| Electronics > Beginners |
| Dumb question about resistors in series/parallel |
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| BravoV:
--- Quote from: Brumby on May 02, 2018, 06:49:50 am ---[That sort of presentation does make it easier to 'get it' ... and once you do understand, your ability to see these things will improve as your experience grows. --- End quote --- Yes, lots of practices and once you get enough experience, handling circuit like that can be easily solved by virtually moving the nodes around in the brain. Of course, when I started this method, I used to sketch a lot on paper, but it will become natural once practiced enough. That circuit of yours took me for about maybe just 10 seconds to come out that the whole resistors network equal to 10 Ohm, as there are multiple parallel resistor pairs are using identical values and integer, hence the instantaneous result as there is no need to go into fractional calculation that may take a while. |
| Brumby:
Yes - the values do work out very neatly ... and it is quickly solvable. In the real world, however, it is almost never that easy. |
| Zero999:
--- Quote from: Kirr on May 02, 2018, 03:27:20 am --- --- Quote from: TheN00b on May 02, 2018, 02:57:12 am ---Is that why, when simplifying circuits, we simplify the parallel elements into a single resistance and then add all of the resistors together like it were a series to find a single equivalent resistance? --- End quote --- We only add them together if they actually are in series. Some circuits can't be simplified by just parallel and serial replacements. The simplest example is a bridge circuit. --- End quote --- That can still be solved using simple resistors in series/parallel calculations, along with Thevenin's theorem and Kirchhoff's law. It can be treated as two potential dividers, which are equivalent to two voltage sources, each with a series resistance equal to both resistor values in series: Thevenin's theorem. Remove R5 from the circuit. Calculate: V(pd1_open) and V(pd2_open) V(pd1_open) = V1×R2/(R1+R2) = 5×100/(100+100) = 2.5V V(pd2_open) = V1×R2/(R3+R4) = 5×100/(100+200) = 1.667V The output impedance of pd1 and pd2, which is simply equivalent to the resistances in parallel. R_pd1 = R1×R2/(R1+R2) = 100×100/(100+100) = 50R R_pd2 = R3×R4/(R3+R4) = 200×100/(200+100) = 66.67R Reinstate R5. The current can now be calculated using Ohm's law. The voltage is equal to the open circuit voltages of the potential dividers. It's the voltage difference, which is important, so they're subtracted. V = V(pd1_open) - V(pd2_open) = 2.5-1.667 = 0.833V To get the current, through R5, the impedance of the potential dividers needs to be added to R5: R(total) = R(pd1) + R(pd2) + R5 = 50 + 66.67 + 50 = 166.67R I = 0.833/166.67 = 0.005 = 5mA We know the current through R5 and therefore pd1 and pd2 is 5mA, so the voltages with R5 in place can be calculated using Ohm's law. V(pd1) is higher than V(pd2) so we know that the voltage on V(pd1) will fall (hence the - sign) and the voltage on V(pd2) will rise (hence the + sign) when R5 is connected: Kirchhoff's voltage law. V(pd1) = V(pd1_open) - R_pd1×I(R5) = 2.5-50×0.005 = 2.25V V(pd2) = V(pd2_open) + R_pd2×I(R5) = 0.833+166.67×0.005 = 2V The voltages on R2 and R4 equal v(pd1) and V(pd2) respectively, so the currents can be calculated: V(R2) = V(pd1) I(R2) = V(R2)/R2 = 2.25/100 = 0.0225 = 22.5mA V(R4) = V(pd2) I(R4) = V(R4)/R4 = 2/100 = 0.02 = 20mA The total supply current is simply the sum of the currents in R2 and R4: Kirchhoff's current law. I(V1) = I(R2)+I(R4) = 22.5mA + 20mA = 42.5mA The total resistance can be calculated using Ohm's law: R = V1/I = 5/0.0425 = 117.65R If V1 is unknown, just replace it with 1. The end resistance value will be the same. There are other ways to solve this, but this is my favourite one. |
| Brumby:
To TheN00b - Don't let that ^ ^ ^ scare you. You will be able to cope with that a bit further down the road. When you get there, it will be another challenge to learn, but it will provide you with the ability to solve seemingly impossible problems. |
| Zero999:
--- Quote from: Brumby on May 02, 2018, 11:26:40 am ---To TheN00b - Don't let that ^ ^ ^ scare you. You will be able to cope with that a bit further down the road. When you get there, it will be another challenge to learn, but it will provide you with the ability to solve seemingly impossible problems. --- End quote --- Apologies if I got carried away. I wanted to demonstrate that this can be used to solve more complex problems. The other methods such as nodal and mesh analysis work too and arguably are easier to apply, once one is familiar with them, but I prefer my method which I find more intuitive. Indeed, I'd have to consult a textbook if I had to perform mesh and nodal analysis, as I've not done them since college. |
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