Confused by electricity!!

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.

 

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)

 

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

 

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

 

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

 

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.