Hi guys, I'm just secondary school. I've attached a circuit for us to discuss. Do you guys know what is the voltage value for A?
How do you guys calculate?

 

ignore the 1 meg resistor, there is no current thru it ( untill you try to measure the voltage ) then it is a basic voltage divider, you do know the equation for it?

 

ignore the 1 meg resistor, there is no current thru it ( untill you try to measure the voltage ) then it is a basic voltage divider, you do know the equation for it?

Like what you say, there is no current through it until I try to measure the voltage using a multimeter. I've measured on point A and it is around 1.31V. And the voltage for the 40k resistor is 1.44V. Does that means the 1M resistor is 1.44V-1.31V = 0.13V?

 

In fact, I've investigate further by changing the value of the 40k resistor and the values are shown in the table i attached.
Seems like the lower the resistance, the difference of the voltage of the resistor and point A becomes lower and lower.
Do you guys know why?

 

Like what you say, there is no current through it until I try to measure the voltage using a multimeter. I've measured on point A and it is around 1.31V. And the voltage for the 40k resistor is 1.44V. Does that means the 1M resistor is 1.44V-1.31V = 0.13V?

you are trying to calculate the voltage drop across the 1meg resistor, and this is incorrect. there is NO voltage drop across the 1 meg resistor.
ok, you have a voltage divider with the 30k and the 40k resistor. this divides the 2.5V. 2.5V * 3/4 = 1.875V. Between GND and the center dot, there is 1.875V. The 1 meg resistor is not a part of this voltage divider, it is merely connected to the center of the voltage divider and floating freely in air. it is not part of the circuit. This means that the voltage on either side of the 1 meg resistor is the same, so your answer, the voltage at point A, is 1.875V. Now, when you connect your meter between GND and point A, the 1meg resistor becomes part of the circuit, via your meter, and becomes a resistance in parallel with the 40K resistor, which changes the effective resistance, which changes the ratio of the voltage divider, which throws off your measurements.

 

Ω

Like what you say, there is no current through it until I try to measure the voltage using a multimeter. I've measured on point A and it is around 1.31V. And the voltage for the 40k resistor is 1.44V. Does that means the 1M resistor is 1.44V-1.31V = 0.13V?

Your multimeter also behaves like a resistor, so that some current flows through it when it makes a measurement. This will case a very slight drop in the voltage across the 40k resistor when the meter is connected there, and a much bigger drop in the voltage at the end of the 1MΩ resistor when the resistor is connected at that point.

The 1MΩ resistor does drop about 0.13V (or a little bit more, can you see why?), but only when the meter is connected to it. In fact, the data you have collected would allow you to estimate the meter input resistance. Perhaps when you have mastered the basic potential divider, you might try to do that.

 

you are trying to calculate the voltage drop across the 1meg resistor, and this is incorrect. there is NO voltage drop across the 1 meg resistor.

You are incorrect.

Our original poster is using a meter that has an internal impedance of approximately 10.077 MegOhms.
As they started out saying, 1.44v at the junction of the 30k and 40k resistors, 1.31v from the 1 Meg to ground, leaving 0.13v across the 1 Meg resistor - assuming that the 1 Meg resistor is actually 1 Meg.

That's 0.13uA or 130pA current flow through 1 Meg, and also across that 1.31v to ground - through the meter.
R=E/I, so 1.31v / 130pA = 10,076,923 - or 10.77 MegOhms; plenty close enough.

 

you are trying to calculate the voltage drop across the 1meg resistor, and this is incorrect. there is NO voltage drop across the 1 meg resistor.
ok, you have a voltage divider with the 30k and the 40k resistor. this divides the 2.5V. 2.5V * 3/4 = 1.875V. Between GND and the center dot, there is 1.875V. The 1 meg resistor is not a part of this voltage divider, it is merely connected to the center of the voltage divider and floating freely in air. it is not part of the circuit. This means that the voltage on either side of the 1 meg resistor is the same, so your answer, the voltage at point A, is 1.875V. Now, when you connect your meter between GND and point A, the 1meg resistor becomes part of the circuit, via your meter, and becomes a resistance in parallel with the 40K resistor, which changes the effective resistance, which changes the ratio of the voltage divider, which throws off your measurements.

NO! the ratio is given by the potential divider equation, Vout = Vin * R1/(R1+R2), in this case 2.5V*40kΩ/(30kΩ+40kΩ) = 1.143V. This is close enough to the OPs result for the discrepancy to be explained by reasonable errors, your figure is not.

The lower voltage measured at the end of the 1MΩ resistor does result mainly from voltage dropped in the 1MΩ. When the meter is connected at this point, it effectively forms a secondary potential divider. There is a reduction in the voltage across the 40kΩ, but this is almost negligible. From the OP's data it is clear that his DMM has a very typical input resistance.

Normally I would not give an explicit answer, but what you have said is not right, and it is important not to give beginners innaccurate information.

 

What he wants is the voltage at point A, which if the meter is not part of the equation, would be the same as the divided voltage. It looks like he's trying to use some measured voltage drop across the 1meg resistor to arrive at the voltage between point A and ground. I was trying to point out that, with the meter not in the circuit, there is no voltage drop across the 1meg resistor. I don't think that the purpose of the exercise is to calculate the internal resistance of the meter.

 

NO! the ratio is given by the potential divider equation, Vout = Vin * R1/(R1+R2), in this case 2.5V*40kΩ/(30kΩ+40kΩ) = 1.143V. This is close enough to the OPs result for the discrepancy to be explained by reasonable errors, your figure is not.

yes, you are right. sorry if I caused any confusion. brain fart.

 

What he wants is the voltage at point A, which if the meter is not part of the equation, would be the same as the divided voltage. It looks like he's trying to use some measured voltage drop across the 1meg resistor to arrive at the voltage between point A and ground. I was trying to point out that, with the meter not in the circuit, there is no voltage drop across the 1meg resistor. I don't think that the purpose of the exercise is to calculate the internal resistance of the meter.

Since this circuit will have been defined by the OPs teacher or in a text he was referred to, it seems very likely that the 1MΩ is included to illustrate the effect of meter loading. In general, it is useful for technicians and engineers to have a clear picture of how meter readings are affected by source resistance.

Had I been dealing with an older student I might have referred to to Thevenin's theorem, and the fact that the voltage across the 40kΩ behaves as a ≈17kΩ resistance source, but at the end of the 1MΩ the effective resistance is 1017kΩ.
Thus connecting a roughly 10MΩ meter to the 40kΩ depresses the voltage at that point by only about 0.17%, but connecting the meter to the 1MΩ drops the voltage at that point by over 9%.
Not only is this a useful point in its own right, it can be calculated as a secondary stage of potential division.

 

Since this circuit will have been defined by the OPs teacher or in a text he was referred to, it seems very likely that the 1MΩ is included to illustrate the effect of meter loading. In general, it is useful for technicians and engineers to have a clear picture of how meter readings are affected by source resistance.

Had I been dealing with an older student I might have referred to to Thevenin's theorem, and the fact that the voltage across the 40kΩ behaves as a ≈17kΩ resistance source, but at the end of the 1MΩ the effective resistance is 1017kΩ.
Thus connecting a roughly 10MΩ meter to the 40kΩ depresses the voltage at that point by only about 0.17%, but connecting the meter to the 1MΩ drops the voltage at that point by over 9%.
Not only is this a useful point in its own right, it can be calculated as a secondary stage of potential division.

That's what I was getting at, albeit less eloquently here:

Now, when you connect your meter between GND and point A, the 1meg resistor becomes part of the circuit, via your meter, and becomes a resistance in parallel with the 40K resistor, which changes the effective resistance, which changes the ratio of the voltage divider, which throws off your measurements.

But I don't think anybody made it that far into my post, after seeing the blaring mistake in the first half.

EDIT: I should restate that anyways: The error comes not so much from the 1meg resistor + 10meg meter being in parallel with the original 40K resistor, but more from the 1 meg resistor forming it's own new voltage divider with the 10meg meter.

and yeah, you guys are right, it was probably to illustrate the effect of meter loading.

 

It was quite a bit more dicey in "the good old days"; using a Simpson 260 meter - you had to remember that the meter had an impedance of 10k Ohms per volt, which had an impact on even moderate impedance measurements.