Physics Homework 3