in this video we're going to go over series and parallel circuits so consider the following two circuits here we have a battery and a few resistors which one of these circuits represent a series circuit and which one is parallel by the way you can view the resistors as light bulbs a light bulb acts as a resistor the one on the left is a series circuit the reason being is there's only one path for the current to flow in a parallel circuit which is what we have on the right there's multiple paths for the current to flow the current can flow in this direction or it can flow in that direction the longer side of the battery is the positive side and the shorter side is the negative side conventional current flows from positive to negative conventional current represents the flow of positive charge but electrons they flow in the opposite direction from negative to positive in a series circuit if you connect two light bulbs in series together the light bulbs will appear to be dim relative to the parallel circuit if you connect two light bulbs in parallel to each other the light bulbs usually will appear very bright that's because as you add resistors in a series circuit the total resistance increases which decreases the current so let's say if you put three light bulbs in series it's going to be even dimmer because there's less current flowing through it whereas in a parallel circuit the voltage is the same so it doesn't matter how many light bulbs you add as long as you have a strong enough battery that can dish out enough current each light bulb will remain bright as you add resistors in a series circuit the resistance goes up but as you add resistors in a parallel circuit the overall resistance goes down but more current is drawn from the battery to power those extra light bulbs or resistors now let's solve a series circuit so let's say if we have two resistors and let's say the first one has a resistance of 4 Ohms and the second one 6 ohms let's say we have a 30 volt battery how can we calculate the current that flows in a circuit by the way in a series circuit the current that flows through the resistors that are connected in series is the same current in a parallel circuit the voltage across resistors that are connected in parallel is the same now in this problem the first thing you want to do is find the total resistance in a series circuit it's simply R1 plus R2 and if you have a third resistor plus R3 so it's 6 + 4 the total resistance is 10 now the next thing you need to do is use Ohm's law V is equal to IR if you use the total voltage of 30 volts and the total resistance of 10 ohms the current that flows through the circuit is 30 ID 10 so it's 3 amps so 3 amps of current leaves the battery and flows through each resistor now what does the voltage drop across each resistor so let's start with the 6 Ohm resistor what is the voltage across it so we have three amps of current flowing through the 6 Ohm resistor to find a voltage across it use ohms law V equal I so 3 amps time 6 ohms gives us a potential difference of 18 volts now what about the 4 ohm resistor using the same equation we have a current of 3 amps flowing through a 4 ohm resistor so the voltage drop is 12 so we have 18 volts across the 6 Ohm resistor and 12 volts across the 4 ohm resistor now let's make some space let's just get rid of some of the stuff that we have here notice that the voltage drop across each resistor adds up to the voltage of the battery and it makes sense according to kof's voltage law the voltages around any closed Circle or any closed loop always add up to zero so let's say if this point has a potential of Z 0 volts that means the voltage potential at this point is 30 the positive side of the battery is 30 volts higher than the negative side and this would be 12 so if you go around the circuit the voltages should add up to zero so let's go from Let's Travel clockwise so if we go this way the voltage increased from 0 to 30 so that's positive 30 and it makes sense because the battery IT Supplies energy to the Circuit voltage is defined as electric potential energy per unit charge one volt is one Jewel per colum so the battery it increases the voltage of the circuit now as we go through the resistor from 30 to 12 we experience a voltage drop and it makes sense because the resistors they could consume energy in a circuit so we have a voltage drop of 18 volts which is the voltage across the resistor now going from 12 back to zero we have a voltage drop of 12 and so these numbers add up to zero which is the sum of the voltages around any closed loop and that's related to Kirk off's voltage law the next quantity that we need to calculate is the power the power delivered by the battery and the power absorbed by the resistors so power is equal to voltage time current now there's two other forms of this equation that you might find useful now now we know that V is equal to IR so if we replace V with ir then we'll get the second form of the equation which is i^ 2 R now since V equals I if we solve for I V ID R is I from Ohm's law so we can replace I with v R so another form of the equation is v^2 / R so we're going to use the different forms of that equation to calculate the power so let's find the power delivered by the battery the battery has a voltage of 30 volts and it delivers a current of 3 amps so 30 * 3 is 90 so the battery delivers 90 watt of power now what about the 6 Ohm resistor how much power does it absorb let's use the second form of the equation p is equal to i s r so the current that flows to the 6 Ohm resistor is 3 amps so 3^ 2times the resistance of 6 3^ 2ar is 9 that's 3 * 3 and 9 * 6 is 54 so the 6 Ohm resistor absorbs 5 4 watts of power now what about the 4 ohm resistor how much power does it absorb let's use the third form of the equation V ^2 / R so the voltage across the 4 ohm resistor is 12 and it has a resistance of four so we can do 12 * 12 / 4 or we can do 12 / 4 which is 3 time the other 12 which is 36 12 2 is 144 144 / 4 is 36 either case the 4 ohm resistor absorbs 36 watts of power notice that the total energy absorbed or the total power absorbed by the two resistors adds up to the power delivered by the battery which is 90 54 + 36 adds up to 90 now what exactly is power how can you define power electrical power or power in general is work / by time one watt is one Jewel per second so power is a rate is the rate of energy transferred so for example let's say if you have a system that could transfer 100 jewles of energy in 1 second and if you have another system that can transfer 100 jewles of energy in 10 seconds which one has more power well it's going to be the system on the left now granted both systems can do the same amount of work they can perform 100 jewels of work however the system on the left can do it at a much faster rate it could transfer or it could perform the same amount of work of 100 jewels in simply 1 second the other system is slower it performs the same work but over a longer period of time so because it's slower it has less power the power is 10 watts the other one it's 100 Watts this system can transfer energy at a much faster rate it can get the job done in 1 second as opposed to 10c so as you can see Power represents the energy transferred the rate of energy transferred like how fast energy is being transferred so power is the rate of change of energy per unit time now going back to the last example that we had where the battery delivered a power of 90 Watts so if the circuit is allowed to run for 10 minutes how much energy is transferred by the battery in a time of 10 minutes so we know that work is equal to power multiplied by time if power is work over time if you multiply both sides by T you get this equation but let's solve it as a unit conversion let's start with time time is 10 minutes now 90 Watts represents 90 juw per second so we need to convert minutes to seconds in 1 minute there are 60 seconds now you want to set it up in such a way that the unit minutes cancel and 90 Watts represents 90 jewles of energy being transferred every second so 60 * 90 what is the value of 60 * 90 we know 6 * 9 is 54 and if you add the two zeros is 5400 and then time 10 that's 54,000 so 54,000 jewles of energy will be transferred in 10 minutes if the battery is delivering 90 watts of power now let's solve another series circuit problem this time we're going to have three resistors instead of two so let's say the resistance of the first circuit is I mean the first resistor is 5 ohms the second one is three and the last resistor has a value of 2 ohms and let's say the voltage across the battery is 60 volts calculate the current that flows through the circuit so first you got to find a total resistance which is R1 + R2 plus R3 so that's 5 + 3 + 2 which is 10 ohms next we need to use V equals IR to calculate the current in the circuit so the voltage is 60 the resistance is 10 so therefore 60 ID 10 will give us the current of 6 amps so 6 amps of current flows through the circuit in this particular example now what do the voltage drop across each resistor so we need to use V equals I it's going to be the current that flows to the resistor multiplied by the value of the resistance so across the 5 Ohm resistor it's 6 * 5 which is 30 Watts I mean 30 volts and across the 3 Ohm resistor it's 6 * 3 which is 18 volt and across the 2 ohm resistor it's 6 * 2 which is 12 Vols notice that 30 + 18 + 12 adds up to 60 the sum of the voltage drops of the battery I mean the sum of the voltage drops of the resistor must add to the voltage of the battery which is in harmony with kirov voltage law now let's calculate the power delivered by the battery and absorbed by the resistors so for the battery let's use V * I so we have a voltage of 60 volts across the battery and it delivers 6 amps of current 6 * 6 is 36 so 6 * 60 is 360 so the battery delivers 360 watts of power now the first resistor it absor absorbs how much power let's see let's use the equation i^ 2 R so the current that flows to it is 6 amps times the resistance of five so 5 * 6 is 30 and 30 * another 6 is 180 so this resistor absorbs 180 watts of power now what about the 3 Ohm resistor how much power does it absorb let's use P equals VI the voltage across the 3 Ohm resistor is 18 and the current that flows through it is six so what is 18 * 6 10 * 6 is 60 8 * 6 is 48 60 + 48 is 108 now what about the last one let's use the third form of the equation so the voltage across the resistor is 12 and the resistance itself is 2 12 / 2 is 6 6 * the other 12 is 72 so if we add these two values that's 180 Watts plus another 180 adds up to 360 so the power delivered by the battery equals the sum of the powers absorbed by the resistors so energy is conserved in that particular example now let's move on to a parallel circuit a circuit with two resistors remember a parallel circuit is a circuit where the current has multiple paths to flow now the voltage of the battery is going to be 24 volts and we're going to use a 6 Ohm resistor and an 8 Ohm resistor now how can we calculate the current that flows through each resistor now what you need to know is that the voltage across two resistors connected in parallel is the same in this case 24 volts both resistors are connected across the same battery so if the potential at the bottom is zero at the top is 24 so let let's use the equation V equal I so let's start with the 6 Ohm resistor there's 24 volts across the 6 Ohm resistor so 24 / 6 will give us a current of 4 amps current flows in a direction from high voltage to low voltage much in the same way as water flows from a high position to a low position water is not going to flow from a low position to a high position gravity is going to pull it down so current in a similar sense flows from a high potential to a low potential voltage can also be called potential difference or electric potential some textbooks describe it as electromotive force or electron moving Force now what about the 8 Ohm resistor what is the current that flows through it so using the same equation V equals I ohms law the voltage is 24 volts and the resistance is eight so all you got to do is divide 24 by 8 and you can get the current that flows to it so 24 ID 8 is 3 so 3 amps of current flows through the 8 Ohm resistor now what is the current that leaves the battery how much is it the current that leaves the battery is equal to the sum of the currents that flow through the two resistors it's four + 3 which is 7 amps so 7 amps flows in this direction now if we focus on this point right here s amps of current approaches that point 4 amps goes down and 3 amps Trel this way this is related to Kirk off's current law the current that enters the point must equal the total current that leaves the point which is true seven amps of current approaches the point and seven amps of current moves away from the point the 4 amp current and the 3 amp current which adds up to seven they both move away from that point so that's in harmony with Kirk off's current law now now the last thing that we need to do in this circuit is calculate actually there's two things we need to do we need to find the equivalent resistance and the power delivered by the battery and absorbed by the two resistors now to find the total resistance of the circuit there's two methods that we can use first let's use ohms law V is equal to IR the total voltage of the battery is 24 volts and the current that leaves the battery is 7 amps so then this network of resistors can be viewed as one equivalent resistance if you use the total voltage and the current that leaves the battery so that equivalent resistance is basically 24 / 7 which as a decimal is about 3.43 so the equivalent resistance is about 3.43 ohms notice that the equivalent resistance or the total resistance of the circuit is less than six or eight in a series circuit as you add resistors the total resistance increases but in a parallel circuit as you add resistors the total resistance decreases because as you add a new Branch more current can flow through the circuit if the current goes up the resistance Goes Down And if if the resistance goes up current goes down so these two are inversely related if the voltage is held constant now there's another way in which you can calculate the equivalent resistance you can use this equation 1 / R let's say R total which is the same as RQ is equal to 1 R1 + 1 R2 so if you raise both sides to the minus one then you get this equation RT is equal to parentheses 1 6 + 1 8 raised to the minus one so if you type that in the calculator you're going to get the same answer 24 over 7 which is 3.43 ohms so you can find the total resistance both ways and now let's calculate the power delivered by the battery so p is equal to VI the voltage is 24 and the current is 7 so 24 * 7 that's 168 Watts delivered by the battery now what is the power absorbed by the 6 Ohm resistor so let's use this form of the equation i^ s r so we have a current of 4 amps that flow through it and the resistance is 6 4 * 4 is 16 16 * 6 is 96 so 96 watts is absorbed by the 6 Ohm resistor for the last one let's use V if we use v^2 R we will have to square 24 actually we could do it a different way let's use v^2 R so what is 24 ID 8 in this case let's divide first 24 / 8 is 3 so 3 * the other 24 is 72 notice that 96 is equal to I mean 168 is equal to 96 + 72 the total power delivered by the battery equals the total power absorbed by the two resistors now let's try one more example with a parallel circuit so let's say the voltage is 100 volts maybe that's too much let's say 20 and let's choose a 2 ohm resistor a 4 ohm resistor and a 5 Ohm resistor so to find the current that flows through each resistor we need to use ohms law V equal IR the current is going to be the voltage divided by the resistance so remember the voltage across each resistor connected in parallel is the same it's 20 volts so 20 / 2 is 10 so 10 amps of current flows through the two oral resistor 20 divid 4 is five so 5 amps of current flows through the 4 ohm resistor and 20 / 5 is four so 4 amps of current flow through the 5 Ohm resistor which means that a total of 10 + 5 + 4 or 19 amps leaves the battery now here's a question for you one the 19 amp of current reaches this point how much current flows in this direction so 10 amps goes in a downward Direction 19 minus 10 is 9 so 9 amps of current flows this way now once the 9 amps of current reaches this point five is going to go down so four travels this way because 9 minus 5 is four so we have four amps of current flowing in this branch then it's going to meet up with the 5 amp current 4 + 5 is 9 so nine is going to go this way and then that's going to meet up with the 10 amp current to add up to 19 and 19 amps of current flows through the battery so you can see how it works according to kirov current law now that we have all the currents in the circuit what about the power delivered by the battery and the power absorbed by each resistors go ahead and find the answer for that so p is equal to VI for the battery it has a voltage of 20 volts and it delivers 19 amps of current so the power delivered by the battery is 380 Watts now what about the power absorbed by the 2 ohm resistor so what we could do is simply use the same equation P equal VI so the voltage is 20 volts across it times the current of 10 amps that flow through it so 20 * 10 that's 200 so the 2 ohm resistor absorbs 200 watts of power now for the 4 ohm resistor it's going to be the voltage of 20 volts * 5 amps so 20 * 5 that's 100 watts of power and for the last resistor it's 20 volts * 4 amps which is 80 watts of power so if you add 200 Watts plus 100 that's 300 plus 80 that's 380 which equals the total power delivered by the battery now what about the equivalent resistance let's find it using both techniques as we uh did in the last last example so using ohms law the voltage of the battery is 20 the total current is 19 so the total resistance is 20 / 19 which is 1.05 and it makes sense the total resistance is going to be less than any of the three resistors so it has to be less than a 2 ohm resistor if we have a parallel circuit now we can also calculate it using equation 1 / R1 + 1 R2 + 1 / R3 in close in parentheses and raise to the 1 power so it's 1 over 2 + 1 4 + 1 5 raised to minus1 if you type it in you should get 1.05 as your answer so now you know how to solve a series and a parallel circuit just for review in a series circuit there's only one path for the current to flow and in a parallel circuit it contains multiple paths for the current to flow in a series circuit whenever you have resistors connected in series the current that flows through those resistors is the same current whenever you have resistors connected in parallel the voltage across those resistors is the same voltage so that is it for this video thanks for watching and have a great day