How Do You Calculate Voltage Across a Resistor Using V1 and V2?

[rajohns08]

If I know the voltages on both immediate sides of a resistor...lets say v1 is on voltage on left side of resistor, and v2 is voltage on the right side of the resistor, is the votage of the resistor = v2 - v1 if the positive side of resistor is to the right? Or can you not even calclulate the voltage of a resistor just based on the two voltages on either side of it.

 

[tiny-tim]

Welcome to PF!

Hi rajohns08! Welcome to PF!

If by voltage you mean the potential difference between one side of the resistor and a fixed point, then yes V_A - V_B = (V_A - V₀) - (V_B - V₀).

 

[gnurf]

The voltage on the right hand side relative to the left hand side of the resistor is (V2 - V1). The voltage on the left hand side relative to the right hand side is (V1 - V2).

The voltage is simply the potential difference between two points. Often one of these points will be chosen to be ground (at 0 volts), against which all other measurements are taken. But in general you can measure a voltage somewhere relative to any other point, including the two sides of a resistor.

EDIT: Sorry when I said right I meant left of course, and vice versa -- fixed now!

 

[Adjuster]

If I know the voltages on both immediate sides of a resistor...lets say v1 is on voltage on left side of resistor, and v2 is voltage on the right side of the resistor, is the votage of the resistor = v2 - v1 if the positive side of resistor is to the right? Or can you not even calclulate the voltage of a resistor just based on the two voltages on either side of it.

Could you please explain what you mean by the "left" and "right" sides of a resistor. It would be most helpful if you could post a diagram showing what you mean. Resistors are not always mounted horizontally, nor are they always drawn horizontally on diagrams, so it is better not to refer to "left" and "right" except in relation to a drawing.

However, it seems likely that "left" and "right" refer to the terminations of the resistor, and that the voltages are measured from each "side" to a common point such as ground. In this case it does follow from Kirchhoff's Voltage Law that the voltage across the resistor is given by subtracting the terminal voltages measured to a common point.

This method can be useful when one terminal of a measuring instrument cannot be connected to high potential, but any value found by subtracting two measurements is likely to be less accurate than a direct measurement - especially where the measurements have the same sign and similar values.
This may be complicated by voltage drops caused by the current drawn by measuring instruments. There may be further errors if the voltages are varying and the two measurements are not obtained at the same time.

The bottom line is that in theory knowing two of the voltages defines the third, but that some limitations apply to using this result to make practical measurements.

 

[zgozvrm]

Yes. You can measure each side of the resistor in reference to ground, then take the absolute value of the difference, or you can simply measure across the resistor. Either way, you get the voltage drop across that resistor.

Assuming the resistor is undamaged and still has the resistance value indicated by its markings, you can then determine the current through the resistor by dividing the voltage drop by the resistance.

 

[Proton Soup]

Could you please explain what you mean by the "left" and "right" sides of a resistor. It would be most helpful if you could post a diagram showing what you mean. Resistors are not always mounted horizontally, nor are they always drawn horizontally on diagrams, so it is better not to refer to "left" and "right" except in relation to a drawing.

However, it seems likely that "left" and "right" refer to the terminations of the resistor, and that the voltages are measured from each "side" to a common point such as ground. In this case it does follow from Kirchhoff's Voltage Law that the voltage across the resistor is given by subtracting the terminal voltages measured to a common point.

This method can be useful when one terminal of a measuring instrument cannot be connected to high potential, but any value found by subtracting two measurements is likely to be less accurate than a direct measurement - especially where the measurements have the same sign and similar values.
This may be complicated by voltage drops caused by the current drawn by measuring instruments. There may be further errors if the voltages are varying and the two measurements are not obtained at the same time.

The bottom line is that in theory knowing two of the voltages defines the third, but that some limitations apply to using this result to make practical measurements.

you can get some common-mode measurement errors, but unless the voltage drop across the resistor is very small compared to how far it is away from "ground", it shouldn't be noticeable.

for theoretical, textbook problems and computer simulation, there should be no problem that I'm aware of. in fact, i think pspice does all calculation using nodal analysis wrt whatever you selected as ground.

 

[Adjuster]

you can get some common-mode measurement errors, but unless the voltage drop across the resistor is very small compared to how far it is away from "ground", it shouldn't be noticeable.

for theoretical, textbook problems and computer simulation, there should be no problem that I'm aware of. in fact, i think pspice does all calculation using nodal analysis wrt whatever you selected as ground.

Yes, it's OK for simulation. It's also OK for most practical measurements most of the time, but there are possible problems which can fool a novice (I don't like the word "noob"!). I think it is better to be aware of these things from the start.

I did have in mind measuring the voltage across a high-side current monitor resistor, using a ground-referenced meter of modest accuracy. Not long ago someone at my work became confused doing just that. He thought that the negative terminal of the meter should always connect to ground, which was not true for the instrument he was using. The voltage across the monitor resistor was barely resolvable on the meter range necessary to measure the full supply voltage, so he got a hopelessly inaccurate answer.