Why is my calculated resistance higher than the resistor used

[Daniel2244]

Homework Statement

Why is my calculated total resistance of the circuit higher than the resistor used in series and the theoretical total resistance? In a series circuit I used a 100Ω resistor and measured the current at 1.0v intervals from 1.0-5.0v.
At 1.0v the current was 8.70mA using a 100Ω resistor
2.0v - 17.60mA
The 100Ω resistor measured 97Ω using a DMM
DC circuit was used.

Homework Equations

R=V/I, Rt=R1+R2+R3+Rn...

The Attempt at a Solution

The theoretical total resistance of the series circuit is equal to the sum of resistor(s) (97Ω).
Calculating total resistance at different voltage outputs: 8.70mA x 0.001=0.0087A
at 1.0V total resistance = 1/0.0087=114.9Ω
17.60x0.001= 0.0176 at 2.0V total resistance = 2v/0.0176A=113.63Ω

I have no explanation on how the total resistance can exceed the 98Ω resistor.

 

[gneill]

Did you measure the actual resistance of the resistor (with an ohmmeter), or did you assume that its value was exactly its labeled value? What tolerance was it labeled to have?

 

[Daniel2244]

Did you measure the actual resistance of the resistor (with an ohmmeter), or did you assume that its value was exactly its labeled value? What tolerance was it labeled to have?

It was a 100Ω resistor and I used a digital mulimeter on ohmnmeter setting to measure the resistance of 98Ω.

 

[gneill]

Okay. How were the voltage levels established? Did you measure them each time? If so, with what instrument?

 

[Daniel2244]

Okay. How were the voltage levels established? Did you measure them each time? If so, with what instrument?

voltage were measured in 1.0v steps - 1.0v, 2.0v, 3.0v, 4.0v, 5.0v

 

[gneill]

voltage were measured in 1.0v steps - 1.0v, 2.0v, 3.0v, 4.0v, 5.0v

With your DMM? How about the current?

 

[Daniel2244]

With your DMM? How about the current?

yes with the DMM and the current for 1.0v - 8.70mA and 2.0v -17.60mA

 

[gneill]

You had a single DMM to measure all the values? So you couldn't measure the voltage and current at the same time?

 

[Daniel2244]

You had a single DMM to measure all the values? So you couldn't measure the voltage and current at the same time?

I used two DMM one for current and one for voltage

 

[gneill]

I used two DMM one for current and one for voltage

Did you measure the voltage across the supply or across the resistor? In other words, was the current-measuring DMM included as part of the "load resistance"?

 

[CWatters]

Have you got a make and model for the dmm?

 

[Daniel2244]

Did you measure the voltage across the supply or across the resistor? In other words, was the current-measuring DMM included as part of the "load resistance"?
View attachment 221043

I measured the voltage across both power pack and resistor

 

[Daniel2244]

Have you got a make and model for the dmm?

no. I am not sure why both of you are talking about the DMM.

 

[gneill]

no. I am not sure why both of you are talking about the DMM.

There are four "variables" in your circuit: The power supply, the resistor, and two meters. One meter is inserted in series to measure the current. If it happened to be a really poor meter, it might present more resistance than you expect.

I measured the voltage across both power pack and resistor

You took both readings in each case? Were the two readings the same in each case?

What did you find for the other voltage values (3,4, and 5 volts)?

 

[Daniel2244]

There are four "variables" in your circuit: The power supply, the resistor, and two meters. One meter is inserted in series to measure the current. If it happened to be a really poor meter, it might present more resistance than you expect.You took both readings in each case? Were the two readings the same in each case?

No, the voltage across the resistors decrease (isn't this potential difference?) at 1v- 0.89across component (ac for short) 2v- 1.20v, 3v-2.80v, 4v-3.72v, 5v-4.65v

What did you find for the other voltage values (3,4, and 5 volts)?

3v- 28.2mA, 4v- 38mA, 5v-48.3mA

 

[gneill]

No, the voltage across the resistors decrease (isn't this potential difference?) at 1v- 0.89across component (ac for short) 2v- 1.20v, 3v-2.80v, 4v-3.72v, 5v-4.65v

Okay. So there's an additional potential drop occurring somewhere else in your circuit. The candidates include unexpected connector resistance, and the current meter. Did you ever measure the potential across the current meter for any of the setups?

Are you certain that both of your DMMs were set on DC ranges and not AC by accident?

 

[Daniel2244]

Okay. So there's an additional potential drop occurring somewhere else in your circuit. The candidates include unexpected connector resistance, and the current meter. Did you ever measure the potential across the current meter for any of the setups?

No, I didn't measure p.d across DMM (used to measure current)

Are you certain that both of your DMMs were set on DC ranges and not AC by accident?

No, I positive that they were set of DC

 

[gneill]

No, I didn't measure p.d across DMM (used to measure current)

Too bad. It would be interesting to track down where the unexpected potential drop occurred.

 

[Daniel2244]

Too bad. It would be interesting to track down where the unexpected potential drop occurred.

This is how I set up my circuit: https://gyazo.com/72f50f46557719cabcdd93b9a31cc3dd.
So, Have I calculated the total resistance right? using the voltage output from power pack and dividing it by the current Rt=V1/A1? And the reason the calculated total resistance is higher than the theoretical total resistance and the resistor used is because there is added resistance from the DMM and power pack?

 

[gneill]

This is how I set up my circuit: https://gyazo.com/72f50f46557719cabcdd93b9a31cc3dd.
So, Have I calculated the total resistance right? using the voltage output from power pack and dividing it by the current Rt=V1/A1? And the reason the calculated total resistance is higher than the theoretical total resistance and the resistor used is because there is added resistance from the DMM and power pack?

Yes, the total load resistance is $R_t = V_1/A_1$. This would include any resistance in the connectors, wires, and current meter DMM. It would not, however, include the internal resistance of the battery (power pack) as presumably $V_1$ is being measured at its external terminals. Any voltage drop due to its internal resistance takes place where you can't see it. All you can "see" is the voltage at its terminals, which presumably you are manually adjusting to read 1 V, 2 V, 3 V, ...

 

[Daniel2244]

Yes, the total load resistance is ##R_t = V_1/A_1

This can't be correct as for the 2nd resistor I used 220Ω with the measured value of 218Ω on the ohmmeter (in series). If a voltage output 1.0v the current was 03.1mA. Using the equation R=V/I changing mA to A 03.1 becomes 0.0031. Putting the numbers into equation R=1/0.0031= 322.58Ω which cannot be correct that's 104Ω higher than the actual value

 

[gneill]

This can't be correct as for the 2nd resistor I used 220Ω with the measured value of 218Ω on the ohmmeter (in series). If a voltage output 1.0v the current was 03.1mA. Using the equation R=V/I changing mA to A 03.1 becomes 0.0031. Putting the numbers into equation R=1/0.0031= 322.58Ω which cannot be correct that's 104Ω higher than the actual value

If you are measuring the voltage across the power supply and the current in the circuit, then the load is everything connected to the power supply:

So something must be fishy with that DMM you're using for the current readings. Is it autoranging or do you have to select particular ranges manually for the current?

 

[Daniel2244]

So something must be fishy with that DMM you're using for the current readings. Is it autoranging or do you have to select particular ranges manually for the current?

Manually, I selected milliamps because the current was very small.

 

[gneill]

I don't suppose you snapped any pictures of the lab setup? It's a good idea to do so. Sometimes one forgets to note a detail during the lab and can refer to the pictures. I think it would be helpful to know more about the make and model of the DMM.

 

[CWatters]

no. I am not sure why both of you are talking about the DMM.

An "ideal" voltmeter would have an infinite resistance so that it didn't act like another load resistor.

An "ideal" current meter would have zero resistance so that there was no voltage drop across it as that would also affect the circuit you were measuring.

Most modern DMM are close to "ideal" but aren't perfect. That's why we were asking about your DMM. However the size of the error makes this unlikely to be the cause of the discrepancy

Was the battery in the DMM ok?
Does it look like a normal commercial product or has someone put it in a wooden box?

 

[Daniel2244]

Was the battery in the DMM ok?
Does it look like a normal commercial product or has someone put it in a wooden box?

Yeah, everything looked fine. The used was a standard commercial one.
Would the resistor becoming un-clipped cause such a high resistance?

 

[gneill]

Would the resistor becoming un-clipped cause such a high resistance?

Yes, or if the contact was poor. Corroded contacts can have this effect.

What sort of connections were made between components?

 

[Daniel2244]

I don't suppose you snapped any pictures of the lab setup?

No I didn't, however I will in future

 

[Daniel2244]

Yes, or if the contact was poor. Corroded contacts can have this effect.

What sort of connections were made between components?

Crocodile clips

 

[CWatters]

On my meter poor contact at the probes gives an error of a few ohms. I suppose that sort of thing might account for your 13ohm error in 100 =13%. It's harder to explain a 100 ohm error in 200= 50% error??

 

[Daniel2244]

On my meter poor contact at the probes gives an error of a few ohms. I suppose that sort of thing might account for your 13ohm error in 100 =13%. It's harder to explain a 100 ohm error in 200= 50% error??

+ If my component did un-clip the circuit would become broken and no current would flow

 

[gneill]

Does your DMM require you to move one of the probe connections on the meter to a different jack for measuring current ? Some DMMs require this.

 

[Daniel2244]

Does your DMM require you to move one of the probe connections on the meter to a different jack for measuring current ? Some DMMs require this.

I don't use probes I use wires connecting into the circuit

 

[gneill]

I don't use probes I use wires connecting into the circuit

Even so, you must make the connections to the correct inputs on the meters.

 

[Daniel2244]

Even so, you must make the connections to the correct inputs on the meters.

Yes, sorry, I misread. They is 3 sockets connection socket, V,mV,micro-V, mA,micro-A and the last socket for 10 amps