# Confused by electricity!!

Discussion in 'General Electronics Discussion' started by flypadre, Jan 30, 2013.

14
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Jan 30, 2013
I made two really simple circuits because I wanted to understand how my multimeter worked. The first was just a 6 volt source and a 100k resistor. I was surprised when I touched the probes to both sides of the resistor. I was expecting to see a reduced voltage, but I saw 6 volts which confused me.

2nd circuit was the same as the first except I put another 100k resistor in series with the first. Now when I placed my probes on each side of the first resistor I got my 3 volts that I was looking for.

What am I not getting??

Last edited: Jan 30, 2013
2. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,497
2,839
Jan 21, 2010
6 volts across a 100k resistor will result in 6V across the resistor. What is the problem with this?

When you have two 100k resistors in series, with 6V across them both. The voltage drop across each of them will be proportional to the fraction of each resistor's value to the total resistance.

In this case, both are the same value, so the fraction of one to the sum is 100/(100+100) = 1/2.

So you expect that the voltage you see across either is 6 * 1/2 = 3V and this is what you see.

In the first case, we can do the same thing. Since there is one resistor, the individual resistor and the sum or resistances is the same. So the voltage you see is 6 * (100/00) = 6 volts.

14
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Jan 30, 2013
Well I was expecting to see some type of drop in the voltage because of an increase in resistance.

I don't think I am understanding something properly. Here is where my thinking/understanding is coming from:
I have set up simple circuits lighting an led. To keep the led from burning out I used a resistor and used ohms law to determine what size. If the led was rated for 2.3 volts forward at 20mA and I was using a 6v power source I would subtract 2.3 from 6 = 3.7 and multiply that by .02A and get 185 ohms. I just assumed that using the resistor resulted in both a drop in the amperage and a drop in the voltage.​

So, when I put that resistor into my simple 6 volt circuit I thought that I would see this drop.

Would you happen to have any links to something that explains how/why those calculations you mentioned in your post work?

BTW All help appreciated!
Thanks

4. ### poor mystic

1,072
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Apr 8, 2011

The resistor will only allow a certain amount of current to pass for any given voltage. In this case, for every volt, the resistor allows 1/100000 Amps to flow. When 2 resistors are put in series, the resistance doubles, so only half the 1/100000 Amps can flow, if 1V potential is supplied.
For 2V, double the current figure; for 3V you can expect triple the current, and so forth.
To see what voltage can be measured across a resistor, multiply the resistance by the current. In a simple case current might be 10 microAmps. 1000000 x 10uA = 1V.
This is called "Ohms Law" and it is the most used calculation in electronics.
http://www.wisc-online.com/Objects/ViewObject.aspx?ID=dce11904

MarkGood luck - call back again!

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

25,497
2,839
Jan 21, 2010
Anything placed across a perfect voltage source will result in the voltage from that voltage source appearing across it.

So any resistance from 1 ohm to a million, million ohms will act the same.

In this case the resistor drops *all* the voltage.

You will only see a resistor with less voltage across it if there is something else in series with it, so the resistor is not directly across the voltage source.

Note that perfect voltage sources do not exist. So a 1 ohm resistor may behave totally differently to a 100,000 ohm resistor if connected across a battery.

In practice, a battery (or indeed any voltage source) behaves very much like a perfect voltage source in series with a low value resistor. We generally aim to keep the resistance of the circuit high in comparison to this series resistance (ESR) so that we can ignore it.

To understand this, you need a partial mastery of the black art of "Ohm's Law". (It's not that hard really)

6. ### davennModerator

13,865
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Sep 5, 2009
you ARE seeing a drop in voltage across the single resistor.
you have 6V on one side and 0V on the other side ... you have a 6V drop!!

yes you will get a voltage drop across the resistor in series with the LED
BUT the primary purpose of the resistor is to limit the current through the LED
Without the resistor, the low forward biased resistance through the LED will mean that it will draw as much current as the powersupply ( battery etc) can supply. This will result in a very brief, bright and illustreous life of the LED.

Dave

7. ### davennModerator

13,865
1,956
Sep 5, 2009
OK some diagrams relating to Voltage, Voltage drop, current and resistance

Ohms Law Triangle for simple DC circuits

The first one ... a simple battery and resistor circuit, 100V battery and 100 Ohm resistor
Current flowing = 1 Amp

I = V / R = 100 / 100 = 1 Amp

Next a 2 resistor circuit (resistors are the same value)

what Voltage do you think you will measure across points A and B or points B and C ?
NOTE ... that we have doubled the resistance and that has halved the current flowing in the circuit.

Example 3 ... a 3 resistor circuit

here we know the supply voltage and the voltage drop across each resistor and the current flowing in the circuit.

NOTE ---- the current flowing in a simple DC circuit like this will be the same regardless of where in the circuit its measured!

We can work out the unknown values of the resistors using Ohms Law

R = Vx / I ( where x = the voltage drop across the resistor you are working with)
lets label the V's from top to bottom V1, V2, V3 and resistors from top R1, R2, R3
Note that the voltage drop, in this case, is the same across each resistor. 4V
This tells you that the values of the resistors must be all the same ... so....

R1 = V1 / I = 4V / 1A = 4 Ohms

Now to confirm all this .... add all 3 resistor values 4 + 4 + 4 = 12 Ohms total

confirm the current flowing I = V / R

12V supply across 12 Ohms = 12 / 12 = 1 Amp

how's that grab ya ?

Dave

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8. ### davennModerator

13,865
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Sep 5, 2009
Ohhh ... NOTE in reference to the last diagram ( 3 resistors)

voltage measuring .....

If you put the negative of your meter on point D and positive on point A
you measure 12V the Voltage from the battery and the total voltage drop across the 3 resistors
Now ... keeping the negative lead on D and moving the positive down to point B
you only measure 8V ... ie ... there is only 8V at that point cuz you have already dropped 4V across the top resistor 12 - 4 = 8

Now moving the positive lead down to point C and you only measure 4V ... cuz you have dropped 8V across the first 2 resistors ... 12 - 8 = 4

Measuring across any one of the 3 resistors ...
between points A and B, or B and C or C and D, will show 4V ( the Voltage drop across each individual resistor)

Dave

14
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Jan 30, 2013
You guys are the greatest.
All of the responses helped to clear up some of my confusion. I just have one more question

In measuring voltage
It seems like the meter should be reading zero because voltage is passing through all those points A,B,C,D.

It seems backwards to me from what is happening. If like your 3rd diagram shows 12 v dropping to zero it seems like the measurement from A to B should be 8v, A to C 4v and A to D 0v to get to that 0v you have in the diagram, but just the opposite is happening.

Thanks again everyone

10. ### BobK

7,682
1,688
Jan 5, 2010
It is you who is looking at it backwards.

The voltage from D to A has to be 12V because the battery is connected to these two points, and the battery is a voltage source, i.e. the two terminals always have 12V between them.

The current (not voltage, voltage does not flow) flowing through the top resistor causes a voltage drop from A to B of 4V. This comes from Ohms law.

So the voltage from D to B has be 4 less than the voltage from D to A or 8V.

And so on...

Bob

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Jan 30, 2013
Ok so I think that I got this now. The reading on the meter taken at A and D that I see is showing me the voltage between the positive and negative terminals. Basically what I am doing then is touching the two battery terminals.

The meter is showing me everything in the circuit counter-clockwise according to the diagram. If I connected the red probe to B and D the meter is going to show me the voltage affected by the resistor between A and B not the voltage affected by the resistors between B and D.

Do I have this right now?

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

25,497
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Jan 21, 2010
Maybe...

Your term "voltage affected" isn't one that indicates you know what's going on. Resistors don't really "affect" voltages. There is a relationship between resistance, voltage, and current (with any two defining the other).

Think of voltage like pressure. The voltage of a battery is like the pressure of a pump.

Resistors are like narrow pipes that resist (i.e. impede) the flow of water. The flow of water is a current -- just like in a river.

(As an aside, if you imagine a pipe that impedes the flow of water, the rate it flows is determined by both the pressure, and how much it impedes it then this is like ohms law).

The voltmeter is like a pressure measuring device, it measures the differential of pressure (voltage) between 2 points.

Each of the resistors (restricted pipes) will have a pressure difference across them when water is flowing. Intuitively, you can tell that the sum of those pressure differences will be the pressure difference across the pump.

If we're measuring voltage (pressure) with respect to point D, then Point A represents the battery voltage (pump pressure) and also the total voltage (pressure) dropped across all of the resistors (restricted pipes).

Point B represents the battery voltage (pump pressure) less that voltage (pressure) dropped across the top resistor (restricted pipe). It also represents the voltage (pressure) dropped across the bottom 2 resistors (restricted pipes).

So connecting the meter between B and D shows you the voltage dropped across the bottom 2 resistors, which is exactly the same as the battery voltage less the voltage dropped across the top resistor.

13. ### CDRIVEHauling 10' pipe on a Trek Shift3

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May 8, 2012

Chris

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Jan 30, 2013
Your drawing totally helped me, along with the post before yours by Steve.
Thanks everyone for sharing your knowledge with a noob like me.

Hopefully at some point I will be able to pay it forward to someone else!

Sean

15. ### poor mystic

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Apr 8, 2011

There is a useful circuit simulator at http://www.falstad.com/circuit/ which illustrates the movement of electric current very nicely.
Once the simulator applet is installed, you can set up a circuit with a voltage source, resistors and conductors. The applet will show you current moving around the circuit and you'll be able to measure voltages.
For simple circuits the applet is pretty good really, though it has its limitations where more complex circuits are involved.
Mark