Download Quiz 4 - Rutgers Physics

Solutions to Quiz 4
May 1, 2010
1
Lightning and the Tree
Lightning of current I is traveling down the surface of a tree. As shown in the
figure, a portion of the lightning jumps a distance d over the air, and travels
down a person of height h. Another portion travels through the air, along
the surface of the tree. If the person has negligible resistance compared to
that of the air, find the current traveling through the person. Your answer
should be expressible in terms of I and d/h.
1.1
Solution
From the circuit diagram, we see that the current I splits up into two parts.
For the purposes of the problem, we will model these two parts as wires with
equal cross section area but of different lengths.
The part that travels through the air can then be thought of as a wire
with resistance
R1 = ρair ∗ d
(1)
where we have lumped the cross sectional area into the constant ρ.
The other part travels down the air along the tree.
R2 = ρair ∗ h
(2)
We want to find the current i1 through the person. Since the current
splits in two parts, the other wire has current I − i1 . Now that you have the
current in terms of one unknown, and you know the resistances, you could use
Kirchoff’s law. But there’s a slightly easier way. Note that the voltage drop
1
across both parallel wires has to be the same (since the ends are connected),
so that
V1 = V2
(3)
or
i1 R1 = (I − i1 )R2
or
(4)
i1 = I
R2
R1 + R2
(5)
i1 =
I
1 + d/h
(6)
which simplifies to
so if d/h is small, the person will get a sizable zap!
2
An Electrical Short
A 10 km long underground cable extends from west to east and consists
of two parallel wires, each of which has resistance 13 Ω/km. An electrical
short develops at a distance x from the west end when a conducting path of
resistance R connects the wires. The resistance of the wires when measured
across the west ends is 100 Ω, and when measured across the east ends is 200
Ω. Find x and R.
2.1
Solution
An electrical short means that all current travels through that channel. If
we measure the current along the west ends, then our circuit will consist of
the two left branches of length x and the short circuit. The total resistance
in this circuit is
100 = 13x + R + 13x
(7)
Similarly, when the resistance is measured across the east ends, our circuit
consists of the two right branches of length 10 − x and the short circuit. The
total resistance in this circuit is
200 = 13(10 − x) + R + 13(10 − x)
2
(8)
Adding the two equations,
300 = 260 + 2R → R = 20Ω
and
x=
80
m
26
3
(9)
(10)