Resistors in Series - Lab data confusion

In summary, the conversation was about data discrepancies in a lab on resistors in series. The individual resistances and voltage values were given, and the total measured current did not match the sum of the separate currents measured at each resistor. It was discovered that the circuit was most likely set up as a parallel series instead of a series circuit. The participants discussed different methods for calculating the total current in a series circuit and the voltage drops across each resistor. Finally, they discussed the formula for calculating total resistance in a parallel circuit and how it relates to the total resistance in a series circuit.
  • #1
shmoop
Hi guys!

(see my attached photo to better understand where I am coming from!)

So after some research, I've discovered that the current at different points in a simple series circuit is supposed to be the same value, and that the voltage is supposed to be different values.

I performed a lab on resistors in series the other day, and I believe my data is incorrect.
The majority of the data I got tells me that I may have set my circuit up as a parallel series - however the fact that the total measured current is not equal to the sum of the separate currents measured at each resistors conflicts with this. (see attached image of my lab data for clarity!)

Furthermore, due to the value of the total current as well as the value of the total voltage, the total R calculated using Ohm's law is equal to the Req that was calculated through adding the individual resistances up (which were also calculated through Ohm's law).

I am wondering, what did I do to obtain this kind of data information? As it does not conform what I would designate as a series circuit, nor a parallel circuit.

Does anyone have any idea what I have done to obtain this lab data? What kind of circuit did I create?

Thanks a lot in advance.

Edit: Here are the resistor values:

R1=100 ohms
R2= 220 ohms
R3= 470 ohms
 

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  • #2
I think you wired it in series but wrote the current values in the voltage column and the voltage values in the current column.

Edit: oh darn that can't be the only problem as I doubt you were playing with 50volts.
 
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  • #3
CWatters said:
I think you wired it in series but wrote the current values in the voltage column and the voltage values in the current column.
I considered that too. But I followed the formula R=V/I, and the values worked out that way. If I divide the current values by the voltage values instead, the resistance doesn't work out like it should...
 
  • #4
shmoop said:
If I divide the current values by the voltage values instead, the resistance doesn't work out like it should...

you SHOULDNT be dividing current by voltage ... .it will never give you the correct answer
 
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  • #5
davenn said:
you SHOULDNT be dividing current by voltage ... .it will never give you the correct answer
I know haha! But I was responding to CWatters who had mentioned that it looked like my current and voltage values were reversed. In which case I would do the opposite of what I had previously done, therefore dividing the current by the voltage.
 
  • #6
shmoop said:
(see my attached photo to better understand where I am coming from!)
your attached image is VERY difficult to read

What is the battery voltage ?
are the resistance values given ... even at least one of them ?
 
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  • #7
davenn said:
your attached image is VERY difficult to read

What is the battery voltage ?
are the resistance values given ... even at least one of them ?

Here is a zoomed in version of the photograph!

The supply was approximately 5 volts.

R1=100 ohms
R2= 220 ohms
R3= 470 ohms

Sorry about that!
 

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  • #8
shmoop said:
The majority of the data I got tells me that I may have set my circuit up as a parallel series - however the fact that the total measured current is not equal to the sum of the separate currents measured at each resistors conflicts with this. (see attached image of my lab data for clarity!)

you only need to work out the current through 1 resistor to work out the total current in the circuit
you DONT add it up as you do with the voltage drops across each resistor
 
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  • #9
davenn said:
you only need to work out the current through 1 resistor to work out the total current in the circuit
you DONT add it up as you do with the voltage drops across each resistor

How would you mathematically determine the total current in the circuit with the current through 1 resistor?
I know that you don't add it up for a series circuit - I was trying to say that that is what I would have done for a parallel circuit. I wrote series accidentally! Sorry. I meant to say the majority of my data makes it look like I set up the situation as a parallel circuit. (all the different current values, and the fact that the voltage stays the same)
 
  • #10
21397305_10212281321761481_848215488_n-jpg.jpg


OK that is easier to read

the figures in your current column cannot be correct
and I don't know why they are asking you to calculate resistor values for column 3 when the values are already given to you ??
 
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  • #11
davenn said:
View attachment 210432

OK that is easier to read

the figures in your current column cannot be correct
and I don't know why they are asking you to calculate resistor values for column 3 when the values are already given to you ??

I assume they ask so that you can then use the values to perform the Rtotal=1/R1+1/R2+1/R3

To prove that it is a series circuit
 
  • #12
shmoop said:
How would you mathematically determine the total current in the circuit with the current through 1 resistor?

you treat all series resistors as 1 resistor ... add up those 3 values for a total resistance value

give me that answer

then divide the voltage (5V) by the total resistance to get your current

give me that answer

then you can look at the voltage drops across each resistor

Vdrop R1 = resistance x current =

Vdrop R2 = resistance x current =

Vdrop R3 = resistance x current =

adding up those 3 voltages should equal your supply voltage = 5V
 
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  • #13
shmoop said:
I assume they ask so that you can then use the values to perform the Rtotal=1/R1+1/R2+1/R3

no, that formula is for parallel resistors
 
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  • #14
davenn said:
no, that formula is for parallel resistors

Oopsies. I meant to make sure it matches the total R with the Rtotal=R1+R2+R3
 
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  • #15
shmoop said:
Oopsies. I meant to make sure it matches the total R with the Rtotal=R1+R2+R3

OK

so copy and paste that post #12 of mine and fill in the answers
and show your working, not just the answer
 
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  • #16
davenn said:
you treat all series resistors as 1 resistor ... add up those 3 values for a total resistance value

give me that answer

then divide the voltage (5V) by the total resistance to get your current

give me that answer

then you can look at the voltage drops across each resistor

Vdrop R1 = resistance x current =

Vdrop R2 = resistance x current =

Vdrop R3 = resistance x current =

adding up those 3 voltages should equal your supply voltage = 5V

My voltages are 5V throughout. Which is confusing. Because they equal 15V when summed, as opposed to 5V. However, when calculating my individual resistance values, when summed they equaled the total resistance initially calculated. This is what confuses me
 
  • #17
shmoop said:
My voltages are 5V throughout.

they cannot be 5V throughout because that is your total voltage

do the sums that I have asked you to do and get your working and answers
 
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  • #18
have you copied and pasted this ?

you treat all series resistors as 1 resistor ... add up those 3 values for a total resistance value

give me that answer

then divide the voltage (5V) by the total resistance to get your current

give me that answer

then you can look at the voltage drops across each resistor

Vdrop R1 = resistance x current =

Vdrop R2 = resistance x current =

Vdrop R3 = resistance x current =

adding up those 3 voltages should equal your supply voltage = 5Vjust a note I didn't state when I typed that out originally
should be ( for clarity)

Vdrop R1 = resistance R1 x current =

Vdrop R2 = resistance R2 x current =

Vdrop R3 = resistance R3 x current =
 
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  • #19
davenn said:
have you copied and pasted this ?

you treat all series resistors as 1 resistor ... add up those 3 values for a total resistance value

give me that answer

then divide the voltage (5V) by the total resistance to get your current

give me that answer

then you can look at the voltage drops across each resistor

Vdrop R1 = resistance x current =

Vdrop R2 = resistance x current =

Vdrop R3 = resistance x current =

adding up those 3 voltages should equal your supply voltage = 5Vjust a note I didn't state when I typed that out originally
should be ( for clarity)

Vdrop R1 = resistance R1 x current =

Vdrop R2 = resistance R2 x current =

Vdrop R3 = resistance R3 x current =

The values that were measured were the voltage and the current. The resistances were calculated from these. For me to perform these calculations seems aimless, as I will just get ~5V for each voltage drop. Which is part of my main issue.
 
  • #20
shmoop said:
The values that were measured were the voltage and the current. The resistances were calculated from these. For me to perform these calculations seems aimless, as I will just get ~5V for each voltage drop.

it's not aimless as you will find your mistake
AGAIN, there is NOT a 5V drop across each resistor --- it's impossible to have a 5V drop across each resistor

Copy and Paste ( DONT QUOTE IT) my post and do the sums as I have laid them out and you will see where you have gone wrong :smile:Dave
 
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  • #21
davenn said:
it's not aimless as you will find your mistake
AGAIN, there is NOT a 5V drop across each resistor --- it's impossible to have a 5V drop across each resistor

Copy and Paste ( DONT QUOTE IT) my post and do the sums as I have laid them out and you will see where you have gone wrong :smile:Dave

IMG_6720.JPG


I am confused
 
  • #22
shmoop said:
I am confused

Im not surprised

what is Ohms law ? as you didn't use it correctly to find the current

and since you got the incorrect current, everything following that is incorrect

I also notice you didn't use the resistor values you stated in post 1

100 Ohms 220 Ohms 470 Ohmsyou didn't do your working on the post so I couldn't copy and paste individual parts
 
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  • #23
davenn said:
Im not surprised

what is Ohms law ? as you didn't use it correctly to find the current

and since you got the incorrect current, everything following that is incorrect

I also notice you didn't use the resistor values you stated in post 1

100 Ohms 220 Ohms 470 Ohmsyou didn't do your working on the post so I couldn't copy and paste individual parts

Woops. Huge mistake there. I will redo it. I thought I was supposed to use my experimental values.
 
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  • #24
we will get you there, you just need to get the basics correct :smile:

If I give you the answers, you won't learn anything, because you still won't understand where you went wrong :wink:
 
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  • #25
shmoop said:
Woops. Huge mistake there. I will redo it. I thought I was supposed to use my experimental values.

that's OK we can work with the actual values rather than the idea values you originally posted
it's just your answers will be a little different, that's all :smile:
 
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  • #26
davenn said:
that's OK we can work with the actual values rather than the idea values you originally posted
it's just your answers will be a little different, that's all :smile:

IMG_6721.JPG


I see! So these are the voltage drops!
 
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  • #27
shmoop said:
I see! So these are the voltage drops!

YES ! well done :smile:

none of the individual drops can = 5V, they all must be smaller
and when added together will = 5V ( or whatever the supply voltage is)
 
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  • #28
Now you could redo the math using your actual resistor values
you should again end up with 5V ( as a total drop) give or take a couple of decimal points

each individual drop will be a little different to what you just worked out for the ideal resistor values
 
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  • #29
Now that we have the resistor values I think I can see what you did in the experiment.

It looks to me like you started with the resistors in series and measured the overall current correctly (but forgot the current values were in mA not A).

For example the total resistance is 790 Ohms so the current should be about 6.37mA but you wrote down 6.37A. Then when calculating the total resistance you did R=V/I (which is correct method) but you used:
R = 5.03/6.37 = 0.789 Ohms
instead of
R = 5.03/(6.37 * 10-3) = 789 Ohms (very close to 790 Ohms).

Then when you came to measuring the current through each resistor you connected each resistor to the power supply on it's own, which is incorrect. So the voltage across each resistor was the same about 5.03V.

The currents should have been...

For R1... I = 5.03/100 = 50.3mA (You recorded 50.2A)
For R2... I = 5.03/220 = 22.8mA (You recorded 23A)
For R3... I = 5.03/470 = 10.7mA (You recorded 10.9A)

So again it just looks like you forgot the values were in mA.

Again this made all your resistance calculations out by a factor of 100. Your calculations should have given you..

R1 = 100 Ohms
R2 = 218 Ohms
R3 = 463 Ohms

These are all within tolerance.

Of course you should have left all the resistors in series and measured the voltage across each.
 
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  • #30
How did you calculate the resistance with those current values?

In a series circuit, the current value is supposed to stay the same, and the voltage value is supposed to vary.
 
  • #31
shmoop said:
How did you calculate the resistance with those current values?

In a series circuit, the current value is supposed to stay the same, and the voltage value is supposed to vary.

That's what got me thinking that you didn't have all three resisters in series when measuring the current through each resistors . If you changed the circuit and connected each resistor on it's own to the power supply (or put them all in parallel) then the voltage would be the same each time and the current values would change. The calculated current values are very close to the figures you measured if you allow for the mA vs A error.

So using your data with the current corrected to mA...

R1 = 5.06/(50.2 * 10-3) = 100.8 Ohms
R2 = 5.03/(23.0 * 10-3) = 218.7 Ohms
R3 = 5.05/(10.9 * 10-3) = 463.3 Ohms

Which are close to the correct values.
 
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  • #33
CWatters said:
That's what got me thinking that you didn't have all three resisters in series when measuring the current through each resistors . If you changed the circuit and connected each resistor on it's own to the power supply (or put them all in parallel) then the voltage would be the same each time and the current values would change. The calculated current values are very close to the figures you measured if you allow for the mA vs A error.

So using your data with the current corrected to mA...

R1 = 5.06/(50.2 * 10-3) = 100.8 Ohms
R2 = 5.03/(23.0 * 10-3) = 218.7 Ohms
R3 = 5.05/(10.9 * 10-3) = 463.3 Ohms

Which are close to the correct values.

Yes, I believe you are very correct. I must have disconnected the series circuit in the process of attempting to measure the voltage drops.

Thanks a lot for your input!
 
  • #34
davenn said:
Now you could redo the math using your actual resistor values
you should again end up with 5V ( as a total drop) give or take a couple of decimal points

each individual drop will be a little different to what you just worked out for the ideal resistor values

Thank you very much! You were a great help in having me figure out what my voltage drops were. I appreciate it
 
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  • #35
You can read the resisters by their color codes or you can measure them one at a time, stand alone. That will help greatly. For resisters in series there are three important things to consider that your data has completely wrong:
1) The current through all resisters will be equal.
2) The voltage drop over any resister is proportional to the resistance.
3) The total voltage drop is the sum of the individual voltage drops across the resisters.

You should make sure that your measurements agree with these principles for resisters in series before you precede.
 
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Related to Resistors in Series - Lab data confusion

1. What is the purpose of conducting a lab experiment on resistors in series?

The purpose of this lab experiment is to understand the behavior of resistors when connected in series, and to measure their total resistance using different methods.

2. What is the difference between resistors in series and parallel?

Resistors in series are connected one after the other in a single path, while resistors in parallel are connected side by side in multiple paths. In series, the total resistance is equal to the sum of individual resistances, while in parallel, the total resistance is less than the smallest individual resistance.

3. How do you calculate the total resistance of resistors in series?

To calculate the total resistance of resistors in series, you simply add up the individual resistances. This is because the current has to pass through each resistor, and the total resistance is equal to the sum of all the individual resistances.

4. What are some potential sources of error in a lab experiment on resistors in series?

Some potential sources of error in this lab experiment include using incorrect values for the resistors, having loose connections in the circuit, and not measuring the resistances accurately. Other sources of error could include variations in the power supply or temperature changes during the experiment.

5. How can the results from a lab experiment on resistors in series be applied in real-life scenarios?

The concept of resistors in series is commonly seen in electronic circuits, such as in household appliances, computers, and cars. Understanding how resistors behave in series can help in designing and troubleshooting these circuits. Additionally, the knowledge gained from this experiment can be applied in other areas of science and engineering, such as in understanding the flow of electricity in a series of wires or pipes.

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