Berkley Review power loss question...

[pineappletree]

For a fixed potential difference across a resistive wire, the power loss through the resistor can be increased by increasing the:
a.) Neutron density in the wire
b.) length of the wire
c.)Temperature of the wire
d.) conductivity of the wire




Can anyone tell me the answer and explain it please??

Thank you!!

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[PiBond]

For a fixed potential difference across a resistive wire, the power loss through the resistor can be increased by increasing the:
a.) Neutron density in the wire
b.) length of the wire
c.)Temperature of the wire
d.) conductivity of the wire




Can anyone tell me the answer and explain it please??

Thank you!!

my guess would be answer D. increasing conductivity allows it to conduct more current. P = I^2R

 

[lorenzomicron]

p=v^2/r
so to maximize the power, we need voltage to be maximized and resistance to be minimized.

a) this speaks to the resistvity of the wire, but it is kinda random so should not be worried about because electricity is about electrons. they're just trying to trick you. neutron density has no effect really. electron density matters.

b) the length of the wire correlated to resistance. the longer the wire, the more the resistance. so, this would not increase power.

c) this one is sort of hard to say. in real wires, resistance increases with increasing temperature. thus, this would not increase power.

d) increasing conducitivity leads to a lower resistance, as resistivity is the inverse of conductivity. thus, this option decreases resistance. if we decrease resistance while keeping voltage constant, then the power will be maximized.

cheers

 

[BerkReviewTeach]

For a fixed potential difference across a resistive wire, the power loss through the resistor can be increased by increasing the:
a.) Neutron density in the wire
b.) length of the wire
c.)Temperature of the wire
d.) conductivity of the wire




Can anyone tell me the answer and explain it please??

Thank you!!

I concur with Pi-bond.

Choice A has nothing to do with anything.

Choices B and C both result in the resistance of the wire increasing while choice D results in the resistance of the wire dropping. Because choice D is exclusive, it should be picked as the best answer.

But the more correct answer is what Pi-bond posted, in that P = IV = I^2R. Anything that increases current will increase the power drain.

 

[Stephen2009]

I concur with Pi-bond.

Choice A has nothing to do with anything.

Choices B and C both result in the resistance of the wire increasing while choice D results in the resistance of the wire dropping. Because choice D is exclusive, it should be picked as the best answer.

But the more correct answer is what Pi-bond posted, in that P = IV = I^2R. Anything that increases current will increase the power drain.

If p=v^2/r and p= I^2R, how is it so that one eq. has R inversely proportional and another have it directly prop?

And per p= I^2R, wouldn't increasing both I and R increase the power drain? I'm confused.

 

[lorenzomicron]

I concur with Pi-bond.

Choice A has nothing to do with anything.

Choices B and C both result in the resistance of the wire increasing while choice D results in the resistance of the wire dropping. Because choice D is exclusive, it should be picked as the best answer.

But the more correct answer is what Pi-bond posted, in that P = IV = I^2R. Anything that increases current will increase the power drain.

I think the best way to look at is in with p=v^2/R, because in p=i^2*R, both i and r are variables, where as with the previous only R is variable. it makes sense to use the equation that has only one variable, to test the effect of the variable on power.

 

[PiBond]

i think of conductivity as the movement of electrons, which by definition is current. i used to use P = V^2/R when I was comparing the power dissipated by two different resistors when they were in parallel (and thus had the same voltage drop) but had different resistances. i'm not sure if we can think of resistance in the same way for this scenario, but i could be wrong

 

[fizzgig]

If p=v^2/r and p= I^2R, how is it so that one eq. has R inversely proportional and another have it directly prop?

And per p= I^2R, wouldn't increasing both I and R increase the power drain? I'm confused.

P=IV and V=IR

P = I (IR) = (V/I) V

now look at P=I^2*R. R goes up, P goes up. not that i have assumed in doing this that I is NOT changing as I mentally scale P using R! i'm increasing resistance but the current is staying the same?? i must have a larger motor pushing this current through - V has gone up. if you increase resistance and keep I constant, increasing V to whatever value is necessary to do so, then yes it makes sense that power goes up with R. you're sending the same charge/second by a 'tighter' spot.

now P = V^2/R. R goes up, P goes down. when I mentally do this scaling i assume V is not changing. ok, here i have the same driving force and a narrower pipe for the charge to get shoved through. so here it makes sense that as R increases, since i'm not 'pushing any harder', less current goes through, and power dissipation goes down.