# What exactly Is voltage drop and how Is It detected?

Discussion in 'General Electronics Discussion' started by Reprovo, Jan 5, 2014.

1. ### Reprovo

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Jan 5, 2014
Hi.
I have trouble understanding what voltage drop Is.Is It a change In energies? If It Is,how Is It detected by a voltmeter If a voltmeter Is based on a solenoid / permanent magnet.How can It detect a drop In energy?
Excuse my poor syntax/simplistic Ideas/overuse of smilies

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3. ### Reprovo

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Jan 5, 2014
I've been there.It was a dark place.

4. ### duke37

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Jan 9, 2011
V = I * R
where V = voltage across the resistor, the voltage drop.
I = current
R = resistance.
The voltage can be measued with various instruments, modern meters will use very little power from the circuit and the affect on the circuit will be negligible.

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Dec 18, 2013
I tend to use the term potential difference and not voltage drop. For instance when a current passes through a resistor it will develop a potential difference across that resistor. This is what you call voltage drop. It is the difference in voltage between two points. Voltage is not energy, energy has the units of joules and electrical potential has the units volts or just V.
If we take the input as A and the output of the resistor as B then A-B = Work/Charge. So 1 Volt is 1 Joule per Coulomb. So if we have a potential difference of 1 volt between two point then 1 coulomb of charge will gain 1 joule of energy between these two points
Thanks

7. ### Reprovo

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Jan 5, 2014
That was Informative but you've confused me a bit at the end! Why will energy be gained.I thought that energy was lost as heat etc.

8. ### davennModerator

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Sep 5, 2009
agreed

Dave

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Dec 18, 2013
For anything to move it has to gain energy even electrons, what happens to this energy is dependant on what it passes through. To move an amont of charge in a circuit it has to get energy from say a battery. So a battery of 1V producing a current of 1A will have to gain 1 joule of energy. If this happens over a second then this is 1J per second which is 1W.

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Dec 18, 2013
Does it make it clearer if I said to move between point A and B 1C needs 1J of energy in an electric field of 1V. I assumed you would realise it was from start point flowing to end point.

11. ### Reprovo

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Jan 5, 2014
I understand the math but I don't understand what occurs when you place a voltmeter In parallel to a resistor.How does the voltmeter detect a drop In voltage ?

12. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
The voltmeter doesn't detect a drop in voltage, it simply detects a difference in voltage.

The thing you're measuring is either providing power to the circuit or it's dissipating power.

If it's doing the latter, then that difference in voltage is a voltage drop.

Essentially it is the difference in voltage caused by a current passing through a resistance.

13. ### Reprovo

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Jan 5, 2014
What I don't understand Is the path of the electrons when you connect the voltmeter.
I assume a very small current flows (due to the high resistance of the voltmeter) before the circuit resistance being tested and comes out the other side after the circuit resistance.This current causes a deflection induced by the moving solenoid.The way I'm understanding It In my head Is wrong obviously because I'm Ignoring the load somehow.

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Dec 18, 2013
Because you have put the voltmeter across the resistor the internal resistance of the voltmeter will effectively reduce the value of the resistor ever so slightly. And you will have two paths for current. One through the resistor and one through the meter. The voltage meter will have on it's data sheet the internal resistances of the meter so you can ensure you are using the correct voltmeter for your measurement task. .

Thanks

15. ### Reprovo

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Jan 5, 2014
Thanks for all the Information.
What I mean Is I envision current flowing through one terminal of the voltmeter and coming out the other.If I place the voltmeter In series wouldn't current still flow through the voltmeter and cause deflection ?

16. ### KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
Hi Reprovo and welcome to Electronics Point

As folks have already said, voltage "drop" is a difference in voltage between two points, due to current flow through a non-zero resistance, such as a resistor, an electronic component, or even a piece of wire.

This combination of current flow through a component and voltage difference across the component causes power to be lost (converted to heat, usually).

What's voltage? That's where it gets messy. I know what it is, but I can't describe or define it. Folks here have described it in terms of other quantities, but it's not easy to imagine what it really IS.

You know the hydraulic analogy - water flows in pipes; flow is current; pressure is voltage. When water flows through a narrow pipe (resistance), the water coming out of the pipe has less pressure than the water going in. Personally, I don't find the hydraulic analogy very useful, because voltage is measured BETWEEN two points, like distance. I don't think of it as pressure.

I've created an analogy called the DTS Model which is included in the thread I linked above. It MAY (or may not) be helpful to you.

Good luck!

17. ### BobK

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Jan 5, 2010
Maybe this will help. Think of holding a bowing ball above your head. It has a lot of potential energy. If you drop it, it will do a lot of damage (work). Now lower it to 1 inch off the floor. It now has much less potential to do work (damage) if dropped.

This is just what happens to electrons as they go through a resistor. They lose some of their ability to do work. This loss shows up as heat in the resistor. A voltage, in electronics, is a difference in the potential (ability to do work) between two locations in a circuit.

It is easy to measure potential in electronics, becuase where there is a difference in potential, current will flow from the higher potential to the lower potential. Your meter is measuring the difference in potential (voltage) by noting how much current flows when it its leads are connected to the two places in a circuit that have different potentials.

Bob

18. ### Reprovo

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Jan 5, 2014
Thanks for your time first of all.
I was thinking of something along those lines and it makes sense but then i dont see why current wouldnt flow through the voltmeter if it were in series.i know theres no potential difference but current flows through resistors in parallel.why not through the voltmeter.
If it does flow through why is the reading 0 or close to it in series.

Last edited: Jan 6, 2014
19. ### Reprovo

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Jan 5, 2014
Ok i understand it now.it just clicked!
Thanks for bearing with me.
It always helps to step away and do or think about something else even for a short time.

20. ### KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
Yes.

Say we have a 9V battery (let's assume it's a perfect voltage source) with a 100k (kilohm) resistor connected across it. The current that will flow through the resistor can be calculated using Ohm's Law: I = V / R where V is 9V and R is 100,000 ohms. 9 / 100,000 is 0.00009 amps, or 90 µA (microamps).

If you connect a voltmeter across the resistor, you now have a 100k resistor and the voltmeter, in parallel, connected across the battery. The voltage across each component is 9V. The current through the resistor doesn't change. Current will flow through the voltmeter. If it's a digital multimeter with a 10 megohm input resistance, this current will be 0.9 µA (by Ohm's Law). So the total current coming out of the battery will be 90.9 µA.

If the voltmeter is an analogue one (I'm mentioning this because you mentioned "deflection"), and has an input impedance of 20 kilohms per volt, and is set to the 10V range, it will have an input impedance of 200 kilohms. With 9V across it, 45 µA will flow through it, and the total current drawn from the battery will be 135 µA.

Now, if we break the circuit and connect a voltmeter in SERIES - in other words, battery positive, through resistor, through voltmeter, back to battery negative, the 9V from the battery is applied across the series combination of the resistor and the voltmeter, and will be divided according to the relative resistances of the resistor and the voltmeter.

If it's a digital multimeter, the total resistance (resistors in series add together) will be 10.1 megohms. Therefore the current will be 9 / 10,100,000 which is 0.89109 µA. This current flows though both resistances, and the voltage across the resistances can be calculated from V = I R. For the 100k resistor, V will be 0.089109 volts, and for the multimeter, V will be 8.9109 volts. So the mutimeter will read 8.9109 volts.

You can say that there is 9V coming from the battery, and a voltage drop of 0.089109 volts across the resistor, so the remaining voltage, 8.9109 volts, appears across the multimeter.

If it's the analogue voltmeter, the total resistance will be 300 kilohms. The current will be 30 µA and the voltages across the resistor and the voltmeter will be 3V and 6V respectively. These voltages are in proportion to their resistances. So there will be a voltage drop of 3V across the resistor, and the analogue voltmeter will read 6V.

Last edited: Jan 7, 2014