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Transcript
Electromagnetism
G. L. Pollack and D. R. Stump
What is the resistance of this network?
Four stepped exercises.
Exercise 1. Consider the unit cube, shown in Fig. 1(a). Suppose
each edge is a conductor with resistance r. What is the total resistance
between diagonal corners of the cube?
General Strategy The surefire way to find the total resistance R
between two terminals in any network is to let current I enter at one
terminal and exit at the other. Use Kirchhoff’s laws (about which
more below) together with any symmetries available to find the current
through each of the conductors of the network. Finally, calculate the
total voltage change ∆V for any path between the original terminals
and obtain R from R = |∆V |/I.
Kirchhoff ’s first law states that the current entering any junction
equals the current leaving it. This law is a consequence of conservation
of charge.
Kirchhoff ’s second law states that the net emf, i.e., the algebraic
sum of the voltage changes, around any closed path in the circuit is zero.
This law is a restatement of the conservativeness of the electrostatic
field—a consequence of the field equation ∇× E = 0 for electrostatics.
Exercise 1 has sufficient symmetry so that only Kirchhoff’s first
law need be applied. What is the resistance R between the diagonal
corners, say, (0, 0, 0) and (1, 1, 1)?
Let current I enter the network at (0, 0, 0) and leave at (1, 1, 1).
From symmetry it is clear that three equal currents, each I/3, flow away
from (0, 0, 0) along the x, y, z axes, repesectively. When these current
reach a corner, they divide symmetrically into two equal currents. Thus,
upon reaching the corner (0, 0, 1) the current along the z axis divides
into two currents, each I/6, one from (0, 0, 1) to (0, 1, 1) and the other
from (0, 0, 1) to (1, 0, 1). The currents along the x and y axes behave
similarly. Consider next what happens at the junction (0, 1, 1): Two
b
equal currents bjI/6 and kI/6
flow into it so that current biI/3 must flow
out from (0, 1, 1) to (1, 1, 1). The cube with all currents is shown in
Fig 1(b).
Finally then for the path (0, 0, 0) → (0, 0, 1) → (0, 1, 1) → (1, 1, 1)
we have the voltage drop
I
I
I
∆V = − r − r − r = IR
3
6
3
so that R = 5r/6.
Exercise 2. Consider now the planar array of 12 resistors, each with
resistance r, shown in Fig. 2(a). What is the total resistance R between
the corners A, at (−1, 1), and A0 , at (1, −1)?
In solving this exercise, Kirchhoff’s first law and symmetry will
take us only part way. We will also need to apply the second law. Let
current I enter the network at A and leave at A0 . From symmetry it is
clear that two equal currents I/2 flow along AB and AD. But it is not
clear, i.e., there is not enough symmetry, to determine how the current
divides at junction B. Therefore call I1 the current that flows along
BE and, because Kirchhoff’s first law must be satisfied, we must have
I/2 − I1 flowing along BC and CF . Similarly the current along AD
divides into currents I1 along DE, and I/2 − I1 along DC 0 . Figure 2(b)
shows the array with labeled currents in each resistance of the network.
Note the symmetry at junction E—two equal currents I1 flow in and
two currents I1 flow away.
Now to determine the unknown current I1 , apply Kirchhoff’s second law to a counterclockwise loop around F CBEF . Counting the
voltage changes against the current direction as positive and those in
the current direction as negative, we have
I
I
− I1 r +
− I1 r − I1 r − I1 r = 0,
∆V =
2
2
from which I1 = I/4.
Finally then for the path A → B → C → F
drop is
I
I
I
I
I
∆V = − r −
−
r−
−
r−
2
2 4
2 4
so that R = 3r/2.
→ A0 , the voltage
I
r = IR,
2
The next two exercises refer to the 24 resistor array shown in Figure
3. Each segment has resistance r.
Exercise 3. What is the resistance between terminals F and F 0 ?
(Hint: First convince yourself that no current flows past the corners A
or A0 . However, your solution should also show that I/14 flows past
the corners D and D 0 .)