Proteus voltage across resistor, simulation

When connected a voltmeter across a resistor in parallel in proetus, it gave no change in voltage from a 12V,
But when i made it series with the voltmeter it gave change in voltage, why this happens ?

and what is load resistance in the case of voltmeter in proteus ?

Right-click the voltmeter and select edit properties. In there is 'Load Resistance' - mine was set to 100M. Play with that setting. I get the correct reading with it as part of the potential divider formed.

what should be Load Resistance in the case of a 3k resitance in 12V ?

The 100 Mohm is fine , the higher the better but 100M is already high, the typical value for a real volt meter is about 10M.

I'm not sure what you are trying to do, you want to measure the voltage across the resistor, where is this resistor connected, is there another resistor in the circuit, maybe you can post a schematic.

Alex

this is the simulation in proteus




we can see no voltage change in parallel mode and in series when using 100M
but in 1k load resistance the voltage changes

You must read the basics about Ohm's law, do a search in google and read a couple of tutorials.

When you place the meter across the resistor then it is in parallel and has the same voltage as the resistor.

When you place the meter in series then it is like having two resistors (the second one in the meter resistance) and you create a voltage divider.

In a voltage divider the voltage across a resistor is Vrm= Vbat * Rm/(R+Rm) where Vrm is the voltage across the meter, Vbat the 12v of the battery, Rm is the meter resistance and R the the 1k resistor on your circuit.

When you have the meter set to 100M then you get 12v * 100000000/(100000000+1000) = 11.999v
When you have the meter set to 1K then you get 12v * 1000/(1000+1000) = 6v

That is correctly simulated operation, with the voltmeter resistance acting as one half of a potential divider. One normally uses a voltmeter to measure the potential difference across (in parallel with) something, not in series. Ammeters go in series.

As Alex said, what exactly are you trying to do?

yes,
i want to check the potential across the resistor,
but in parallel , it does not give a voltage drop

sorry, i didnt noticed alex's post sorry

That is correct - it should not show a voltage drop because you are connecting it effectively straight across the supply; all three are in parallel: supply, resistor, voltmeter. In parallel, all items see the same voltage, but the current through them will vary according to their properties.

Do some google research on series and parallel circuits.

Oh, and ideally, a voltmeter would have infinitely high resistance so that it does not take any current away from the circuit. 100MΩ is good. With a low resistance to the voltmeter, as you have seen, the voltmeter itself becomes part of the circuit and causes changes. That is not what you want.

When you connect 1,2,10,100 or any number of resistors in parallel they all get the same voltage , it is the current of each one that depends on the resistance value.

On the contrary when you have the resistors in series then the same current flows through all of them but each one has a different voltage across drop it, if you add all the voltage across the resistors it equals the source voltage.

Alex

Here's a quick illustration of what Alex and I are talking about, for the voltage at least.

There are three sub-circuits in the diagram - each is powered from the 12V source. Note first that each of the three sub-circuits is in parallel with the others, so each sub-circuit sees the full 12V supply across its extents.

Sub-circuit A has two resistors in parallel. So, each resistor sees the full 12V, as shown.

Sub-circuit B has two resistors in series. Now, the potential difference is shared across each resistor, in the ratio of their resistances. Since they are equal (10KΩ each), each gets an equal share. Thus, 6V across each series resistor. Note that the sum of the voltages across each resistor is 6V+6V=12V. The voltage must always add up to the supply - one cannot lose or gain voltage overall (not with passive components anyway).

Sub-circuit C also has two resistors in series. This time though, the resistors are not equal and you will see that the larger resistor gets the larger voltage share, according to their ratio. Again, note that the sum of the voltages is still 12V.

I hope that helps.