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Transcript
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)