# Voltage measurement using multimeter

Discussion in 'Homework Help' started by lippyf, Dec 9, 2015.

1. ### lippyf Thread Starter New Member

Sep 26, 2015
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0
A circuit with DC voltage of 5 V with an output resistance of 0.1 MΩ. What voltage do you measure when you
using a digital multimeter APPA62 having 10 MΩ input impedance? If the defect is small or large in relation to the error specified in the instrument's specification? See attached APPA62 electrical specifications.

When using APPA62 to measure the output resistance, I assume the ciruit will be like a DC volage U=5V connected with two parallel resisitor R1=0.1MΩ and R2=10MΩ. So I just have to calculate the voltage over R2 which is the value the multimeter will be showing?
Before I start to attempt to calculate, am I thinking this right?

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2. ### RBR1317 Active Member

Nov 13, 2010
232
48
Can you explain why you would need to 'calculate' the voltage across two parallel resistors?

3. ### crutschow Expert

Mar 14, 2008
13,033
3,241
The two resistors are in series, not parallel.
They form a voltage divider. For 0.1MΩ and 10MΩ the measurement error will be about 1% low.

4. ### lippyf Thread Starter New Member

Sep 26, 2015
15
0
Oh ok, so the multimeter will be showing:
5*((0.1*10^6*10*10^6)/(0.1*10^6+10*10^6))=495049.505≈4950
4950/495049.505=0.009999≈1%
is this right?

5. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,373
1,159
I don't think so. Your question was:

If a 5 volt source with the source resistance of 0.1 megaohm resistance has a a 10 megaohm resistor load, a 10 megaohm input resistance meter connected across the load will read _____ volts.

6. ### lippyf Thread Starter New Member

Sep 26, 2015
15
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Yes, you're right. How should I do then?

7. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,395
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When voltage source is modeled, it has two components, the source of voltage and a resistor, the resistor is in series with the source of voltage. Therefore when a voltmeter is introduced into the circuit, the resistor inside the voltmeter will be in series with the resistor inside the voltage source.

8. ### lippyf Thread Starter New Member

Sep 26, 2015
15
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Not sure if I understand it correct, but doesnt it form a circuit with a voltage source connected with two resisitor in series?

Feb 19, 2010
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10. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,373
1,159
You need to DRAW the circuit as you see it.

You have two questions to answer. What is the calculated voltage across the 10 megaohm resistor and what voltage will the meter read?

11. ### Bordodynov Active Member

May 20, 2015
643
188
Vmeas=5*(10*10^6/(0.1*10^6+10*10^6))=5*(10/(0.1+10))=5*(1/(0.01+1))=5/1.01=4.950495049505
and another error of +/-1.5% --->Vmeas will be in the range [4.876 ; 5.025].

Last edited: Dec 10, 2015
12. ### crutschow Expert

Mar 14, 2008
13,033
3,241
I thought the 10MegΩ resistance was the meter impedance, not a separate resistor.

13. ### MrAl Well-Known Member

Jun 17, 2014
2,439
492
Hello,

This is simply a voltage divider problem. The top resistor is 100k and the bottom resistor is 10Megs.
The 5v ideal part of the supply is connected to the 100k, and the voltmeter sees the voltage only across the 10Meg resistor.
An approximation is to say that the voltage will read 1 percent low because 100k is 1 percent of 10Megs, but there is a more exact calculation which you should do too which takes it down very slightly more than 1 percent.

14. ### MikeML AAC Fanatic!

Oct 2, 2009
5,450
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What does the "+x digit" mean in the accuracy specification (x is 2, 3 or 5 depending which spec you look at)?

15. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,373
1,159
The first part was measuring the power source.

The second part I assumed was measuring the voltage across a 10 megaohm load resistor, R2.

Anyways, the TS is to take into account the meter specifications. I'm waiting for him to return to answer Mike's question.

Last edited: Dec 10, 2015
16. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,373
1,159
Which question did you answer, the first one or the second? And you shouldn't be taking it out so far because the meter doesn't support that many places to the right of the decimal place.

Where did you get the error was +/- 1.5%?

17. ### Bordodynov Active Member

May 20, 2015
643
188
Yes, I have to clarify my statement. It should read:

Vmeas=5*(10*10^6/(0.1*10^6+10*10^6))=5*(10/(0.1+10))=5*(1/(0.01+1))=5/1.01=4.950495049505 (measure ideal voltmeter)
and another error of +/-0.5% + +/-delta --->Vmeas will be in the range [4.9257-delta ; 4.9752+delta].

If the display after the dividing point 3 digits, the delta = 2 * 0.001 = 0.002V -->
Vmeas will be in the range [4.923 ; 4.977].

18. ### lippyf Thread Starter New Member

Sep 26, 2015
15
0
4.950495049505-delta=4.923742574
how do the multimeter digit rounding? will it not round up to 4.924?

19. ### lippyf Thread Starter New Member

Sep 26, 2015
15
0
"If the defect is small or large in relation to the error specified in the instrument's specification?"
also can someone explain what do they mean with that?

20. ### Bordodynov Active Member

May 20, 2015
643
188
Lippyf. The It belongs to the interval error of the instrument 0.5% + 2digit. How many digits after the dividing point you see in the measurement?
delta=2digit !
Vmeas It belongs to the interval [Videal-0.5%-2digit; Videal+0.5%+2digit]