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capa 8 - Phys 1120 CAPA#8 solutions 1 The key thing to work...

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Phys 1120, CAPA #8 solutions. 1) The key thing to work out for yourself is that if you have two resistors in parallel, and one of them is decreased, the effective R of the pair also goes DOWN. (I find it helpful to exaggerate - if one of them becomes a superconductor, it's like a short circuit, all the current goes that way, there's now zero R for the pair as well, effective R went down) Similarly, if R of one of them goes up, the effective R of the pair goes up. I know this can be a little confusing, so think it through (maybe do some calculations of specific cases? Whatever it takes for you to understand why and how this works!) Look back at the first circuits Tutorial, it might help you with this intuition. A) If the voltage V increases, the total current from the battery INCREASES. After all, all those resistors have some R(equiv), and V=I*R(equiv). If V goes up, I out of the battery goes up. B) If R6 increases, the power output from the battery DECREASES. If R6 goes up, the overall R(equiv) goes up, because R6 is in SERIES with the others. Bigger overall R means overall power=V^2/R goes down. (Or equivalently, bigger overall R means less current comes out of the battery, and power = V*I. V is FIXED for the battery,so less I means less power) If you're not convinced, exaggerate. If R6 goes to infinity, no more current can flow through it, and so no current can flow ANYWAY because it's in series, it's "blocking up the works". No current from a battery means no power at all. C) If R1 increases, the current through R6 DECREASES. Why? If R1 goes up, the R of the1-2 pair increases. (That was my discussion at the start of the problem!) Since that PAIR is in series with the rest, if the R of the pair increases, the whole effective R of the circuit increases. And, just like the last one, if R of the circuit goes up, the current coming out of the battery goes down. (The current through R6 is the same as the current of the battery, do you see that?) D) If R1 increases, the current through R3 DECREASES. The logic is just like the previous part. There is less current flowing everywhere. Less current coming into the "3-4-5" part means less for each. E) If R3 increases, the current from the battery DECREASES. Similar logic to part C. If R3 goes up, the effect R of the 3-4-5 pair goes up. (Convince yourself, it was true for 2 in parallel, it's still true with 3 in parallel!) So, overall R is up, which means overall current goes down. F) If R3 decreases, the current through R4 DECREASES. This one I find the trickiest to think about. But look, suppose R3 goes down all the way to 0, i.e. it becomes an ideal wire. Then, all the current will flow through R3, and there will not be ANY VOLTAGE DROP across R3 any more (there's never a voltage drop across an ideal wire!) So, there's no voltage difference between the top and bottom of R3, which means no voltage difference across R4: no current flows through it at all. (It's all going through R3) Here's a rather different way to get the right answer: if R3 decreases, overall circuit resistance decreases, which means there's more current coming out of the battery. That means a bigger
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