Electric Current - What Am I Doing Wrong?

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Homework Help Overview

The discussion revolves around calculating electric current through a modeled cylindrical tube representing a person's arm and chest when a potential difference is applied. The problem involves concepts of resistivity, resistance, and current in the context of electric circuits.

Discussion Character

  • Mathematical reasoning, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between resistivity, resistance, and current, questioning the calculations and unit conversions involved. There is confusion regarding the conversion of area from cm² to m² and its impact on the final current calculation.

Discussion Status

Participants are actively discussing the calculations and unit conversions. Some suggest that the original poster may need to check their unit conversions, particularly regarding the area, while others confirm the calculated current value. There is no explicit consensus on the correct answer, but multiple interpretations of the problem are being explored.

Contextual Notes

There is a noted discrepancy in unit conversions, particularly with the area measurement, which may be affecting the calculations. The original poster is receiving feedback from a computer system regarding unit correctness, indicating potential issues with how the answer is formatted or expressed.

evilempire
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By modeling the arm and the chest to be a cylindrical tube with a total length 2.0 m, cross-sectional area 10 cm^2, and resistivity 1.5 ohm*m, you can calculate the current in amperes through the person when a potential difference of 110 V is applied across the two hands. Assume that the current flows only through the modeled cylindrical tube. What is the current flow through the body?


The equation for resisitivity is R=rho*(L/A), and the equation for current is I=V/R. So for the R equation I plugged in the numbers, changing cm to .1 m, and got 30 ohms. I then divided 110 volts by 30 and got 3.7 (rounded to two sig figs), and my answer is not being taken as correct. The computer feedback system is saying 'check your units'. What have I done wrong?

Thanks.
 
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One idea I had - is the proper answer done such that I=V/((rho*L)/A), which would make I equal to 367 A?
 
the cross section area is in cm... and the rest of your problem is in SI ...
 
vincentchan said:
the cross section area is in cm... and the rest of your problem is in SI ...

Yeah, so I converted it to m, 10 cm = .1 m.
 
and do you still got 367... if you do... change battery for your calculator, or do the calculation by hand...
 
vincentchan said:
and do you still got 367... if you do... change battery for your calculator, or do the calculation by hand...

So it's not 367 or 3.67? Hmm...
 
it is 3.67
 
vincentchan said:
it is 3.67

That's what I calculated initially, but the computer program won't take it, which led me to believe there was error in my calculation. Thanks for the help, I greatly appreciate it. I'm not sure what I'm going to do about answering the problem, but it's nice to know I had it right to begin with.
 
maybe you have to enter 3.67A... since the computer said check your unit
 
  • #10
vincentchan said:
maybe you have to enter 3.67A... since the computer said check your unit

It says 'A' outside the text box to the right, so I'm pretty sure they just want a value.
 
  • #11
10 cm^2 = .o1 m^2...did you do the convertion correctly?
 
  • #12
vincentchan said:
10 cm^2 = .o1 m^2...did you do the convertion correctly?

It does? I thought 1 cm was .01 m, so 10 cm would be .1 m.
 
  • #13
but 10 cm^2 = 10 (.01 m) ^2 = 10 * .001 m^2 = .01 m^2
 

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