solving series parallel circuits

this tutorial is on how to solve series parallel combination circuits let's begin with this circuit it's got five resistors a DC circuit and we'll use this as an example of how to solve for these to find resistance total resistance current throughout the circuit and voltage drops across those resistors okay the first thing that we're gonna do is is take a look and draw the current yeah we're gonna use conventional current and realize that it travels from the positive of the voltage supply to the negative goes all the current travels through resistor 1 then it gets to this point where it splits some of the current goes this direction and some of it goes the other direction meeting back together at this point where all the current flows through resistor 5 and back to the voltage supply we're drawing our current path does forces help us realize which components are in series and which are in parallel so right away we can see that when the current splits we're gonna have two in parallel we've got resistor two and three that are in parallel with resistor four gave that combination of two and three and four is in series with resistor one and resistor five okay so the first thing we're gonna do is take a look and see are there any series resistors that we can combine together remember the resistors in series you can just add the total resistances together resistor two and three look like that they're just in series so we're gonna we're gonna say that we can put an equivalent resistance here and just add these two together 10k ohms plus 4.7 km ziz 14.7 km/h and ooh this is by doing a redraw method like this so we could simplify our circuit to to show resistant resistor equivalent of two plus three is 14 point seven K ohms then it's easy to see that that's in parallel with resistor four resistors in parallel you have to use one of two formulas to calculate the resistance you can use the product over the sum formula if there's just two resistors that would be resistor 1 times resistor 2 divided by resistor 1 plus resistance 2 or you can use a formula that will work no matter how many resistors you have in parallel the one over one over method I'll use that one so we're gonna say the resistance equivalent for these two in parallel would be one over one over resistor two plus 3 which is fourteen point seven km or fourteen thousand seven hundred ohms plus one over eight thousand nine hundred ohms or eight point nine K ohms and if we do that math we get five point five four four K ohms or five thousand five hundred forty four ohms all right now we could also simplify this circuit with a redraw and show that now it's easy to see here that all three of these resistors are in series because the current only has one way through all of those resistors so we can actually do a redraw of that and add those resistors together the 10k plus the 5.54 4k plus the 500 ohms adds up to a total resistance of sixteen point o4 for K ohms or sixteen thousand forty four ohms at this point we can use Ohm's law to calculate the current and here's a graphical representation of Ohm's law voltage over current times resistance or V over IR so since we already know the source voltage and the total resistance we can calculate the current and doing that calculation gives us one point eight seven zero milliamps again that's thirty volts divided by sixteen thousand forty four ohms we get one point eight seven zero milliamps now that we know the total current let's work backwards through our redraws since all the current has to flow through resistor one and resistor five we know those are also one point eight seven zero milliamps we also know that the one point eight seven zero milliamps flows through what would be the equivalent resistance of that parallel combination we'll solve that a little bit later let's justtake recurrent passing through resistor 1 and resistor 5 and redraw it clear back to the very beginning diagram so we can redraw that on our second redraw and back to our first redraw as well now on this redress since on resistor 1 we already know the resistance and the current again we can use Ohm's law to calculate the voltage since we know current is 1.8 7-0 milliamps and resistance is 500 ohms we'll calculate the voltage 935 millivolts we can use the same Ohm's law to calculate the voltage for resistance by the voltage drop on resistance five and it's eighteen point seven volts now let's go back to one of our other redraws now there's two ways we can find the the voltage drop on the equivalent resistance of that parallel combination one way is we know our source voltage is 30 volts so if we subtract 935 millivolts for resistor one and eighteen point seven volts from resistor five were left over with what has to be on that combination the other way to do it is also using Ohm's law since we know the resistance and we know the current we can calculate the voltage and the voltage would be ten point three seven volts once we know the voltage on this this combination represented as one resistor let's go backwards work backwards through our redraws since we know of ten point three seven volts is on the both of these in parallel and we remember the voltage drops on resistors in parallel have to be the same we know that we have a voltage drop of ten point three seven volts on resistor four now that we know the voltage and resistance for resistor four let's use you got it Ohm's law and we'll calculate the current for resistor four now to calculate the current going through resistor four we have to remember that the current splits going through resistor four others going through the equivalent of two and three but since we know the voltage drop on resistor 4 and its resistance we can calculate the current using Ohm's law the current ends up being 1 point 1 6 5 milliamps since we know that 1 point 1 6 5 milliamps is going through resistor 4 and we have a total current of 1 point 8 7 zero milliamps then whatever is left over must be going through resistor 2 & 3 so we just do that subtraction and we find out that the current through resistors 2 & 3 is 705 micro amps now that we know the current going through those resistor 2 and resistor 3 we have again two out of the three things we need for Ohm's law so let's calculate the voltage drop on resistor 2 that would be 10,000 ohms times 705 micro amps which gives us seven point zero five volts we can do Ohm's law for resistor 3 also getting three point 3 1 4 volts and using kirchoff's voltage law we can add up any voltages in series and we come up with 30 volts give or take a few rounding errors and that's how you solve series parallel combination circuits again it's it's just a matter of reusing Ohm's law over and over and finding what you need you can use redraws as a technique to do that

Channel: Ron Call

Views: 676,567

Rating: 4.6839728 out of 5

Keywords: MHS, electronics, circuit analysis

Id: 0MKcxCHkT1c

Length: 8min 3sec (483 seconds)

Published: Mon Apr 01 2013