in this video we're going to talk about the voltage divider circuit so here's an example and our goal is to calculate the current flowing through the circuit and determine the voltage across r2 so we have a series circuit because it's only one path for the current to flow and the first thing we need to do is calculate the total resistance so it's going to be 10 plus 20 which is 30 ohms next we need to calculate the current flowing in the circuit so that's going to be based on ohm's law v equals ir so the current flowing in the circuit is going to be the voltage of the battery divided by the total resistance so that's 12 volts divided by 30 ohms and so the current flowing in a circuit is 0.4 amps now to calculate the voltage across r2 we could use v equals ir so v2 is going to be the current in a circuit times r2 so that's 0.4 amps multiplied by 20 ohms and so v2 is 8 volts so that's the voltage across r2 now there's another way in which you can calculate that same answer now you can use a formula to calculate the voltage across r2 we'll call it the output voltage so the output voltage is going to be the voltage of the battery which i'll use vt for that times r2 divided by the total resistance r1 plus r2 so the voltage of the battery in this example is 12 volts and r2 is 20 and r1 plus r2 that's 10 plus 20 so that's 30. it's going to be two-thirds of 12 volts so 12 times 20 divided by 30 is 8 volts so that's another way in which you can calculate the output voltage try this example problem design a voltage divider circuit with an output of 6 volts and a current of 60 milliamps using a 12 volt battery so take a minute and try this problem well let's begin by drawing the circuit so the circuit is going to look the same now this circuit consumes 60 milliamps so we need to calculate the total resistance of the circuit so we can use v equals ir so the voltage of the battery is 12. the current is 60 milliamps but if we divide that by a thousand that's 0.06 amps and so r is going to be 12 divided by .06 so the total resistance is 200 ohms so what that means is that r1 plus r2 has to add up to 200. now notice that the output voltage is 6 volts it's basically half of the voltage of the battery if you want to decrease the voltage if you want to cut it in half what you need to do is need to make sure that these two resistors are equal in value so what two numbers add up to 200 and are equal to each other so this has to be a hundred and this has to be a hundred in that case we can have a current of 60 milliamps and an output voltage of six volts so if the input voltage was 20 volts the output voltage will be 10. it's going to be half of whatever the input voltage will be now this problem is similar but a little different so go ahead and design a voltage divider circuit with an output of 12 volts and a current of 50 milliamps using a 20 volt battery just take a minute and try that problem so here we have a 20 volt battery and there's a current of 50 milliamps flowing in this circuit so just like before the first thing we need to do is calculate the total resistance so let's use ohm's law to do that so the voltage is 20 volts the current is 50 milliamps if we divide that by a thousand that's 0.05 amps and so r is going to be 20 divided by 0.05 so the total resistance is 400 ohms so we know that r1 plus r2 has to be equal to 400 now the output voltage across r2 is 12 volts so how can we calculate r1 and r2 individually now let's use this equation the output voltage is equal to the input voltage multiplied by r2 divided by r1 plus r2 so the output voltage of this circuit is 12 volts the input voltage is the voltage of the battery and that's 20. we don't know the value of r2 as of yet however we do know that r1 plus r2 is 400 so we can replace this with 400 so this equation allows us to solve for r2 so now it's all algebra at this point so what's the first thing we should do the first thing i would like to do is multiply both sides by 400. let me see if i can fit in here by doing this i can get rid of the fraction so on the left side it's going to be 400 times twelve which is forty eight hundred and so that's equal to twenty times r2 now to isolate r2 i need to divide both sides by twenty so r2 is going to be 4 800 divided by 20 and so r2 is 240 ohms so now i can use this equation to calculate r1 so i'm going to replace r2 with 400 so it's r1 plus 240 is equal to 400 now i need to subtract both sides by 240. so r1 is 400 minus 240 so that's 160. so this is 160 ohms and if this is 240 ohms then the voltage across r2 will be 12 volts and the current will be 50 milliamps in a circuit number four calculate the output voltage across r2 in the circuit shown below now because these two resistors are equal in value the output voltage will be half of the input voltage so we know it's going to be 10 volts but now here's the important part what will be the voltage across r2 if a device with an internal resistance of 1k ohms is connected across it so without a device the voltage is 10 volts so keep that in mind now it's important to understand that if you connect this with a device so it has its internal resistance of a thousand ohms which is one kilo ohm so i'm going to enclose this because it's a some sort of device you need to understand that when you connect a device across r2 the voltage across r2 will decrease and the reason for that is some of the current that's flowing through r1 will go in this direction and so less current will be flowing through r2 than before and so if you decrease the current flowing through a resistor the voltage across that resistor will decrease now as this resistor increases in value the voltage drop across r2 will be less significant so if you decrease the resistor then this voltage across r2 will drop to a much greater extent so let's calculate the voltage across r2 which is the same as the voltage across the 1000 ohm resistor when this is a thousand ohms and let's say when it's 10 ohms the first thing that we need to do is we need to find or calculate the equivalent resistance since these two resistors are in parallel and so you could use this formula the equivalent resistance is going to be 1 over r1 so that's i mean 1 over r2 that's 100 and 1 over the internal resistance value so that's 1 over 1000 and then raise it to the minus one so the equivalent resistance is 90.9 ohms so now we can calculate the output voltage using the same formula it's going to be the input voltage but instead of r2 we're going to use the equivalent resistance of that circuit so it's going to be the equivalent resistance of those two resistors divided by the equivalent resistor plus r1 so you can treat the equivalent resistance as the new r2 value so the input voltage is 20 the equivalent resistance is 90.9 and it's going to be divided by 90.9 plus 100 so it's 20 times 90.9 divided by 190 190.9 and so the voltage is now 9.52 volts so notice that it's less than 10 volts so you need to take that into consideration whenever you're connecting a device across a voltage divider circuit now let's see how much the voltage will decrease by if we decrease the internal resistance of this device to 10 ohms so let's change this number and make that 10. so first let's calculate the equivalent resistance of r2 and the internal resistance so it's going to be 1 over r2 plus 1 over r raised to the minus 1 so that's 1 over 100 plus 1 over 10 raised to negative 1. and so the equivalent resistance is 9.09 ohms so now to calculate the input i mean the output voltage rather it's going to be the input voltage which is 20 times the equivalent resistance of the two resistors on the bottom that's 9.09 divided by the equivalent resistance plus r1 so it's 9.09 times 20 divided by 109.09 so notice that the voltage greatly decreases to 1.67 it's much less than 10. so if you want to keep the voltage as close as 10 as possible you should only use a device where the internal resistance of the device is a lot greater than r1 and r2 if it's significantly greater then the voltage won't be they won't change as much so for example let's do one more problem so we see that if this voltage i mean if this resistance is 10 times the value of one of these the voltage decreased by almost five percent of its original value so what we're going to do let's see what happens if the internal resistance is a hundred times greater than the value of r1 or r2 so this time i'm going to make it 10 000 ohms or 10 kilo ohms so let's start by calculating the equivalent resistance so it's going to be 1 over 100 plus 1 over 10 000 raised to the minus one over ten thousand is a very very small number so the equivalent resistance will still be approximately a hundred so the bigger this value is the less significant um one over that value will be but let me just show you how it works so one over a hundred plus one over ten thousand raised to the negative one this is going to be ninety nine zero which is very close to 100 and so the higher the internal resistance of the device the better because the equivalent resistance won't be very far from r2 which is what we want so that means the voltage across r2 will be very close to 10. so now let's calculate it the output voltage is going to equal the input voltage of 20 times the equivalent resistance of 99.01 divided by that value plus 100. so 20 times 99.01 divided by 199.01 is 9.95 volts so notice that the voltage decreased by only 0.5 percent which is good enough so if you want to design a voltage divider circuit and you don't want the voltage to change more than one percent i recommend that you should design your circuit in such a way that the internal resistance is at least 100 times the value of r2 because when you add a resistor in parallel to r2 the equivalent resistance will decrease so make sure it's 100 times the value of this resistor you don't really have to worry about r1 much but r2 is what you need to compare the internal resistance to number five a device with an internal resistance of five kilo ohms requires a voltage of three volts using a 12 volt battery design a voltage divider circuit that meets such requirements and also calculate the current that the circuit draws from the battery and the amount consumed by the device so let's start with this circuit again so this is r1 r2 and then we're going to have the device across r2 so we know that this is 5 kilo ohms and if we want the voltage to be approximately 3 volts r2 is going to have to be 100 times less than this value so 5000 divided by 100 this means this has to be approximately 50 ohms now if the voltage across r2 is approximately 3 volts and the total voltage of the battery is 12 volts then the voltage across r1 must be 9 volts because they have to add up to 12. now this may not be the exact answer but it's simply an approximation so if the voltage across r1 is three times as great as the voltage across r2 then this resistance must be approximately three times more than r2 so it has to be 150 now let's calculate the output voltage across the device so first we need to calculate the equivalent resistance so it's going to be 1 over r2 plus 1 over the internal resistance which is five thousand raised to the minus one and so that's going to be 49.505 now to calculate the output voltage it's going to be the input voltage which is 12 times the equivalent resistance of 49.505 divided by the equivalent resistance plus r1 which is 150 so that's 12 times 49.505 divided by 199.505 and so the output voltage is going to be 2.98 volts which is approximately 3 volts so that's a quick and simple way to design a circuit where the voltage across this device will be approximately three now there is one significant issue with the circuit how much current does the device consume so the actual voltage across these two points we realize is 2.98 volts so to calculate the current flowing through this resistor we can use the equation v equals ir so the resistance is 5000 ohms so 2.98 divided by 5000 that's 5.96 times 10 to the negative four now let's multiply that by a thousand so that we can convert it to milliamps so the current that flows in this branch is 0.596 milliamps now how much current flows through this resistor it's going to be 2.98 divided by 50. and so you should get .0596 amps if you multiply that by a thousand that's equal to 59.6 milliamps now if we add those two numbers 59.6 plus 0.596 that's equal to 60.2 milliamps so that's how much power is delivered by the battery so as you can see in this circuit the device only consumes a fraction of the current compared to the amount that's delivered by the battery in terms of percentages 0.596 divided by 60.2 times 100. it's only consuming about one percent of the current that's being delivered from the battery and so it's not very efficient so how can we design a circuit where we have three volts across the device and that is highly efficient because 99 of the energy delivered by the battery is not going into the device and that's a problem now perhaps you know the answer already but if you don't the most simplest way to make the circuit more efficient is simply to get rid of r2 just to replace the device with r2 so let's draw a new circuit so we're still going to have a voltage divider circuit however the device is going to be placed here we're going to replace r2 with the device now the calculations will be a lot more simpler in this example so the voltage across the device which will still call it the output voltage that's going to equal the input voltage times the internal resistance of the device divided by the sum of the internal resistance plus r1 so our goal is to get exactly 3 volts so that's the output voltage the voltage of the battery is 12 and the eternal resistance we know it's 5k and then we could solve for r1 if we need to but because the numbers are simple we really don't need to use the equation we know that 9 volts will be across r1 because 9 plus 3 has to add up to 12. and so since the voltage drop across r1 is three times greater than r we know that r1 has to be three times this value so it's going to be five times three so r1 is going to be 15 kilo ohms and so this is a simple circuit in which the voltage across the device will be exactly three and at the same time the current flowing through the device will be three volts divided by five thousand which is 0.6 milliamps so notice that all of the current that is delivered by the battery flows into the resistor so no energy is wasted and so this is the ideal circuit if you want to create a voltage divider for device you want to replace r2 with the device itself you