DC Voltage Divider Circuits (OLD LECTURE)

good day and welcome to Big Bad Tech I’m your instructor Jim Pytel today’s topic of discussion is voltage divider circuits our objective today is to take a close look at a voltage divider circuit in both the unloaded and loaded condition voltage dividers are common series parallel circuit configurations in this lecture assumes you have a passing familiarity with series parallel circuit analysis if you haven’t watched the series parallel circuit analysis lecture available at the big bad tech channel or only dimly recall its contents by all means do so now as anyone that has ever organized or meticulously planned an event can tell you no plan survives first contact this is true for the biggest of battles all the way down to a toddler’s birthday party success and sometimes survival means you often have to toss that beautiful plan out the window and deal with a crumbling reality before you the same is true for a voltage divider circuit in the unloaded and the loaded condition if you can walk away from this lecture with a simple understanding that you cannot use any data obtained during circuit analysis of a voltage divider in the unloaded condition to solve for the same quantities while on the load of condition I will consider my time having not been spent in vain voltage dividers in the unloaded condition are totally totally different circuits than voltage dividers in the loaded condition and must be analyzed as such voltage dividers often take the form of a pure series circuit with several accessible nodes that electrical loads can be connected to in parallel if nothing is connected to these nodes or the switches between the nodes and electrical loads are open this is an unloaded voltage divider and can be analyzed using the simplest of series circuit techniques because there is but one path for current to flow an unloaded voltage divider as the name implies is a perfect opportunity to apply the voltage divider role here’s an unloaded voltage divider with three resistors in series pause the lecture and use basic series circuit properties to solve for v1 the voltage drop across R 1 ohm wait to perform the analysis of v2 and v3 in other quantities until after I’ve reviewed the answer for v1 the VTR setup to solve for v1 is as follows substituting in the necessary values we find v1 equals roughly seven point one volts if you’ve been watching the basic electronics one DC circuit analysis playlist and its intended sequence this should have been the most trivial of tasks the only reason I had you stopped here is to consider the nodal voltages with reference to ground while the voltage divider is in the unloaded condition let’s start by assigning the nodes a b c and ground if the source makes node a 50 volts higher than ground and v1 drops seven point one volts from A to B this means node B is only 50 minus seven point one or forty two point nine volts higher than ground while in the unloaded condition if this voltage divider circuit was inside a box and only node B and the ground reference was sticking out of the box a voltmeter would read the same forty two point nine volt differential you might incorrectly assume any load attached between node B and ground would experience the same forty two point nine volt differential as if it was a constant voltage source but it isn’t it’s a voltage divider and there’s a resistor in series between the 50 volt source and the accessible node B this will affect output voltage as we’ll learn in a moment when we discussed loaded conditions continuing on I could use the regular voltage divider rule again or a combination of Ohm’s law and Kirchhoff’s voltage law to solve for v2 and v3 however I’m gonna throw a twist than the voltage divider rule to solve for one of these quantities here’s the twist notice I’m only including r2 and r3 in the vdr formula and I’m not using the source voltage I’m using the voltage at node B with reference to ground this is to imply that r2 and r3 those elements between node B and ground should drop representative portions of this voltage you can do this this is the beauty of the vdr as long as you’re flexible it’s flexible substituting in the necessary values we find v2 equals approximately fourteen point three volts again if this source makes node a fifty volts higher than ground and r1 dropped seven point one

volts from A to B and our two drops an additional fourteen point three volts from B to C this means node C is only fifty minus seven point one minus fourteen point three or twenty eight point six volts higher than ground while in the unloaded condition if this voltage divider circuit was inside a box an only node C in the ground reference was sticking out of it a voltmeter would read the same twenty eight point six volt differential then you might incorrectly assume any load attached between nodes seeing ground would experience this same twenty eight point six volt differential as if it was a constant voltage source but it isn’t it’s a voltage divider circuit and there are two resistors in series between the source and the accessible node C this will affect output voltage as we’ll learn in a moment when we discussed the loaded conditions finally an Ohm’s law analysis of this voltage divider circuit in the unloaded condition suggests that current through any of these three resistors is approximately fourteen point three million peers this current internal to the voltage divider circuit is sometimes known as bleeder current for an unloaded voltage divider circuit source current and bleeder current are one and the same because a pure series circuit has but one path for current if switched loads were attached to the accessible nodes B or C and these switches were open the voltage divider circuit is still in its unloaded condition since no current can flow through these open switches switch X is between node B and load resistor rx a three kiloohm resistor switch Y is between node C and load resistor ry a to kill own resistor in the open condition switch ax would experience a forty nine point two volt drop between its two terminals and no current would flow through it in the open condition switch Y would experience a 28.6 volt drop between its two terminals and no current would flow through it since no current is traveling through the load resistors X or Y these load resistors will experience no voltage drop this all changes when one or both of these switches close because the unloaded voltage divider has changed from a simple pure series circuit to a series parallel circuit absolutely none of the values obtained during unloaded conditions can be used to calculate any value during loaded conditions because these are two fundamentally different circuits and they will behave in two fundamentally different manners consider the radically different behavior this same voltage divider circuit when switch Y is closed this closed switch places the load resistor R Y in parallel with r3 and this additional path for current will decrease the resistance seen by the source and source current will increase source current is now the bleeder current that current internal to the voltage divider and that current bound for load resistor R Y both r1 and r2 carry the combination of both currents however at node C these two currents divide our three carries only the bleeder current and our Y carries only the load current iy since r3 and ry are in parallel the closure of switch Y is in effect replacing the original series configuration of r1 r2 and r3 with a series configuration of r1 r2 and our prime where R prime is the parallel combination of r3 and our Y since r3 and ry are both 2 kilo ohm resistors our prime would be equal to 1 kilo ohm the loaded voltage divider circuit has fundamentally changed from its unloaded configuration total resistance has decreased because of the additional current carrying path source current will increase and voltage drop across each element and nodal voltages will be reapportioned a loaded voltage our circuit is a circuit fundamentally different from unloaded conditions and must be approached as such if you feel up to the task by all means pause the lecture and see if you can solve for the voltage drop across to each component including the switches and load resistors the nodal voltages at node B and C with respect to ground and the current through all elements here’s how I solve for the desired unknowns you may have performed different steps in a different sequence but ultimately our answers should agree I’ll use the pure series simplification of r1 r2 + R Prime and start by solving for V R prime using the VTR the VD are set up to solve for V R prime is as follows substituting in the necessary values we find V R prime to be 20 volts our prime

is the parallel combination of r3 and load resistor ry voltage across components hooked in parallel is the same b3 is 20 volts as is V Y additionally it can be said that the voltage at node C with respect to ground is 20 volts ohms loss and just the current through load resistor y is equal to the voltage across load resistor Y divided by the resistance of load resistor Y substituting in the necessary values we find the current through load resistor Y and switch Y is 10 milli amperes since r3 is equal in magnitude to our Y and is experienced in the same voltage drop across it we can be reasonably confident it too will also draw 10 milli amperes of current an additional Ohm’s law analysis supports this time-saving step given both R 3 and ry are drawing 10 milli amperes of current each it can be said that 10 + 10 or 20 milli amperes of current must have entered node C prior to being divided this means current through r2 is 20 milli amperes as is the current through r1 as is source current since in this configuration there are no other paths diverting or supplying additional current Ohm’s law suggests that b2 equals I 2 times r2 substituting in the necessary we find v2 to be 20 volts we can use KVL to solve for v1 a KVL analysis of a pure series simplification suggests e equals v1 plus v2 plus BR Prime substituting in the given value of e and the calculated value of v2 and B are prime and solving for V 1 we find v1 to be 10 volts in this configuration our KVL analysis also suggests either a minus v1 or v2 plus VR prime equals the voltage at node B with respect to ground there for node B is 40 volts higher than ground in this case the bleeder current that current internal to the voltage divider is 10 milli amperes and the additional 10 mili amperes of current bound for load resistor Y is coming along for the ride through r1 and r2 before it takes a turn at node C to its intended destination the current through switch Y is 10 milli amperes and since we’re assuming switch Y is an ideal closed switch with no internal resistance there will be no voltage drop across its two terminals open switch X experiences no current through it since node B is 40 volts higher than ground the voltage across switch axes to terminals is 40 volts since no current is flowing through load resistor rx there will be no voltage drop across to compare and contrast the values we obtain during unloaded conditions and that one switch Y is closed not one of the values we obtained are identical with the exception of the current through the open switch X and the voltage drop across the unconnected load resistor rx the only reason these values are the same is because in both circumstances load resistor Rx is totally disconnected from the circuit and does not influence it in any way shape or form the larger take-home point is this the connection of load resistor R Y changed everything including points above stream of it the moment contact is made even the best of plans must change consider now what happens when we close switch X this is again a completely different circuit than our previous unloaded analysis and when the voltage divider circuit was loaded only by the load resistor our Y in this case bleeder current still travels through r1 r2 and r3 the current bound for load resistor Rx travels through r1 before getting off at node B the current bound for load resistor ry travels through both r1 and r2 before getting off at node C in this case source current increased because the addition of yet another path in parallel decreased total resistance seen by the source as can be expected everything will change load resistor ry is still in parallel with r3 at node C a simplification I’ll call R prime with a value of 1 kilo ohm our prime is in series with r2 these can be taken together as another imaginary resistor I’ll call our double Prime with a value of 1 kilo ohm plus 1 kilo ohm or 2 kilo ohm note by combining these two resistors in series we’ve effectively lost access to node C however we can always return to this point if we keep track of any simplifications we’ve made in the process of solving for our desired

quantities our double prime is in parallel with a 3 kilo ohm load resistor Rx and node B these two can be combined to parallel to yield another imaginary resistor I’ll call our triple prime since our double prime is a 2 kilo ohm resistor an Rx is a 3 kilo ohm resistor the parallel combination of these two means our triple prime is a one point two kilo ohm resistor our triple prime is in series with a 500 mm r1 we can now apply the voltage divider rule to solve for V our triple prime the VD are set up to solve for V our triple prime is as follows substituting the necessary values we find a V our triple prime to be approximately thirty five point three volts since our triple prime is the parallel combination of our double Prime and load resistor our X we can say that voltage across components hooked in parallel is the same V R double prime is thirty five point three volts as is the voltage across the load resistor rx this also means node B is thirty five point three volts above ground given V one is in series with our triple prime a KVL analysis of this loop suggests that v1 equals fifty minus thirty five point three volts or the remaining 14 point 7 volts Ohm’s law suggests that I 1 equals v1 divided by r1 substituting the necessary values we find the current through r1 is twenty nine point four milli amperes this means source current is twenty nine point four million peers this is the current that enters node B the parallel combination of our double Prime and load resistor rx ohms all suggest that IX the current through load resistor Rx is equal to the voltage across it divided by its resistance substituting in the necessary values we find IX the current through low resistor R X is approximately eleven point eight million peers with our previous understanding that approximately twenty nine point 4 milliamperes of current entered node B and eleven point eight million pairs of it went through load resistor rx it can be said that the remaining seventeen point six million peers traveled through R double prime where our double prime is this series combination of r2 and our single prime returning to the previous level of simplification that suggests that I 2 should equal seventeen point six milliampere Oh massage suggests that V 2 equals I two times R two substituting the necessary values we find v2 should equal approximately seventeen point six volts if the source makes node a fifty volts higher than ground and R one drops 14 point 7 volts from A to B and R two dropped seventeen point six volts from B to C it stands to conjecture the volt at node C with reference to ground so remaining approximately seventeen point seven volts this means b3 is also approximately seventeen point seven volts and V Y is also seventeen point 7 volts since they are in parallel with each other if that node C seventeen point six milli amperes of current enters the parallel combination of our three and our Y we can use either Ohm’s law or the current divider role to determine how the seventeen point six million amperes of current divides between our three and load resistor our Y however we’re aware that both our 3 and our Y our resistors of the same magnitude of two kilohms with the same voltage drop across them current entering node C should split equally between them I 3 equals half of seventeen point six million peers or eight point eight million amperes as does iy since we’re assuming switch ax is an ideal closed switch with no internal resistance there will be no voltage drop across its two terminals the same thing could be said about switch Y assuming as an ideal closed switch with no internal resistance there will be no voltage drop across its two terminals again notice the values we obtained when both switch ax and switch Y are closed are totally totally different than when only switch Y was closed or during unloaded conditions the closure of both switch ax and switch Y makes us a totally different circuit and it must be analyzed as such the moment contact is made even the best of plans must change consider yet another permutation of this same voltage divider circuit what a switch ax was to remain closed and switch Y were to open effectively removing load resistor R Y from

consideration this is another circuit entirely and must be treated as such this is why I love voltage divider circuits so much there are numerous possibilities in each configuration brings with it a new chance to put our series parallel circuit analysis skills to the test in this configuration r2 and r3 can be combined in series with one another into something I’ll call our prime since r2 is a 1 kilo ohm resistor and r3 is a 2 kilo ohm resistor our single prime will be 3 kilo ohms our single prime is in parallel with our X also a 3 kilo ohm resistor these two resistors can be combined in parallel into something I’ll call our double Prime since our single Prime and our X are both 3 kilo ohm resistors our double prime will have a value of 1.5 kilohms the VD are set up to solve for V R double prime is as follows substituting in the necessary values we find V R double prime equal to 37 point 5 volts since our double prime is a parallel combination of our single Prime in our X this means V X the voltage across load resistor R X is also 37 point 5 volts additionally this means no B is 37 point 5 volts higher than ground the KVL analysis of the simplified pure series circuit suggests e equals V 1 plus V R double prime substituting in the given value of E and the calculated value of V R double prime and solving for V 1 we find V 1 equals 12 point 5 volts o mazaa suggests V 1 divided by r1 equals I 1 substituting in the necessary values would find I 1 equals 25 milli amperes given I 1 is the combination of bleeder current the current internal to the original voltage divider and current bound for load resistor R X we can say source current is also 25 milli amperes and use either Ohm’s law or the current divider rule to determine how the 25 million I preserve current entering node B divides between R Prime and load resistor R X however we’re aware that both our imaginary resistor R Prime and load resistor R X our resistors of the same magnitude of 3 kilohms with the same to drop across them current entering node B should split equally between them ir prime equals half of 25 milli amperes or 12.5 million peers as does IX given R prime as the series combination of r2 and r3 it can be said that the current through these two resistors should also be equal to 12 point 5 milli amperes because they are in series with one another I 2 equals 12 point five million amperes as does I three Ohm’s law Kirchhoff’s voltage law or the V er suggests V 2 equals 12 point 5 volts and V 3 equals the remaining 25 volts therefore node C is 25 volts above ground no current will flow through open switch Y and no voltage will be dropped across load resistor R y given node C is 25 volts higher than ground open switch Y will experience a 25 volt drop across its two terminals ideal closed switch X should exhibit no voltage drop across its two terminals again this circuit analysis yields radically different values than any of our previous analyses because this is a radically different circuit than our previous renditions and must be approached as such voltage divider circuits in their numerous configurations present one with numerous opportunities to practice series parallel circuit analysis skills this single circuit presented us with four different opportunities to employ our skills once in the unloaded condition when neither switch ax nor switch Y was closed another one it was loaded by ry only another one it was loaded by both Rx and ry and finally another when I was loaded by rx only each different combination of switch positions presented us with different current paths and results in voltage drops every situation was different and necessitated a completely new analysis because they were completely different circuits notice at no time did I ever have to resort to reducing this circuit to a single total resistance and returned through a tangled mess of simplifications to solve for the desired quantities at all times I use basic series and basic parallel properties and repeatedly made use of kirchoff’s voltage law and Kirchhoff’s current law to quickly and directly solve for the desired quantities without any heart breaking labor visualize circuits at the nodal level from the perspective of flowing current if there is no choice of paths those elements and the same paths

are in series with one another and current through them must be the same if there is a choice in paths those paths are in peril with one another and voltage across them must be the same in this spirit here is not one but rather eight different opportunities to put your skills to the test here’s a voltage divider circuit consisting of a 12-volt source and three resistors r1 at 200 ohms r2 at 400 ohms + r3 at 600 ohms with three switches and three loads Rx is a one point two kilo ohm resistor our Y is one point eight kilo ohm resistor and our Z is a two point four kilohm resistor pause the lecture and solve for the voltage drop across each component the current through each component the source current and the bleeder current recall the bleeder current is the current internal to the voltage divider in eight different switch positions as indicated in the table in the upper right hand corner you may be bummed out about the number of repetitions however a quick Kirchhoff’s current law analysis of one of the switches in the electrical load will allow you to make powerful and time-saving deduction about that load resistors effect on the remaining circuit I’ll not go into details about the calculations the required values since the methods employed can be numerous I will however dispense general advice that may serve to guide you through resolution of each problem don’t wait for me to give you the answers if a hint frees you from a sticky spot by all means pause the lecture and see if you can take it from here our first scenario is one in which the voltage divider is unloaded and all switches are open the simplest of series circuit analysis techniques will suffice all current travels through r1 r2 and r3 and all voltage will be proportionally distributed between these resistors no current travels through rx ry or RZ and no voltage will be dropped across them the voltage drop across each open switch be the nodal voltage with respect to the ground reference when switched Z is closed RZ is placed in parallel with r3 in r3 only source current will travel through r1 r2 and only at node C or bleeder current and current bound for load resistor RZ part ways since r3 is in parallel with RZ the voltage drop across these two components will be the same supply voltage e will be proportionately distributed between r1 r2 and the parallel combination of r3 and RZ r prime no current travels through our X or our Y and no voltage will be dropped across them the voltage drop across each open switch will be the nodal voltage with respect to the ground reference when switch X is open and switch wise closed our wise placed in parallel with something I’ll call our prime the series combination of r2 and r3 source current will travel through r1 and at node B bleeder current and current bound for load resistor ry part ways since our prime is in peril with ry the voltage drop across these two components will be the same internal to our prime since r2 is in series with our three r2 and r3 should have the same current supply voltage e will be proportionally distributed between R 1 and R double prime the parallel combination of our single prime and ry no current travels through our X or our Z and no voltage will be dropped across them the voltage drop across each open switch will be the nodal voltage with respect to the ground reference when both switch C and switch wire closed our Z’s placed in parallel with r3 something I’ll call R prime R prime is in series with our to some I’ll call our double prime source current will travel through r1 and at node B current bound for load resistor our Y parts ways with that current bound for our double Prime at node C bleeder current and current bound for load resistor R Z part ways since our double prime is in parallel with our Y the voltage drop across these two paths will be the same since r3 is in parallel with our Z the voltage drop across these two components will be the same no current will travel through our X and no voltage will be dropped across them supply voltage e will be proportionately distributed between V 1 and V our triple prime where our triple prime is the parallel combination of our Y and our double Prime when switch X is closed and all other switches open the voltage divider could care less what our X is

doing our X has been placed perfectly in parallel with a series combination of r1 r2 and r3 this analysis will be nearly identical to the unloaded condition with a single exception that rx now draws current and experience as a voltage drop and source current has increased this can be extrapolated into the remaining scenarios and which switch X is closed such as when switch ax is closed switch Y is open and switch Z is closed switch X is closed switch Y is closed and switch Z is open and finally when all three switches are closed since our axes placed perfectly in parallel with a voltage divider it does not affect voltage division internal to the voltage divider only those load resistors internal to the voltage divider affect it with their presence if you’ve done the analysis that this voltage divider correctly you should expect to observe the following values for the following conditions in the unloaded condition one would expect to observe the following values our second scenario in which only switched Z is closed one would expect to observe the following values for our third scenario which only switch why is closed one would expect to observe the following values the fourth scenario when both switched y&z are closed one would expect to observe the following values in a fifth scenario which only switch axis closed it’s almost like it’s a repeat of the unloaded voltage divider the minor exception that source current has increased because load resistor Rx is now drawing current in our sixth scenario when both switch X and switch Z are closed we find something very similar to just when switch C was closed the one exception being load resistor R X now draws current and source current will increase our seventh scenario and which both switch X and switch Y are closed is very similar to the scenario in which just switch Y is closed the one exception being that load resistor rx draws current and experiences a voltage drop and source current increases our final scenario which all three switches are closed is very similar to our previous analysis of just one y&z were closed the only exception is that load resistor our axe draws current and source current will increase please take your time to revisit any calculations that may be an error and pause to consider the general guidance that issued before giving you these answers visualize circuits at the nodal level and from the perspective of flowing current elements within the same path must have the same current paths in parallel must have the same voltage drop across them in closing I leave you with this voltage divider circuit to tack on your own as an assessment exercise I’ll post the answers to this problem in the information section associated with this lecture don’t peek try this one by yourself if you can arrive at the same answers you have met the standards for success for this task and can feel free to move on to the next section if not I suggest you revisit this lecture and correct any misconceptions you may have before continuing we have a 40 volt source supplying a combination of r1 400 ohms r2 600 ohms where switch X is switching the series combination of our x1 of 1 point 2 kilo ohm resistor and our x2 another one point two kilo ohm resistor switch Y is switching load resistor our Y a 2 kilo ohm resistor solve for voltage across each component including switches and the current through each component including source current and bleeder current for for different conditions one when both switches are open one when switch Y is open and switch X close another one switch y’s closed and switch X is open and finally when both switches are closed you may find circumstances in which different switch positions yield the same answer for very fundamental reasons in conclusion this lecture took a look at a common series parallel circuit called a voltage divider in both the unloaded and loaded condition we learned that values obtained during circuit analysis of the unloaded condition cannot be used to calculate values during the loaded because the unloaded and loaded condition present radically different circuits and must be approached as such remember to review these concepts and practice his techniques as often as you need to really drive at home imagine how well lab will go if you know what you’re doing thank you very much for your attention and interest and we’ll see you

again during the next lecture of our series remember to tell you lazy lab partner about this resource and be sure to check out the Big Bad tech channel for additional resources and updates