blue cobra wrote:
Could someone please explain the wye and delta arrangements? They seem common on old Circuit Lab tests. Even if they are not those specific arrangements, I'm very insure of how to go about solving for things like that. A fine example would be in Section 2 of this test
, the answer key for which can be found here
. I have no idea how they got those answers. I'm also very unsure about questions about voltages between terminals. Any help in these areas would be appreciated.
And there aren't going to be capacitors in this event, are there?
Wye and Delta generally refer to a three-terminal connection of resistors. In the Wye arrangement, you have three resistors that tie together at a central point. The path between any two terminals takes you through two resistors in series.
The Delta arrangement has resistors directly between adjacent terminals. Now you want to think about the paths between the two terminals you are testing. There are two paths in parallel - one that takes you through the resistor between the two, and one that takes you through the other two resistors in series. Thus, apply the series and parallel methods of calculating your resistance to get your result.
This is a pretty good explanation of the different configurations.http://www.play-hookey.com/dc_theory/delta_wye.html
You can apply this to problems like those in the test you linked. For example, let's look at Section 2, Problem 1. What are the paths from A to B? There is the simple direct path through R1, at a resistance of 5k. Then there is the path from B to C and then C to A and C to D to A (the Delta configuration). So let's break that down.
What is the resistance between points C and A? This is the delta configuration, we further break it down and ask what the paths for current are. We get through R5 (25k) and the series of R2 and R3 (15k + 20k = 35k). So the resistance between C and A is the parallel resistance of 25k and 35k = (25k*35k)/(25k+35k) = 14.58k.
Now we want the resistance between B and A through the "long" path. That's simple - it's a series of R4 plus the delta arrangement, which we found to be 14.58k. So it equals 10k + 14.58k = 24.58k.
Now we've established the resistances of our two parallel paths from A to B. They are 5k, through R1, and 24.58k, going the other way. The parallel resistance then is (5*24.58)/(5+24.58) = 4.15k, which checks out with the answer key.
You would use the same method to calculate the resistance across AD, BC, and CD. AC would be a tad different - but the same principles still apply.
You have not two, but three paths for the current between AC - directly across R5, and through the two other branches, R1+R4 and R2+R3. So let's find the resistance of those branches, which is done by simply adding the resistors as they are in series. R2+R3 = 15k + 20k = 35k and R1+R4 = 5k+10k = 15k. Now we have three resistances in parallel - 25k (R5), 35k, and 15k. The parallel resistance is 1/(1/25+1/35+1/15) = 7.39k, also consistent with the answer key.
For questions 5 and 6, if R2 is shorted, just assume it's not there. Calculate BC as earlier but don't add R2 to R3, as it's shorted - R3 will be the only resistance through that "branch". If R1 is open, no current will flow across it, so the R1-R4 branch will not be part of the circuit - thus just calculate the Delta of C-D and C-A-D.
To calculate voltages, apply Ohm's and Kirchoff's Laws. You have resistance. Know that for each parallel branch, voltage is constant (thus the total voltage flows down each branch). In series, you know that current is constant, so calculate current for the total resistance of that branch (add the series resistors together). That current is the current through each resistor in series in the branch. Once you get resistance and current in a resistor, you can apply Ohm's Law and solve for voltage.