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?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?
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.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.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 incorrect.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.
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.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.
yes, you are right. sorry if I caused any confusion. brain fart.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.
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.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.
That's what I was getting at, albeit less eloquently here: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.
But I don't think anybody made it that far into my post, after seeing the blaring mistake in the first half.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.
Thread starter | Similar threads | Forum | Replies | Date |
---|---|---|---|---|
S | Voltage divider | Homework Help | 13 | |
Z | Having trouble connecting my voltage divider circuit to my Pi | General Electronics Chat | 6 | |
Help with voltage divider | General Electronics Chat | 18 | ||
F | Voltage Divider CE amplifier | Homework Help | 6 | |
K | Voltage divider design questions | General Electronics Chat | 11 |
by Luke James