Electrical Resistance (Easy Question)

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Homework Statement


A person gains weight by adding fat - and therefore adding girth - to his body and his limbs, with the amount of muscle remaining constant. How will this affect the electrical resistance of his limbs?
a. the resistance will stay the same
b. the resistance will increase
c. the resistance will decrease

Homework Equations


[itex]R=\frac{\rho L}{A}[/itex]


The Attempt at a Solution



The answer is that the electrical resistance will decrease, but I don't get why. I get that the question hinted that area will increase ("adding girth"), which would make the resistance go down. But wouldn't resistivity also increase (fat has a higher [itex]\rho[/itex] than muscle)? This would make the resistance go up if it exceeded the effects of the increase in area.

Why is the answer that it will decrease resistance?

Thanks
 

Answers and Replies

  • #2
BvU
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Difficult to give a hint and not spoil the exercise. Think about the amount of muscle tissue remaining the same.
 
  • #3
haruspex
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I think you are only supposed to consider resistance along the limb. Consider the fat and muscle as two separate resistors. What is their relationship in the limb, electrical circuit-wise?
 
  • #4
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I think you are only supposed to consider resistance along the limb. Consider the fat and muscle as two separate resistors. What is their relationship in the limb, electrical circuit-wise?

So if I'm not mistaken you can some them,

[itex]\Sigma R = R_{muscle} + R_{fat}[/itex]

I think I can see why it would decrease now; just because more fat is being added doesn't make resistivity a bigger number. Fat was already present so the only number changing was A.

Thanks guys.
 
  • #5
haruspex
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So if I'm not mistaken you can some them,

[itex]\Sigma R = R_{muscle} + R_{fat}[/itex]
You are mistaken. That formula would mean the resistance increases.
You didn't answer my question: in terms of resistors in electrical circuits, what is the logical relationship between the fat along the limb and the muscle along the limb?
 
  • #6
BvU
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I know this is physics, not democracy. But I'm inclined to agree with coop: more fat means less resistance from the fat, whether the resistances are in series or in parallel. There is simply more fat for the current to flow through.
That's what I liked about this exercise.
Of course our entrails peeker is leading you towards one of the two and there I agree with him.

(Didn't the haruspices divine the future from this gory stuff ? Or do you mean that their future is our past?)
 
  • #7
haruspex
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I know this is physics, not democracy. But I'm inclined to agree with coop: more fat means less resistance from the fat, whether the resistances are in series or in parallel. There is simply more fat for the current to flow through.
If you consider the muscle and fat as concentric cylinders, fat on the outer, and the electrical contacts are on the outer surface of the outer cylinder, you'll see it's not that simple. If the muscle is a very much better conductor then on a lean limb the current flows through a little fat, along the muscle, then through a little more fat. Increasing the fat layer adds resistance in series on that path.
(Didn't the haruspices divine the future from this gory stuff ? Or do you mean that their future is our past?)
You're right that the time order doesn't fit. Much of my life was spent studying the entrails of crashed computer software to determine what had happened.
 
  • #8
BvU
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Poor coop.
Difficult to convince me. On our scale one conducting pad is under the left foot, the other under the right one. More fat means A increases, so R_fat decreases.
Your limbs and electrodes are a bit tougher, but L and A both increase linearly so I have at worst the same resistance ?
(The ? shows I'm not all that certain. Any biomedics around?)
 
  • #9
haruspex
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Poor coop.
Difficult to convince me.
I'm not arguing that Coop's answer was wrong; in my first post on this thread I suggested we are supposed to consider only conduction along the limb, and that's still my position. But Coop's later post proffered the formula for resistance in series, which, if appropriate, should have led to the wrong conclusion.
Then you posted
more fat means less resistance from the fat, whether the resistances are in series or in parallel
which makes no sense to me. The only way I can understand that is if your model for the resistances being in series is a solid cylinder of fat joined end-to-end with a cylinder of muscle. My limbs do not follow that topology.
 
  • #10
BvU
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@haru: With L/A in the expression for R I see a tendency for the denominator to grow faster than the numerator. You know, a cartoon-like bloating of a lumpy fat person. But I concede that omnidirectional growth wasn't in the OP.
 
  • #11
Curious3141
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What's the difficulty here? Fat is subcutaneous, and therefore superficial to muscle. The simplified model here is that of a solid cylinder of muscle (low resistivity) immediately surrounded by a hollow cylindrical shell of fat (high resistivity). The electrodes can reasonably be assumed to be sited on the skin, which makes the current go lengthwise along the cylindrical segment. So the two resistors are in parallel. If one of the resistors changes in value, the overall resistance will also change in the same direction.

So all that remains is to figure out how the lengthwise resistance of the hollow cylindrical fat shell changes. It should be quite easy to demonstrate that it decreases when the outer radius increases while everything else remains constant. Hence the overall resistance decreases.

Note that in real life Bioelectrical Impedance Analysis (BIA), an increased body fat percentage is associated with an increased resistance. This is probably because the configuration of the resistors is broadly different - they're measuring from one foot pad to the other. With increased subcutaneous fat, there will be a greater fat pad thickness in series with the lower resistivity lean mass, so an overall higher resistance.
 

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