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Suppose you have a circuit Electric circuit

,you are told that the voltage at point 2 relative to point 1 is -90V, the voltage across R2 is -30V and V0 is 200V.

What is the meaning of voltage at 2 relative to 1? Does it mean the potential difference between 2 and 1 as in (voltage at point 2) - (voltage at point 1) or is it vice versa?

Does voltage at a point relative to another mean a voltage drop between those two? And what roles do the negative voltages at either points play?

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  • \$\begingroup\$ It's impossible to say, since that's not a self-consistent set of conditions. There's no way to make those voltages add up to zero around the loop. \$\endgroup\$
    – Dave Tweed
    Commented Sep 14, 2012 at 4:28
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    \$\begingroup\$ @pyler: "...Does it mean the potential difference between 2 and 1 as in (voltage at point 2) - (voltage at point 1)..." Yes, that is exactly what it means! \$\endgroup\$
    – Shamtam
    Commented Sep 14, 2012 at 4:35

2 Answers 2

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Voltage is always relative. When you just hold one probe of your DMM to the + of a battery, for instance it will display 0 V, like it's saying "this is nothing, what do I have to compare it with?".

Any circuit should have a ground (yours hasn't). Ground is your reference everything else is compared to. Usually that's the negative terminal of the power supply. It's only by calling ground 0 V that the supply's +12 V makes sense: it's 12 V higher than ground. If you measure 2.7 V in your circuit the "referenced to ground" is implied, unless indicated otherwise. That's the case when you say "relative to X", then you don't use ground as a reference, but the voltage at X.

For DC normal arithmetic applies: if A is +5 V relative to ground and B = +12 V relative to ground then A is -7 V relative to B.

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    \$\begingroup\$ One could define absolute voltage, based on the energy required to move a charged object an infinite distance (the total energy would be bounded as the integral of 1/r^2 approaches an asymptotic limit). On the other hand, even though a standing person's height (stature) could theoretically be determined by measuring the absolute elevation at the heat and feet, it's much simpler and more accurate to measure the distance between the head and feet. \$\endgroup\$
    – supercat
    Commented Sep 14, 2012 at 6:10
  • \$\begingroup\$ @supercat - Practical objection: all voltages on my PC would have the same absolute voltage, while the relative voltages may be different from zero. \$\endgroup\$
    – stevenvh
    Commented Sep 14, 2012 at 6:15
  • \$\begingroup\$ The absolute voltage would not be measurably different without some incredibly (impossibly?) accurate equipment, but that doesn't mean it would be the same. To use a gravitational analogy, voltage is equivalent to gravitational potential energy. One cannot measure such energy as what would be released by letting an object free-fall to the center of the planet (which would be infinite), but one could measure it by calculating the amount of energy required to escape the planet's gravitational pull. If an object has 10 joules more gravitational potential energy than another, then... \$\endgroup\$
    – supercat
    Commented Sep 14, 2012 at 6:35
  • \$\begingroup\$ ...the former object would take ten fewer joules to send an infinite distance from the gravitational field than would the latter. Note that if the furthest things in the universe are within a finite distance, the energy required to send something an infinite distance could be computed without regard to any reference point. \$\endgroup\$
    – supercat
    Commented Sep 14, 2012 at 6:36
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It's like this, When you measuring relative voltages like this, suppose for a instance you connect the reference point to ground. And also,there is a voltmeter which it's ground is connected to ground. Then you could measure the relative voltages on points.

I hope you got it.

  • when the current come into ground through a branch, it's negative.
  • When current goes out from ground to the branch , we say it's positive.
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