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xiantom
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Homework Statement
A man was found laying unconscious beneath high tension power lines conveying electricity. He was conveyed to the hospital where he was pronounced dead upon arrival. An autopsy showed that he died from cardiac arrest. The investigation finds that there was indeed an electrical fault at the time of the incident, for 1.0s, a current of 100 amperes leaked onto the ground from a vertical conducting rod whose tip was found below the ground 10m away from the man. The resistivity of the moist ground was about 100Ωm.
1. Under any suitable approximation obtain a reasonable estimate of the potential difference between the man's feet.
2. It is reasonable to suppose that the current went up one leg of the man, across the torso and down the other leg. The commonly accepted value for resistance of a leg is 300Ω and that of the torso is 1000Ω. From these data, estimate the current across the victim's torso.
3.Using the expression for E(r), obtain the potential difference from the tip of the rod and a point at distance r. Take the tip of the rod to have radius a. If the integral is a problem, you can get the answer by noticing the similarity between E(r) and a well known charge configuration.
4. Currents ranging from 0.10A and 1.0A across the torso can trigger fibrillation. Did the accident kill the man?
Homework Equations
[itex]A_(sphere)=4\pi r^2[/itex]
[itex]R=ρl/A[/itex]
This is one of the equations I calculated for one of the earlier questions:
[itex]E_(r)=Iρ/(\pi r^2)[/itex]
The Attempt at a Solution
1. I was told by my physics teacher that current travels in a hemisphere since we are calculating it in ground, where it spreads out. I based this on that.
Assuming the difference between the man's feet as 1m. 1 foot is 10m away, the other is 11m away from the rod.
[itex]V=IR=Iρl/A[/itex]
I assumed that for l, I can use the distance between the man's foot and the rod as it essentially is the distance between one end of the resistor compared to the other if it was in a wire.
My calculations:
[itex]v_10=100A*100Ωm*10m/(2*10m^2*\pi)=159V[/itex]
[itex]v_11=100A*100Ωm*11m/(2*11m^2*\pi)=145V[/itex]
[itex]ΔV=V_10-V_11=14V[/itex]
2.
Total Resistance: 300Ω+300Ω+1000Ω=1600Ω
[itex]I=V/R=14V/1600Ω=0.00875A[/itex]
3.I have no idea how to do this. Here's my attempt at the problem.
[itex]E=ΔV/R=ΔV/(ρa/A)=VA/ρa=(V2\pi r^2)/ρa[/itex]
[itex]E=Iρ/2\pi r^2[/itex]
Now I solve these as equal equations.
[itex]Iρ/2\pi a^2=V\pi a^2/ρr[/itex]
[itex]Iρ=V\pi ^2a^2/ρr[/itex]
[itex]Iρ^2r=V\pi^2a^2[/itex]
[itex]V=Iρ^2r/ \pi ^2a^2[/itex]
4. Since 0.00875A<0.10A, he didn't die from the accident.
I'm just wondering if I did these questions right as I couldn't find anything on resistances that are outside wires, nor electric fields on current electricity.