InterviewSolution
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InterviewSolution
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Figure 26-11 shows a swimmer at distance D=38.0 m from a lightning strike to the water, with current I=78kA. The water has resistivity 30 Omega.m, the width of the swimmer along a radial from the strike is 0.70m, and his resistance across that width is 4.00kOmega. Assume that the current spreads through the water over a hemisphere centered on the strike point. What is the current through the swimmer? Swimmer at a distance of 38m from where lightning strikes water. |
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Answer» Solution :Since the current spreads over the hemisphere, the current density at any given radius r from the striking POINTS is `J=I/2 pir^(2)`. From Eq. 26-10, the magnitude of the elctric field at a distance r is `E=p_(w)J=(p_(w)J)/(2pir^(2))," where "p_(w)=30Omega`. is the RESISTIVITY of water. The potential differnce a point at radical distance D and a point at `D+triangler` is `E=-int_(D)^(D+triangler) Edr. =-int_(D)^(D+triangler) (p_(W)I)/(2pir^(2)) dr` `=-(p_(W)I)/(2pir) ((1)/(D+triangler)-1/D)=-(p_(W)I)/(2pir) (triangler)/(D(D+triangler))` which implies that the current ACROSS the swimmer is `i=(|triangleV|)/(R)=(p_(w)I)/(2pir)(triangler)/(D(D+triangler))` Calculations: Substituting the VALUES given, we obtain `i=((30.0 Omega.m) (7.80 xx 10^(4) A))/(2PI (4.00 xx 10^(3) Omega)) (0.70m)/((38.0m) (38.0 m+0.70m))` `=4.43 xx 10^(-2)A` |
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