Power dissipated by resistor -- Right answer, but need insight

[kostoglotov]

imgur: http://i.imgur.com/9VILpYL.jpg

Now, I can easily solve this by applying the formula $P_R = \frac{(\Delta V_R)^2}{R}$ and get all the correct answers.

However, the answers feel strange to me intuitively. In the formula above, lowering the resistance increases the power that is dissipated.

Shouldn't a device with MORE resistance dissipate MORE energy per unit time? After all, isn't the idea behind something like a superconductor that it dissipates no energy while carrying it, and thus, wouldn't convert any of the flowing currents energy into heat.

Or am I misunderstanding the meaning of what it is to dissipate power? I thought it was just heat. Shouldn't something with more resistance get hotter, thus dissipate more power.

Or is it a matter of power vs energy? Is it that a higher resistance converts a higher fraction of every Joule of electrical energy into heat, but doesn't allow as much current, and so makes the conversion more slowly?

 

[Mister T]

Low resistance means high load.

The higher the current the more heat generated. Think of a wire that gets so hot it melts. Or blows a fuse. A superconductor has no voltage drop across it.

 

[dsoodak]

Shouldn't a device with MORE resistance dissipate MORE energy per unit time?

This would be true if you were actively adjusting voltage to maintain a constant current.
In a circuit, the energy an electron has is proportional to voltage, in a similar fashion to how gravitational potential energy is proportional to height.

Is it that a higher resistance converts a higher fraction of every Joule of electrical energy into heat,?

100% of the potential energy of each electron is converted into heat, but a larger resistor slows down the rate at which they are "falling" to 0 volts.

 

[haruspex]

Low resistance means high load.

You seem to be taking load as a measurable quantity and equating it to power consumption. As far as I am aware, the term is mostly used in electrical circuits to refer to a portion of the circuit, without implying any measurable attribute. When used as something measurable, it sometimes refers to the power, but sometimes to the resistance, so I would try to avoid using the term that way.
@kostoglotov , your intuition is taking current as the constant aspect. If that were so, yes, a higher resistance would mean more power. But in most situations it is generated voltage that is (roughly) constant, so higher resistance means less current, and the overall effect is to reduce power consumption. In the extreme case, infinite resistance means no current, no power.

 

[kostoglotov]

This would be true if you were actively adjusting voltage to maintain a constant current.
In a circuit, the energy an electron has is proportional to voltage, in a similar fashion to how gravitational potential energy is proportional to height.100% of the potential energy of each electron is converted into heat, but a larger resistor slows down the rate at which they are "falling" to 0 volts.

"100% of the potential energy of each electron is converted into heat, but a larger resistor slows down the rate at which they are "falling" to 0 volts."

Eventually...some of becomes light or kinetic energy first though :\

"This would be true if you were actively adjusting voltage to maintain a constant current."

Unless I'm misunderstanding the text, the current is (ideally) constant everywhere in the circuit...and since the textbook is teaching this in terms of ideal models, that's the framework from which I'm talking about it.

 

[kostoglotov]

You seem to be taking load as a measurable quantity and equating it to power consumption. As far as I am aware, the term is mostly used in electrical circuits to refer to a portion of the circuit, without implying any measurable attribute. When used as something measurable, it sometimes refers to the power, but sometimes to the resistance, so I would try to avoid using the term that way.
@kostoglotov , your intuition is taking current as the constant aspect. If that were so, yes, a higher resistance would mean more power. But in most situations it is generated voltage that is (roughly) constant, so higher resistance means less current, and the overall effect is to reduce power consumption. In the extreme case, infinite resistance means no current, no power.

Why has the textbook told me to idealize a circuit as having a constant current everywhere? It seems you're telling me the opposite...is the model you're referrng to an idealized one?

 

[Mister T]

Why has the textbook told me to idealize a circuit as having a constant current everywhere? It seems you're telling me the opposite...is the model you're referrng to an idealized one?

You're comparing one resistor to another, each with the same current. In that case the one with the larger resistance consumes more power. This would be the case, for example, when two resistors are connected in series.

Imagine instead comparing two resistors with the same voltage across them, each having a different amount of current flowing through them. The one with the larger resistance consumes less power. It has the smaller current flowing through it. This would be the case, for example, when two resistors are connected in parallel.

 

[kostoglotov]

You're comparing one resistor to another, each with the same current. In that case the one with the larger resistance consumes more power. This would be the case, for example, when two resistors are connected in series.

Imagine instead comparing two resistors with the same voltage across them, each having a different amount of current flowing through them. The one with the larger resistance consumes less power. It has the smaller current flowing through it. This would be the case, for example, when two resistors are connected in parallel.

Ok...is the power equation $P_R = \frac{(\Delta V_R)^2}{R}$ assuming constant current?? If so, increased resistance...should decrease the power...and in the answers to this question it does. Resistor B has the highest power dissipation.

but you say, "In that case the one with the larger resistance consumes more power."...I don't understand :) Thanks for your help btw, I'm doing my best not to be difficult :)

 

[Mister T]

Ok...is the power equation $P_R = \frac{(\Delta V_R)^2}{R}$ assuming constant current??

No such assumption is relevant. It gives you the power consumed by a resistor, given the voltage across that resistor. $I^2R$ gives you the power consumed by a resistor, given the current through it.

If so, increased resistance...should decrease the power...

Depends on whether you keep the voltage or the current the same when you do that. You can't keep them both the same because $\Delta V_R=IR$.

In the examples you were presented with, ΔV_R and R were given, so you used $\frac{(\Delta V_R)^2}{R}$. Had you instead been given I and R you would use $I^2R$ where $\Delta V_R=IR$.

 

[kostoglotov]

No such assumption is relevant. It gives you the power consumed by a resistor, given the voltage across that resistor. $I^2R$ gives you the power consumed by a resistor, given the current through it.
Depends on whether you keep the voltage or the current the same when you do that. You can't keep them both the same because $\Delta V_R=IR$.

In the examples you were presented with, ΔV_R and R were given, so you used $\frac{(\Delta V_R)^2}{R}$. Had you instead been given I and R you would use $I^2R$ where $\Delta V_R=IR$.

Right...so in actual fact, with a decrease in resistance, there was an increase in power dissipated.

"You're comparing one resistor to another, each with the same current. In that case the one with the larger resistance consumes more power. "

Was this a typo (did you mean write "consumes less power"), or am I still misunderstanding this?

 

[haruspex]

"You're comparing one resistor to another, each with the same current. In that case the one with the larger resistance consumes more power. "

Was this a typo (did you mean write "consumes less power"), or am I still misunderstanding this?

You have four equations, and they all apply in all circumstances:
V=IR; P=IV; P=V²/R; P=I²R.
If you increase the resistance, one or both of V and I must also change. If by some means I is held constant then these equations tell you that V and P must increase. Alternatively, if V is held constant then I and P must decrease.

 

[insightful]

Unless I'm misunderstanding the text, the current is (ideally) constant everywhere in the circuit...and since the textbook is teaching this in terms of ideal models, that's the framework from which I'm talking about it.

I think the text probably says the current is the same (not constant) everywhere in a single loop circuit. Those are two different conditions.

 

[Mister T]

"You're comparing one resistor to another, each with the same current. In that case the one with the larger resistance consumes more power. "

Was this a typo (did you mean write "consumes less power"), or am I still misunderstanding this?

No typo. You're still not getting it. Try reading through the comments made in the thread again.

Suppose you had instead been given this problem:

Resistors in series have the same current through them. But that's a special case, not the general case.

Another special case is when resistors are in parallel. Then they have the same voltage across them.

I recommend you work out the solutions to both problems using only $\Delta V_R=IR$ and $P=I\Delta V_R$.

 

[kostoglotov]

You have four equations, and they all apply in all circumstances:
V=IR; P=IV; P=V²/R; P=I²R.
If you increase the resistance, one or both of V and I must also change. If by some means I is held constant then these equations tell you that V and P must increase. Alternatively, if V is held constant then I and P must decrease.

I understand all that. That's how I got the correct answer in the first place.

What I'm asking about is, what's actually happening in these circuits? What is resistance actually? How can less resistance lead to a lower power dissipation? For instance, if I wanted an element in a heater, to heat a room for instance, wouldn't I want to choose an element with a higher resistivity, a higher resistance, to put out more heat, to dissipate more power? How then could choosing a lower resistance give a higher power dissipation, unless I'm missing something crucial in the definition of what all these different terms PHYSICALLY mean!

 

[kostoglotov]

No typo. You're still not getting it. Try reading through the comments made in the thread again.

Suppose you had instead been given this problem:

View attachment 91495

Resistors in series have the same current through them. But that's a special case, not the general case.

Another special case is when resistors are in parallel. Then they have the same voltage across them.

I recommend you work out the solutions to both problems using only $\Delta V_R=IR$ and $P=I\Delta V_R$.

Everyone is just saying, follow the correct equations to get the right answer.

I already did that...I got the right answer the first time, the math is easy...but what I'm crucially asking is, to quote myself above

"What is resistance actually? How can less resistance lead to a lower power dissipation? For instance, if I wanted an element in a heater, to heat a room for instance, wouldn't I want to choose an element with a higher resistivity, a higher resistance, to put out more heat, to dissipate more power? How then could choosing a lower resistance give a higher power dissipation, unless I'm missing something crucial in the definition of what all these different terms PHYSICALLY mean!"

 

[Mister T]

Everyone is just saying, follow the correct equations to get the right answer.

I never said that or anything like it. I went to the trouble of making a new problem for you, and gave you specific directions about which equations to use, and which ones not to use. My goal is not to get you to make the right answers, you already have those. My goal is to get you to make sense of them.

What is resistance actually?

It's the voltage across the resistor divided by the current through the resistor.

$$R \equiv \frac{\Delta V_R}{I}.$$

Describe how you'd go about connecting the resistors given in the statement of the problem to a battery? In each case choose numerical values for the quantities and calculate the power dissipated.

For example, Part A. Connect a 1.5 volt battery to a 3 ohm resistor. The current in the circuit would be 0.5 amps. The power dissipated would be 0.75 watts.

 

[kostoglotov]

I never said that or anything like it. I went to the trouble of making a new problem for you, and gave you specific directions about which equations to use, and which ones not to use. My goal is not to get you to make the right answers, you already have those. My goal is to get you to make sense of them.
It's the voltage across the resistor divided by the current through the resistor.

$$R \equiv \frac{\Delta V_R}{I}.$$

Describe how you'd go about connecting the resistors given in the statement of the problem to a battery? In each case choose numerical values for the quantities and calculate the power dissipated.

For example, Part A. Connect a 1.5 volt battery to a 3 ohm resistor. The current in the circuit would be 0.5 amps. The power dissipated would be 0.75 watts.

There are a score of exercises in the text asking me to do just that, and I'm in the middle of doing it.

Can we get away from "plug and chug" for a second (the text has plenty of that), and can I ask about a theoretical heating device...something we can talk about in terms of a concrete goal?

Let's say I want to make an electrical heater to heat a room. Wouldn't the best materials for that element in the heater to generate and output the heat to the room, be something with a high resistance? Would that element not be a 'resistor dissipating the power' from the source of some current?

We want a fixed power output from the heater.

Ok, since I'd assumed we'd want to use an element with a higher resistance to make heat, one of our equations tells us that this higher resistance decreases the power that the element can dissipate, requiring us to increase the voltage to get that fixed power dissipation.

(i) Why don't we just use an element of lower resistance in our device to get the same power dissipation, instead of upping the voltage?

Ok, we up the voltage, to get the power dissipation we want from the element. $P = I^2 R$ tells us that choosing the material of higher resistance at a given fixed power means...lowering the current?

So, we've chosen the higher resistance material (yes?no?), we want a fixed power output, so we'll need a higher voltage and a lower current compared to using a material of lower resistance? Yes? No?

So...chose a material of lower resistance, and use a lower voltage? That doesn't sound right! Doesn't a material of lower resistance heat up less?

 

[insightful]

Let's say I want to make an electrical heater to heat a room. Wouldn't the best materials for that element in the heater to generate and output the heat to the room, be something with a high resistance?

Compared to a 12ga copper wire, certainly.

Would that element not be a 'resistor dissipating the power' from the source of some current?

All normal rooms have a source of constant voltage. Think about a 240V heater with a knob to adjust the resistance from 0 to infinity. Turn that knob to infinity and what's the heat output? Bingo, NO heat output. Turn the knob down from infinity and heat starts coming out. At some point, you'll have maximum heat just before the circuit breaker trips on over-amps.

 

[Mister T]

Let's say I want to make an electrical heater to heat a room. Wouldn't the best materials for that element in the heater to generate and output the heat to the room, be something with a high resistance?

Certainly not. Let's say you have these choices for a voltage source. A 1.5 volt flashlight cell, a 110 volt outlet, or a 220 volt outlet.

We want a fixed power output from the heater.

Ok. Let's say you want 220 watts. You could draw one amp with your 220 volt source, 2 amps with your 110 volt source, or 147 amps with your 1.5 volt battery.

I'll let you calculate the resistance of the heater in each case. Raise the voltage across anyone of them and it will draw more current and consume more power. Lower the resistance of anyone of them and it will draw more current and consume more power.

 

[tjo5112]

I think that that you may be thinking about this too hard. Instead of thinking of P=V^2/R use the other definition of P=VI, which may be more intuitive. Also remember Ohm's law is V=IR. Remember that I is the current going through the resistor and V is the potential across the resistor. The thing is, if you have a constant voltage and the decrease the resistance, the current will go up therefore the power goes up and vice-versa. Likewise, if you keep a constant resistance and increase the voltage you will also increase the current and therefore the power goes up. You can change the voltage, current, and resistance so long as R=V/I holds, and I think this is what is confusing you. I'll give you a few quick ideal examples similar to the situations you posted.

Take an example circuit with a 1V battery across a 1 ohm resistor. In this case the current would be 1A, leading to a power of 1W.

Now use a 2V battery with a 1 ohm resistor, which makes a current of 2A, leading to a power of 4W.

Now use a 1V battery with a 2 ohm resistor, which makes a current of 0.5 A, leading to a power of 0.5 W.

Lastly, use a 1V battery with a 0.5 ohm resistor, which makes a current of 2A, leading to a power of 2W.

This is a very simple example, but this was just to illustrate how the voltage, current, resistance and power are related. I hope this helps.

 

[kostoglotov]

Certainly not. Let's say you have these choices for a voltage source. A 1.5 volt flashlight cell, a 110 volt outlet, or a 220 volt outlet.
Ok. Let's say you want 220 watts. You could draw one amp with your 220 volt source, 2 amps with your 110 volt source, or 147 amps with your 1.5 volt battery.

I'll let you calculate the resistance of the heater in each case. Raise the voltage across anyone of them and it will draw more current and consume more power. Lower the resistance of anyone of them and it will draw more current and consume more power.

A typical (900 mA.hr) 1.5 V battery would be drained in under 30 seconds at 147 Amps...but I guess that's not the important point here.

That's a lot of Amps...no household appliance comes anywhere near that.

But obviously, in terms of getting power out, that last ~0.01 ohm element offers the best gains if we want to increase the voltage...so I'm going to guess that old fashioned light bulbs with the thin coiled tungsten wires in the vacuum aren't actually putting out their power due to the resistance (that high resistance will be lowering their power output), but due to some effect of the magnetic fields generated in the loops of the wire...?

 

[Mister T]

A typical (900 mA.hr) 1.5 V battery would be drained in under 30 seconds at 147 Amps...but I guess that's not the important point here.

Not to mention that it might explode!

so I'm going to guess that old fashioned light bulbs with the thin coiled tungsten wires in the vacuum aren't actually putting out their power due to the resistance (that high resistance will be lowering their power output), but due to some effect of the magnetic fields generated in the loops of the wire...?

No, it's their (nonohmic) resistance.

Here's one that might help. At the grocery store you buy two bulbs, one is "100 W, 120 V" and the other is "60 W, 120 V". Which one has a greater resistance?

Connect them in series to a 120 volt source. Which one consumes more power?

 

[OmCheeto]

...
Shouldn't a device with MORE resistance dissipate MORE energy per unit time?
...

Vacuum cleaners don't. At least the ones I have.
If I seal the end of the hose (infinite resistance!), the motor speeds up, using, theoretically, zero power.

 

[kostoglotov]

Not to mention that it might explode!
No, it's their (nonohmic) resistance.

Here's one that might help. At the grocery store you buy two bulbs, one is "100 W, 120 V" and the other is "60 W, 120 V". Which one has a greater resistance?

Connect them in series to a 120 volt source. Which one consumes more power?

The explosion? Is that due to the internal resistance of the battery. Batteries would be nonohmic as well.

You wouldn't be able to draw that much current from that battery anyway would you? You could try connecting a 0.01 ohm wire between the terminals, but wouldn't the internal resistance of the battery determine the current?

144 ohm and 240 ohm, the second has higher resistance, but if you put them in series in a single loop (circuit) then the current everywhere in that circuit must be the same, otherwise without a junction you'd have charge accumulating somewhere, so you'd want to pick the lowest value for the current, since you don't have charge accumulating in the circuit. The current that can pass through the bulbs is ~0.83 A and 0.5 A. P = I^2R and we need to pick 0.5 A for both. The one with higher resistance will consume more power. P1 = 36W, P2 = 60W...bulb 2 keeps it's original rating, bulb 1 is reduced to about a third of its original rating.

 

[Mister T]

The explosion? Is that due to the internal resistance of the battery. Batteries would be nonohmic as well.

You wouldn't be able to draw that much current from that battery anyway would you? You could try connecting a 0.01 ohm wire between the terminals, but wouldn't the internal resistance of the battery determine the current?

You are absolutely correct. Everything you say is correct.

But your analysis below is not correct.

144 ohm and 240 ohm, the second has higher resistance, but if you put them in series in a single loop (circuit) then the current everywhere in that circuit must be the same, otherwise without a junction you'd have charge accumulating somewhere, so you'd want to pick the lowest value for the current, since you don't have charge accumulating in the circuit. The current that can pass through the bulbs is ~0.83 A and 0.5 A. P = I^2R and we need to pick 0.5 A for both. The one with higher resistance will consume more power. P1 = 36W, P2 = 60W...bulb 2 keeps it's original rating, bulb 1 is reduced to about a third of its original rating.

Your conclusion about the 60 W bulb consuming more power is correct, but your analysis in not. When resistors are connected in series the larger one has a dominant effect on the current, but you can't just assume that the effect of the other one disappears. You have to add their resistance to determine the total resistance. That total resistance predicts the correct current through both resistors.

 

[kostoglotov]

You are absolutely correct. Everything you say is correct.

But your analysis below is not correct.
Your conclusion about the 60 W bulb consuming more power is correct, but your analysis in not. When resistors are connected in series the larger one has a dominant effect on the current, but you can't just assume that the effect of the other one disappears. You have to add their resistance to determine the total resistance. That total resistance predicts the correct current through both resistors.

Ok, so I'd add their currents to get their total currents? So $I_{total} = V(\frac{1}{R_1}+\frac{1}{R_2})$, and then use $I_{total}$ to calculate the power use of each "resistor"?

 

[kostoglotov]

You are absolutely correct. Everything you say is correct.

But your analysis below is not correct.
Your conclusion about the 60 W bulb consuming more power is correct, but your analysis in not. When resistors are connected in series the larger one has a dominant effect on the current, but you can't just assume that the effect of the other one disappears. You have to add their resistance to determine the total resistance. That total resistance predicts the correct current through both resistors.

So, what specifically in that analysis of mine is incorrect? The part about the current needing to be the same everywhere in the circuit? My textbook quite clearly said that in an ideal circuit the current is the same everywhere, or else charge will accumulate.

 

[haruspex]

So, what specifically in that analysis of mine is incorrect? The part about the current needing to be the same everywhere in the circuit? My textbook quite clearly said that in an ideal circuit the current is the same everywhere, or else charge will accumulate.

The part where you calculated the resulting current. You took a 120V supply and found the currents that would arise if that total voltage were applied to each of the two resistances. You then argued that with the two resistors in series the resulting current would be the lower of those two. That is quite wrong, and you know the equations to show it is wrong.
With the two resistors in series, neither would experience the full 120V. There would be a voltage drop across each, and the two voltages would add to 120V. To find the current, add the resistances: 384 Ohms. Divide that into 120V to get about 0.31A. That is the current in the circuit, so is the current through each resistor. The voltage drops are therefore roughly 0.31*144=45V and 0.31*240=74V.

 

[Mister T]

So, what specifically in that analysis of mine is incorrect? The part about the current needing to be the same everywhere in the circuit? My textbook quite clearly said that in an ideal circuit the current is the same everywhere, or else charge will accumulate.

Right, but the value of that current is not calculated in the way you calculated it. You assumed that the bulb with the higher resistance is all you need consider when calculating that current. You must instead consider both bulbs.

 

[Mister T]

To find the current, add the resistances: 384 Ohms.

I'll never forget embarrassing myself by assuming the bulbs have a resistance that stays the same when you hook them up in series. They don't, by a long shot.

 

[kostoglotov]

Right, but the value of that current is not calculated in the way you calculated it. You assumed that the bulb with the higher resistance is all you need consider when calculating that current. You must instead consider both bulbs.

Ok, so I'd add their currents to get their total currents? So $I_{total} = V(\frac{1}{R_1}+\frac{1}{R_2})$, and then use $I_{total}$ to calculate the power use of each "resistor"?

 

[Mister T]

Ok, so I'd add their currents to get their total currents? So $I_{total} = V(\frac{1}{R_1}+\frac{1}{R_2})$, and then use $I_{total}$ to calculate the power use of each "resistor"?

No, you add their voltage drops to get the total voltage drop. (The currents are the same, if you added them together you'd get 2$I$.).

$I = \frac{V}{R_1+R_2}$

 

[haruspex]

I'll never forget embarrassing myself by assuming the bulbs have a resistance that stays the same when you hook them up in series. They don't, by a long shot.

True, but at the level of the present discussion...