How to Solve Series-Parallel Resistor Circuits: Step-by-Step Guide

in this video we're going to talk about how to solve any type of series parallel combination circuit like this one

now the first thing you want to do is you want to calculate the equivalent resistance

so notice that these three resistors are in series with each other and so the equivalent resistance of

those three is going to be five plus three plus two which is ten so we can replace those three resistors

with a single 10 ohm resistor now every other part of the circuit is

going to remain the same initially so this is 10 ohms and this is still 10 this is still 5

and that's still 10. now notice that we have two resistors in parallel

they're directly across from each other and they have the same resistance whenever you have two parallel resistors

with the same value the equivalent resistance is simply half of that value

so individually they're both 10 so the equivalent resistance is going to be 5. so we can get rid of this

and replace it with a 5 ohm resistor so now these three are

in series with each other so then we could just add them 10 plus 5 plus 5 is 20.

so the equivalent resistance in the circuit is 20 ohms

once you have the equivalent resistance you could find the current that

is delivered from the battery and you could use v equals ir

so if you rearrange this equation to solve for i the current is going to be the voltage

of the battery divided by the total resistance or the equivalent resistance so the voltage of the battery is 60

volts in this example and the equivalent resistance is 20 ohms so 60 divided by 20 is 3 amps so that's

the current that leaves the battery now let's call this position a

let's say this is b c d

e and f now in order to calculate the current

that's flowing through each resistor it's helpful to know the potential at each of these points

and we're going to say at point a the potential is zero volts so now if we travel in this direction

from a to b we're going towards the positive terminal of the battery

and the battery increases the energy of the circuit because it delivers power to the circuit so as you go from a to b

it's going to be a voltage lift of 60 volts so the potential at b

is 60 volts now something else they need to know is that whenever or current

flows through a resistor the current flows from a high potential to a low potential

so we have the current of 3 amps it's flowing through the 10 ohm resistor

so let's call that just 3 amps so that means this is positive

and this is negative so c is at a lower potential than b so to calculate the potential at c

it's going to be the potential at b minus i r

because c is lower than b so the potential at b is 60 volts the current flowing through the 10 ohm

resistor is 3 amps multiplied by 10 so 3 times 10 is the voltage drop of 30 and then 60 minus 30

tells us that the potential at c is 30 volts now let's go from a to f

now this 3 amp current which flows through the battery also flows through the 5 ohm resistor

because this path here they're all in series whenever there's only a single path for the current to

flow you have a series path and the current flowing through that path is the same everywhere along that path

so what this tells us is that 3 amps of current flows through the 5 ohm resistor

and current always flows from a high potential to a low potential

so as we travel from a to f that's going to be a voltage lift because we're going from the negative

side to the positive side so anytime you go against the current the electric potential is increasing

if you're following the current like we did over here the potential decreased so to calculate the potential at f is

going to be the potential at a plus i r because we're going towards a higher

potential so this is going to be plus instead of minus

so the potential at a is zero the current flowing through the 5 ohm resistor is 3 amps times 5

so the potential at f is going to be 15 volts now let's get rid of this

and let's focus on the 10 ohm resistor because we now have the potential across c and

across f so we can calculate the current flowing through that resistor

now voltage is equal to current times resistance based on ohm's law

and voltage is also known as the potential difference between two points so we know that the current is flowing

in this direction and also current flows from a high potential to a low potential

so the potential at c is higher than the potential at a so the voltage across that resistor is

going to be the potential at c minus the potential at f

and that's equal to the current times the resistance so the potential at c is 30 the

potential at f is 15 and we don't know the current yet but the resistance is 10. so 30 minus 15

will give us a voltage of 15 volts across that resistor and so 15 divided by 10

is 1.5 so that's the current that flows through the 5 ohm resistor

so now we could calculate the current that flows through the other branch so let's focus on junction c

at that junction we have a current of three amps that is entering the junction

and 1.5 amps flows in this direction now according to kirchhoff's current law or his junction rule

the total current that enters into a junction is equal to the total current that leaves the junction

so if we call this i1 and this i2 then this must be i3

and so i1 enters the junction and the other two currents they leave it so i1 is equal to i2 plus i3

so i1 is 3 amps i2 is 1.5 so i3 is 3 minus 1.5 so this is also 1.5

amps so now we have the current flowing through the remaining three resistors

so i'm just going to put it here 1.5 amps now since the current flows from c to d

we know this is going to be positive and this is negative and it flows from d to e so

e is at a low potential then d and then it flows from e to f which means e is at a higher potential

than f now let's calculate the electric potential at point d

so the potential at point d is going to be the potential at point c minus ir

because as we go in the direction of the current we're going towards a lower potential

so d is lower than c the potential at c is 30. the current flowing through the 5 ohm

resistor is 1.5 so the voltage drop is 1.5 times 5 which is

7.5 and so 30 minus 7.5 tells us that the potential at d is

22.5 volts now let's calculate the potential at e so we have another voltage drop because

we're going towards a lower potential as we follow the direction of the current so it's going to be vd

minus ir so the potential at d is 22.5 we still have a current of 1.5 but this

time it goes to a 3 ohm resistor so 3 times 1.5 that's a voltage drop of 4.5

and 22.5 minus 4.5 gives us a potential of 18 volts at e

now to confirm the answer let's calculate the current going through the 2 ohm resistor

we already know it's 1.5 but let's confirm that answer because we have the potential at point f

so the potential at e minus the potential at f represents the voltage across the 2 ohm

resistor and voltage is equal to current times resistance

so the potential at e is 18 the potential f is 15 and r is 2.

so 18 minus 15 gives us a voltage of 3 volts and so 3 divided by 2

is 1.5 which is in agreement with this answer so now we've completely solved the

circuit we've calculated the current flowing through every resistor

and we also know the electric potential at every point in a circuit

and so that's it now if we wish to calculate the power absorbed by resistor

let's say if we want to calculate the power absorbed by this resistor we can use this formula i squared times

r so the current that's flowing through it is 1.5 and the resistance is 10. so the power absorbed by that resistor

is simply 22.5 watts now if we wish to calculate the power delivered by the battery

we can use this formula it's voltage times current so we have 60 volts

times the current of 3 amps and so that's going to be 180 watts now the power delivered by the battery

must be equal to the power absorbed by each resistor and that's the second way you can

confirm if you have the right answer so let's calculate the power absorbed by each resistor so we already have this

one and we said that's 22.5 watts

now let's focus on this resistor so it's i squared times r it's 3 squared times 10

and so the power absorbed by that resistor is 90 watts and for this one i squared times r is

going to be 3 squared times 5 and so it's 45 watts for that resistor and now for this one

it's going to be 1.5 squared times 5 and that's 11.25 and then for this one it's 1.5 squared

times 3. and so that's only 6.75 and then for this one 1.5 squared times

2 so that consumes 4.5 watts so now let's add up all the numbers

so we have 90 watts plus 45 plus 22.5

and then plus 11.25 plus 6.75 and then plus 4.5

let's see if that adds up to 180 and indeed it does so the total power absorbed by the

resistors is equal to the power delivered by the battery

and so this circuit the values that we have in this circuit is correct

now let's work on a similar circuit but one that's slightly different in a way

so we're just going to add one more resistor to the circuit that we had before

and the values will change of course so what if we put a resistor right in the middle

so this time we're going to have a 120 volt battery and this resistor is going to be 10 ohms

and this one is 12 ohms and this is going to be an 8 ohm

resistor and then 3 9

5 and in the middle this is going to be a 4 ohm resistor so feel free to pause the video

calculate the current that is delivered from the battery and the current that flows through each resistor in a circuit

so go ahead and try this problem now the first thing we need to do is calculate the equivalent

resistance so we need to know which resistors are in series and which ones

are parallel to each other now a series circuit is one in which the current has only one path to flow so

notice that the three and the nine ohm resistor they're in series with each other

because the current can't leave this junction so there's a single path of the currency

flow so we can replace those two with a 12 ohm resistor

so now the circuit becomes what we have here so this is still 10

8 and 12. and now i can write it like this

so i'm going to replace these two with a 12 ohm resistor and this is five

now we still have this resistor going in this direction now these two points are identical

so i can say that this resistor is going in this direction it's the same as drawing it this way

because they're attached to the same thing so this is four the 12 ohm resistor is

in parallel with the 4 ohm resistor now because these values are different

we're going to have to use a formula to find the equivalent resistance and

we can replace this 12 with the equivalent resistance so the equivalent resistance between

those two is going to be one over r one or one over four plus one over r two or one over twelve

raised to the minus one one fourth plus one twelve is one over three

if you use your calculator or if you like get common denominators

and then if you raise it to minus one it's three now if you don't have access to a

calculator here's what we need to do multiply this fraction by three over three

so it's going to be three over twelve plus one over twelve and three plus one is four to get four

over twelve and whenever you raise a fraction to the negative one power you need to flip the

fraction so it becomes twelve over four which is the same as three

so the equivalent resistance of those two in parallel is 3 ohms now notice that

the 3 ohm and the 5 ohm they're in series with each other there's only one path for the current to flow and three

plus five is eight so we can replace this with a eight ohm resistor

and that eight ohm resistor is in parallel with this eight ohm resistor and because

they're the same we know that the equivalent resistance of those two will be half

of the individual values so we can replace that with a 4 ohm resistor

so now the total resistance is going to be 10 plus 12 plus 4

which is 26 ohms so that's the equivalent resistance

now let's calculate the current in the circuit so it's going to be the voltage of the

battery divided by the equivalent resistance and so that's 120 volts

divided by 26 ohms and so the current that leaves the battery

is 4.615 amps and that's a rounded answer so keep that

in mind now let's define this point as point a

b c d e and f

and we're gonna assign point a a potential of zero volts so let's calculate the potential

everywhere else and at the same time let's calculate the currents and other positions

now going from b to c we know that the resistor i mean the potential decreases from high to low

because the current is flowing in that direction the current always flows from high potential to low

potential now the potential at b if we go from a to b we're going towards uh the positive

terminal of the battery so that's going to be a voltage lift of 120 so the potential at b

is 120 volts relative to a so now that we have the potential at b we can calculate the potential at c

and so it's going to be a voltage drop across this resistor so the potential at c is going to be the potential at b

minus ir so at b is 120 volts and the current flowing through the 10

ohm resistor is 4.615 multiplied by resistance of 10. so the potential at c

is 73.85 volts now let's go from a to f

so the current is flowing in this direction and that current is the same as this

current so this potential is positive and over here this side of the resistor

is negative so f is at a higher potential than a so the potential at f is going to be the

potential at a plus ir because as we go from a to f we're going towards the positive

terminal of the resistor towards the higher potential so a is zero the current is 4.615

and the resistance is 12. so 4.615 times 12 that will give us a potential of

55.38 volts at f now that we have the electric potential

at point c and f we can now calculate the current going through the 8 ohm resistor

now the current is going to be flowing in this direction so that it can reach the negative

terminal of the battery so therefore we know that this part of the resistor is positive

and this part is negative also we can see that c is at a higher potential than f so current always flows

from high potential to low potential so the potential at c minus the potential at f represents the voltage

across the 8 ohm resistor so remember voltage is the potential difference between two points

and that's equal to i times r so the electric potential at point c is 73.85 volts and that point f is

55.38 volts and the resistance is 8 ohms so 73.85

minus 55.38 that gives us a voltage of 18.47 so to calculate the current

it's going to be 18.47 divided by 8. so the current that flows through that resistor

let's see if i could fit it in here it's 2.309 if we round it

so i'm going to highlight that in red amps so

if we have a current of 4.615 going towards point c and a current of 2.309

amps going this way now the current that flows through this

branch and that branch has to be the difference between these two values

and because we have two different branches from c to d and c to

e we don't know what the answer is yet however there is something that we can

do so let's say if we have a picture that looks like this

so from c to e we know that this represents a resistance of 12 ohms it's three plus

nine and the current flowing through that branch is going to be the same

through the 3 ohm resistor and through the 9 ohm resistor so we're going to say this resistance is 12.

and the current that flows through this branch goes to a 4 ohm resistor now the current that is going towards

junction c which is right here over here this is a junction

e the current that flows to c is 4.615 amps

and the current that leaves that branch is 2.309 now the current that leaves junction c

to be split off into these two resistors that's going to be the difference of 4.615

and 2.309 and so that's going to be 2.306

that goes in that direction now notice how we're going to calculate the current that flows through each of

those two branches individually so let me just redraw this so right now we have a current

of 2.306 amps going in

so we'll call it it and this is a a 12 ohm resistor and this is a 4 ohm

resistor so with this information alone how can we calculate the current that flows

through the 12 ohm resistor and the current that flows through the 4 ohm resistor

now let's call the current that flows through the 12 ohm resistor i1 and the current that flows through the 4 ohm

resistor i2 we know that i1 plus i2 has to add up to the total current entering this little circuit

and notice that this resistance is less and this resistance is more so which

branch should have a greater current is i1 greater than i2 or is i2 greater than i1

now whenever you decrease the resistance the current increases so because this resistance is less than the other one

i2 is greater than i1 so because this is three times less than 12

i2 is going to be three times greater than i1 so we need two numbers

that add up to 2.306 and

we know that i2 has to be 3 times the value of i1 so we can replace i2 with 3i1

so it's i1 plus 3i1 and that's equal to 2.306 and so 4i1 is equal to that number

and 2.306 divided by 4 tells us that i1 is

0.5765 amps so to calculate i2 it's going to be 2.306

minus 0.5765 and so this current is 1.7295

as we can see this number is three times greater than that number

now it turns out that there's another way in which we can get the answer because sometimes these two resistors

may not be so nice you may have some other numbers that are just hard to work with

so let's call this r1 the 12 ohm resistor and this is r2 so let's say if we wish to calculate i1

it's going to be the total current times the other resistor r2 divided by the total resistance

so the total current entering this little circuit is 2.306 and r2

is 4. r1 plus r2 is 16. so it's 2.306 times 4 over 16 and that will give you 0.5765

now if you wish to calculate i2 you can use this equation

i2 is going to be the total current times the other resistor r1 divided by the total resistance r1 plus r2

so the total current is still 2.306 r1 is 12 r1 plus r2 is 16. so it becomes 2.306 times 12

divided by 16 and that will give you this current 1.7295 amps

so let's write this information here so we have the current flowing in this branch

that's through the 12 ohm resistor so that's 0.5 amps

we don't need this information anymore so let's just get rid of that

and we can get rid of this now the current flowing through the 4 ohm resistor

we said that's 0.5765 amps i should have made the circuit bigger

now let's calculate the potential at d so with the current is flowing through

the 3 ohm resistor in that direction so this has to be the positive terminal of the resistor and this is the negative

terminal of the resistor as it flows from c to d we know that d is a lower potential than the potential at c

so we can say that the potential at d is the potential at c minus i r

the electric potential at c is 73.85 and the current flowing through the 3 ohm resistor

we now know it to be 0.5765 so let's multiply that by 3. and so the potential at point d is

72.12 volts so now let's calculate the electric potential at e

so it's going to be the potential at e minus the potential at d

actually let's not do it that way the potential at e rather is the potential at d minus i r

we're still flowing in the direction of the current as we go from d to e so we still have a voltage drop across

this resistor so anytime we follow the direction of the current

if we move from d to e in the direction of the current that's going to be a voltage drop if we move from e to d

against the current that's a voltage lift so the potential at d we said it's 72.12

and the current is 0.5765 times the resistance of 9. and so the potential at e

is 66.93 volts now i need to correct this current

the current flowing through the 4 ohm resistor had to be the difference between

this current this one and that one so i believe it was like 1.7 something so if we take 4.615

subtract that by 2.309 and subtract that by 0.5765 this current was supposed to be 1.7295

amps now let's confirm this answer because we now have the electric potential between

c and e so we have the voltage across the 4 ohm resistor so current is going to flow from c to e

from high potential to low potential so vc minus ve that's going to be the

voltage across the 4 ohm resistor so vc is 73.85 the e

is 66.93 and that's gonna equal i times four so 73.85 minus 66.93

that's a voltage of 6.92 and divide that by 4 that will give us a current of 1.73 amps

which if you round it that's about the same as 1.7295 so we know that the numbers that we have

so far are correct now we have the potential at e and f so let's calculate the current that flows

from e to f so it's going to be v e minus v f and that's equal to i r so

the potential at e is 66.93 the potential at f is

55.38 and the resistance is 5. so 66.93 minus 55.38

that gives us a voltage of 11.55 so if we divide that by 5

then the current flowing through the 5 ohm resistor is 2.31 amps to confirm this answer

notice that the current that flows from d to e and the current that flows from c to e

must equal the current that flows from e to f which is this current so if we add 0.5765

plus 1.7295 that gives us a current of 2.306

which if you round it that's approximately 2.31 so we have confirmed every value in the

circuit so all the potentials and the currents flowing through each resistor that you

see is correct now we can calculate the power absorbed by each resistor and the power delivered

by the battery but you know how to do that at this point so you can check it that way as well

the power delivered by the battery has to equal to the total power absorbed by all of the resistors

and that could be the final check if you want to make sure your work is correct so that's it for this video

and now you know how to solve any type of series parallel combination circuit

where you have one battery and just a bunch of resistors so thanks again for watching

you