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Open circuit voltage is not what I expected.

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arjob

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I did this circuit in real and found that that the voltmeter is not showing want I expect.
I tough the voltmeter should be showing 5.1 volts.
Why is it showing 4.6 volts?
Note: I am using a digital multimeter. And I tried with a 3.3 v battery, then I am getting 2.8 v.
 

This is due to the input impedance of the multimeter. It should be around 1Mohm, then it will read

Vmeter= Vin*1Mohm/(1Mohm+R10), since R10=100 kohm, then

Vmeter≈0.9*Vin
 

Thank you so so much..
Should a more expensive multimeter solve the issue?
 

There are multimeter with higher input impedance, f.i. 10 Mohms. I don't know about the price.
 

You can use a large resistor Rs in series to the voltmeter.
When your voltmeter has a resistor of 1Mohm and it shows a voltage of 5V you have a current through the voltmeter of: Imeter =5V/1Mohm
The same current Imeter flows also through the serises resistor Rs. Then the overall voltage is Vsum = Vmeter + I *Rs
 

flanello said:
You can use a large resistor Rs in series to the voltmeter.
When your voltmeter has a resistor of 1Mohm and it shows a voltage of 5V you have a current through the voltmeter of: Imeter =5V/1Mohm
The same current Imeter flows also through the serises resistor Rs. Then the overall voltage is Vsum = Vmeter + I *Rs
Click to expand...

No, the current will be I=5V/(Rmeter+Rs).
If R10 is the series resistance of the generator (in you case 100 k), Rm is the resistance of the multimeter, Rk the extra serie resistor and Vm the reading of the multimeter, using simple divider equation:

Vi=Vm*[1+(R10+Rk)/Rm] if Rk>>Rm>>R10 then

Vi≈Vm*(1+Rk/Rm)

In order to have a reasonably accurate measurment you have to know quite good Rm
 

Tanks for the correction, albbg.
 

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