Register to reply

Temp flow and electrical flow

by TSN79
Tags: electrical, flow, temp
Share this thread:
TSN79
#1
Jun7-05, 07:23 AM
P: 357
1)
Let's say you have two items that lie next to eachother sharing a surface. A is 10 degrees C, and B is -10 degrees C. If this system sits there until A is down to perhaps 5 degrees C, does that automatically mean that B is now -5 degrees C? If not, why is it so? It would probably depend on whether or not both A and B were made of the same material?

2)
One often says that electricity takes the way of least resistance, but if a wire splits up to form let's say 5 parallel wires of varying resistance, doesn't the electricity need to flow through all 5 in order to find out which one is the easiest way? And if it then had some kind of "intelligence" it could - based on that - then choose the way of least resistance?
Phys.Org News Partner Physics news on Phys.org
New complex oxides could advance memory devices
Nature's designs inspire research into new light-based technologies
UCI team is first to capture motion of single molecule in real time
Daminc
#2
Jun7-05, 07:47 AM
P: 157
I can't remember all the formula that this invloves (it's been 17-18 yrs since I did the theory).

I=V/R

I believe it's something like

R1 = first resistor
R2 = second resistor

I1 = current going through the first resistor
I2 = current going through the second resistor

I1 = Vtotal/R1
I2 = Vtotal/R2

It's kind of like a big pipe of water leading to 2 smaller pipes (one bigger than the other). The smaller pipes will allow different amounts of water though depending how wide the pipes are (how much resistance they offer). A smaller pipe will give a greater resistance and so there in less water (current) allowed through.

p.s.
Ohm's Law states that the current in a conductor is directly proportional to the potential difference and inversly proportional to the resistance provided the temperature and pressure remain constant.
Daminc
#3
Jun7-05, 07:55 AM
P: 157
Let's say you have two items that lie next to eachother sharing a surface. A is 10 degrees C, and B is -10 degrees C. If this system sits there until A is down to perhaps 5 degrees C, does that automatically mean that B is now -5 degrees C? If not, why is it so? It would probably depend on whether or not both A and B were made of the same material?
It would depend on want they were contained in. If they were in a sealed container so that none of the energy would be dispersed then the heat lost from the 10C object would be directly transfered to the -10C intil both were the same tempurature.

However, if the objects weren't contained then the energy from the 10C object will heat up the surrounding area as well as the -10C object whereas the -10C would have to wait a little bit before it could absorb the energy to heat itself up.

I'm sure there are formulas that would explain it a lot better than I.

ZapperZ
#4
Jun7-05, 08:29 AM
Emeritus
Sci Advisor
PF Gold
ZapperZ's Avatar
P: 29,243
Temp flow and electrical flow

Quote Quote by TSN79
1)
Let's say you have two items that lie next to eachother sharing a surface. A is 10 degrees C, and B is -10 degrees C. If this system sits there until A is down to perhaps 5 degrees C, does that automatically mean that B is now -5 degrees C? If not, why is it so? It would probably depend on whether or not both A and B were made of the same material?
You can actually do this analytically. Figure out the amount of energy given off by A when it goes from 10 C to 5 C, which would be

[tex]\Delta Q_A = m_A c_A \Delta T_A[/tex]

where m and c are the mass and specific heat capacity of A. This energy is then transfered to be absorbed by B:

[tex] \Delta Q_B = m_B c_B \Delta T_B[/tex]

If no heat loss anywhere,

[tex] \Delta Q_A = \Delta Q_B [/tex]

[tex]m_A c_A \Delta T_A = m_B c_B \Delta T_B [/tex]

What this means is that only if both masses and specific heats are identical to each other would the change in temperature be the same, meaning A would have lowered by the same amount that B has increased. If any of the other quantites are different, they need not change temperatures by the identical amount. It also means that under that situation (m and c are different) left to their own devices, the equilibrium temperature that both will reach will not be the "middle" temperature, which is 0C in your example.

Zz.
TSN79
#5
Jun7-05, 08:48 AM
P: 357
So what you're saying is that even through the wires with the highest resistance there would be some flow, just less and less the more resistance? I've always thought that only the wire with the smallest resistance would conduct, and the others would be "ignored"...thanks! Thanks to you too ZapperZ!
Gokul43201
#6
Jun7-05, 11:08 AM
Emeritus
Sci Advisor
PF Gold
Gokul43201's Avatar
P: 11,155
Quote Quote by TSN79
So what you're saying is that even through the wires with the highest resistance there would be some flow, just less and less the more resistance?
Yes, that's right.

I've always thought that only the wire with the smallest resistance would conduct, and the others would be "ignored"...thanks!
This is not an unreasonable thought. However, in reality, the "cost" of putting a lot of current through the smaller resistance is greater than that of diverting a small portion of this current away through a greater resistance. The "cost" here (something known as an 'action' in the general case) is the heat produced by the current flow.

Consider a current I, making a choice between two resistors R1 and R2. Let's say that a current I1 flows through R1 and the rest, namely I - I1, flows through R2.

The heat produced at each conductor is given by Joule's equation, [itex]H = I^2R[/itex]. So, the total heat produced among the two resistors is :

[tex]H = I_1^2 R_1 + (I - I_1)^2R_2 = I_1^2(R_1 + R_2) - 2I_1(IR_2) + I^2R_2 [/tex]

We want this total heat to be minimum, and we find the value of I1 that minimizes this. Basic calculus gives us :

[tex]\frac{dH}{dI_1} = 2I_1(R_1+R_2) - 2IR_2 = 0 [/tex]
[tex]=>\frac{I_1}{I} = \frac {R_2}{R_1+R_2} [/tex]

So, the current through the R2 is given by
[tex]I_2 = I - I_1 = \frac {IR_1}{R_1+R_2} [/tex]

Thus : [itex]I_1/I_2 = R_2/R_1 [/itex]

In other words, the current through each resistor (in a parallel setup) is inversely proportional to the value of its resistance.


Register to reply

Related Discussions
Transition from pipe flow to open channel flow Mechanical Engineering 6
Matlab in applying Finite Difference for Temp. distribution/ rate of heat flow Math & Science Software 0
Can an inviscid flow rotational? Potential Flow? Mechanical Engineering 2
Need help with flow physics Introductory Physics Homework 2
Confusion on electrical current vs charge flow Atomic, Solid State, Comp. Physics 3