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Discussion Session for PHY2049
Physics with Calculus 2 - Electromagnetism
Spring 2012
Week # 7
Problem 2
For the cube like circuit drawn on the blackboard, all the resistors have the
same value R. What is the equivalent resistance between points A and B?
Solution
Recall that the equivalent resistor Req between points A and B means that,
if there’s a current I from A to B (or vice-versa), and we were to measure the
potential difference between these two points VAB created by this current,
we have
VAB
Req =
(1)
I
Thus, imagine that the current I is injected to point A. By looking at the
figure, we see that from point A all the possible three roads that this current
can take look exactly symmetrical. Therefore, all three wires that start from
A must carry the same current, i.e. I/3 since the sum ought to sum up to I.
Take one of these roads. When you hit the next node, you see that the same
thing happens. The two possible ways to continue are again symmetrical
which implies that the current through those wires is 1/2 × I/3 = I/6. Now
you are one wire away from point B, which looks exactly the same as the
starting point. Therefore, the current through this wire is again I/3. Going
along the path just described, the potential difference VAB is then
VAB =
I
I
I
5
R + R + R = IR
3
6
3
6
(2)
Thus, the equivalent resistance is
Req =
VAB
5
= R
I
6
(3)
Problem 3
In the figure shown R1 = 2R, the ammeter’s resistance is zero and the
battery is ideal. What is the current in the ammeter in terms of ξ and R?
Solution
The most salient feature of this problem is to notice that, since the ammeter
has no internal resistance, the two lower resistors are in parallel, i.e. the
voltages across each of them are the same. This is also true for the two
resistors on top. Since the two lower resistors have the same resistance, the
fact that the have the same voltage implies that they also carry the same
current. Name this correct i. Since these two currents join at the bottom
and go to the battery on the left, the full current through the battery is 2i.
Using this fact, and computing the equivalent resistance which turns out to
be Req = 7/6R, we have
3ξ
i=
(4)
7R
Now call the voltage across the two resistors on top V . Since the voltage
across the two resistors in the bottom is iR, we have
ξ = V + iR
(5)
Using this and equation (4), we have
4
V = iR
3
(6)
Now that we have this voltage, we can compute the current through the
resistor in the top left as
itopleft =
V
2
= i
2R
3
(7)
Finally, defining the current through the ammeter iA to be positive going
to the left, current conservation tells us that
iA + itopleft = i
(8)
and we combine this with equation (7) gives
iA =
1ξ
7R
(9)