Parallel Electronically Signed Routing Made Easy

Parallel electronically signed routing

hi so last week we had a look at some serious parallel network problems which were a little bit more complicated in as much as they weren't laid out as you would normally expect them to be this is a sample solution for problem 9 and the reason I've chosen this is because there's a bit of a trick in this question so what we have to do is solve the circuit following the table okay first thing we have to do is to try and simplify the circuit a bit so that we can actually break it down more easily into its series and parallel their components where I suggest you do this is by starting off with the power supply and you don't really have to worry about because it's in the center here you can actually move it so that it's in the on the left hand side so let's start off by doing that we have our power supply at 32 volts and we have a positive and negative connections to it but if we look at the diagram above let's start off with the easier side and we look at our 4 initially and you can see that from the positive side of the curve the power supply he goes to one end of our four on a negative side through this link it goes to the other side so power 4 sits directly across the power supply so let's draw that in so now we have power 4 so the other one we have on this side of the power supply is our 5 and this is a bit of a strange one because if you look at this this is our negative side of the power supply which has a direct link to here so our 5 has in effect 0 volts at this side and it also has 0 volts at this side so our 5 sits underneath this part of the diagram so here we have our 5 and in effect we have 0 volts at this end and we have 0 volts at this end so our 5 has no voltage drop across it and it has no current flowing and as it has no current flowing through it then in actual fact there'll be no power anticipated in that resistor so now we've got the right-hand side of this diagram are sorted out let's look at the left-hand side and what's the best place to start well the top end the positive side there's actually two resistors come off it and the bottom end there's just the one okay and the bottom one that goes to this node here which he said by both of those let's have a look at putting our two in next and it goes from the negative side of the power supply across our two comes across from the negative side of the power supply and that this node here joins up to both those resistors which I'll connect up to the positive side of the power supply to finish off we can stick those two in so now we have a simplified layout showing how the resistors fit together in the various series and parallel networks now it makes it much easier to start simplifying I would recommend when you're doing this that you do draw the diagrams as you go through the simplification process because if you don't do that you're gonna get in a tangle with it okay so first thing we can do is look at these two resistors here which are two resistors in parallel and we can simplify by sticking one resistor in to replace both r1 and r3 what I've actually done is I've replaced r1 and r3 by a single resistor and r5 because we've determined our v has no effect on the circuit at all I've actually omitted that so r1 3 it's going to replace r1 and r3 and because those two are in parallel we use the formula 1 over r1 3 equals 1 over r1 plus one over r3 and we can then substitute in the values r1 is 790 ohms r3 is eight point six times 10 to power 3 so just by calculating the answer 1 over all one two three equals one point three eight two one times ten to the minus three I remember that's 1 over R 1 3 so we need to invert this to get the answer for R 1 comma 3 so 1 comma 3 is equal to 720 3 point 5 3 6 ohms now we've done that we can look at our circuit and see what we can simplify next and these two resistors here are in series so let's replace both of those with one effective resistor R 1 comma 2 comma 3 so we've replaced R 1 3 and R 2 with one resistor with an effective resistance so R 1 2 3 is equal to R 1 3 plus R 2 now I know what R 1 3 is and we know what R 2 is so we can substitute those into the formula like so and then simplify it to find the answer so R 1 2 3 is equal to one thousand seven hundred and twenty three point five three six poems so now we just have the two resistors in parallel the effective resistance R 1 2 3 which replaced the rest of the network that we've taken out and our four two resistors in parallel so let's replace that with one which we're going to call our 1 2 3 4 so if we use 1 over R 1 2 3 4 is equal to 1 over R 1 2 3 plus 1 over R 4 and we can find the effective resistance of the whole circuit so let's substitute in the values substituting in the values we get this then simplifying it we get 1 over R 1 2 3 4 equals this value here and remember this is 1 over so we have to invert it to find the final effective resistance r 1 2 3 4 equals nine hundred and sixty six point four one nine ohms so we worked our way down now and we've simplified this circuit as much as we can so now let's work out what my T the current flowing through circuit start off with V equals that IR we're going to transpose that for I so the total current is equal to V which is 32 volts divided by R on to 3 for the effective resistance of the circuit substituting in the values 32 divided by nine six six point four one nine gives the total current flowing through the circuit as naught point naught 3 3 1 1 amps or 33 point 1 1 milliamps now we can move back up a circuit using that information to help us find some more values in this part it's moving back up the circuit we know that the voltage across R 4 and R 1 2 3 is 32 volts so we can actually work out now the current flowing through R 4 and the current flowing through the rest of the network so let's have a look at that and see what we can work out from that to calculate the current flowing through our 4 we know the voltage we know the resistance so we can just use the formula I 4 equals V over R 4 let's substitute in the values and see what we get I 4 is equal to 32 over 22 kilo ohms and we can simplify that so we get I 4 is equal to point zero one four five amps or 14 and a half millions we can do exactly the same for the current flowing through our one two three network using the formula I 1 2 3 equals V over the effective resistance R 1 2 3 let's substitute in the values and see what we end up with V is 32 volts the effective resistance of our one two three is one seven two three point five three six homes calculating that out we end up with 18 point five seven milliamps now this stays just to make sure we've made no mistakes what we can do is to just do a quick check and we know that the current flowing through the circuit or split here so you have I for going down there and I one-two-three going down there so I total should equal I four plus I 1 2 3 and if you add those together 14 and a half plus eighteen point five seven we should get approximately 31 point one one milliamps which we do so that's good having worked out these two values of current we can now feed those back into the next diagram up and we know now I four and we know I 1 2 3 and I 2 1 2 3 is the same current flowing through the two series resistors so I 1 2 3 flowing through R 2 I 4 flowing through our 4 we can now work out the voltage drop across R 2 so that gives us our fathers directly for the table but also we can work out the voltage drop across R 1 3 which we will need to calculate the final bit of the problem to calculate the voltage drop across R 2 we're going to use Ohm's law so let's have a look at the formula and the substitution the voltage drop across R 2 is equal to the current i-12 3 multiplied by r 2 which is 1 kilo ohms giving us a value the voltage drop across R 2 is 18 point five seven volts for the voltage drop across our one three Network we have the current I 1 2 3 and the effective resistance of our 1 3 which we calculated earlier as 723 point five three six so substituting those in a voltage drop across R 1 3 as fourteen point four 36 and that's a quick check we know that the voltage drop between there and there across that series with a resistor network should be 32 volts and if you add those together that's what you'll actually get so we know the voltage drop across one on our three you know the values if I want on our three so we can now work out the current flowing through r1 and r3 by using the formula I 1 is equal to the voltage drop across R 1 divided by the resistance R 1 so substituting in the values I 1 comes out at 17 point o 4 milliamps and do the same thing for I 3 so doing exactly the same thing I 3 comes out at 1.5 65 milliamps so at this stage we've gone straight away down we've worked out the voltage drops across the various components we've also worked out the current flowing through them so we can now go back and stick those values into the table so we can substitute in to the table all the values that we've calculated so far and we're nearly there so what we have to burn the table now is the power dissipated in the various resistors the formula that springs to mind initially is the power dissipated equals the voltage drop across the resistor times the current flowing through the resistor however there's a slight problem with this because when it comes to the values that we have in the table we calculated all the values of I and the majority of values of V as well so when you're doing the power calculations whenever possible it's best to use known values now the problem is the known values we have are for the resistance so we need to get rid of the current in the formula and use resistance so that we're using as many known values as possible now we know that the current flowing through the resistor is equal to V over R so we can substitute that into the power formula to get rid of I which leaves us with power equals voltage squared divided by resistance so what we're going to do now is use that formula for all these resistors and fill in the table now I don't intend to go through those one by one because you're quite capable of doing that yourself okay so I'll just substitute in the values and see what we get these other values that calculated using P equals V squared over R if you compare that with the answer sheet you'll see that there are some slight discrepancies between the two but this will be down to the rounding errors that we've used throughout the calculations so hopefully you can see that by using this method of simplifying the circuit to start off with and then reducing the number of resistors in the network so you haven't only for one effective resistance at the end then substituting in values as you go back up again that this makes the problem much more straightforward and you're much less likely to make mistakes so it's not difficult it's just time consuming and again draw the diagrams