(Make Electronics SE Book) Help with Chapter 1:The Basics - 2 Questions(Pictures Included)

Thread Starter

Hi_Im_Andy

Joined May 31, 2017
3
Hi all! I'm Andy and holy heck I'm confused!

My first question is about the experiment in Figure 1-54.
Parts List:
9V Battery
1K Ohm Pot
470 Ohm Resistor
1 Basic Led

When I put my DMM on the the pot and resistor to read the resistance levels, I'm coming up with some confusing numbers. I'll include photo of where I am setting my DMM to make sure I'm using the proper setting.

Here are the numbers I'm getting:
(Reading pot, when pot is set to MAX resistance) = 4.76V shown on my DMM
(Reading pot, when pot is set to MIN resistance) = 25.9mV
(Reading resistor, when pot is set to MAX resistance) = 2.34V
(Reading resistor, when pot is set to MIN resistance) = 6.56V <---- **

I don't understand what is going on with the last line. Everything seems to make sense to me until that number goes up! Sorcery I tell you!!!

I was expecting no difference in resistance readings because I thought a resistor is just a static number that it brings current down. Can somebody please explain to me why the pot resistance number goes down, but the resistor resistance is going up when the pot is lowered?

My second question is about Fig 1-58.
It shows what I think is a combined resistance of 4.5V + 4.5V? The picture then shows 9V DC on the '+' and 0V DC on the '-'. Does this mean it's indicating we 'start' with 9V DC, but then because of the resistance it tries to travel through, it 'ends' with 0 DC?

I think what may be causing me all this confusion is I don't really know how resistors work yet, or maybe I'm just missing something else.

Thank you for any help or useful tutorial links you can provide!
 

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Robin Mitchell

Joined Oct 25, 2009
819
In answer to your first question, resistors do limit current but the voltage across a resistor will depend on other resistors in series with it. The voltage across your battery is around 9V and this will never change. If you have a bunch of resistors in series, the voltage across each resistor will be directly proportional to its own resistance. Assuming that your LED requires a forward voltage drop of 2V (and this will barely change), then the voltage "remaining" that must be divided up by your resistor and potentiometer is 7V. Therefore, as the resistance of the pot increases the voltage drop across the resistor decreases. As the resistance of the pot decreases the voltage drop across the resistor increases. The two voltages added up will always be 7V in this case.

Look up kirchoffs EMF law


All the best,
Robin
 

Robin Mitchell

Joined Oct 25, 2009
819
To answer your second question...

Resistance is measured in ohms, not volts.
Those resistors do not have a resistance of 4.5V, they have a resistance of 1000 ohms (or 1k)
 

Thread Starter

Hi_Im_Andy

Joined May 31, 2017
3
Hey Bertus and Robin.

Thank you for the replies :)

I honestly still don't fully understand yet, but I'm going to keep thinking about it until it hopefully clicks! Also, thank you for the link as well Robin. I'm not sure how to read that picture, but I'm guessing my book will eventually give insight into what the parts of that picture are.
 

pmichaud

Joined Jan 25, 2017
5
First, make sure you review Ohms law (V=IR) and know what voltage, current, and resistance are. The water analogy is good.

For the first question (6.56V across the resistor) you have to think of how a voltage divider works. As an example, to get your head around it, using ohms law, calculate the voltage across two resistors in series when connected to a 9V battery. The combined voltage drops are equal to the battery voltage. The voltage across a resistor is inversely proportional to the resistance. You saw this correlation during the first two voltage measurements. When the pot resistance was turned down, the voltage went up. So, if the voltage went up (across the pot) it would have to go down across the other components of the circuit. The total of the voltages will always be 9V. In Figure 1-58, the resistances are the same (1000 Ohms), so the voltage across them will be the same and they will sum to 9 (4.5+4.5=9) If one of the resistors was replaced with a 500 Ohm (500+1000 Ohms) the voltages would be 3V and 6V, respectively. (Again, 3+6=9) Don't forget to recalculate the current in the circuit with new resistance value (1500) before calculating the voltages across those new resistors using Ohms law. I believe that is clear. So, how did the voltage across that resistor change? Well, you changed the resistance in another part of the circuit which caused the current in the circuit to change. Ohms law states that these values are related: V=IR. The total voltage didn't change, it can't. But you changed the R, so the I had to change, which changed the ratio of voltage across the resistances. The ratio always totaling to the battery voltage. I think it should be perfectly clear now if you think it though.

In the Figure 1-58, those are *voltages* ACROSS the resistors as I described above (4.5V + 4.5V = 9V) (You said "combined resistance") The combined resistance is 1000+1000=2000 Ohms. Remember, voltage is the "pressure" or potential difference between two different points. If you measured the voltage from one side of the resistor to the other, you will get 4.5V. They both have the same current though them and the same resistance, so the voltage is the same. Ohms law states this to be true. If you measure voltage from the - to the +, it would be 9V potential difference between those two points. There is NEVER a potential difference at a single point. This is what is shown by the 0 DC. That - point is the reference point to which the 9V is measured. You can think of measuring voltage as "what is the potential difference in reference to the black DMM probe?" If you put the black probe on the bottom of a ramp and the red probe at the same spot, there is no height difference between them. (No potential for a marble to roll) But, if you put the red probe at the top of the ramp, then there is a height difference between them and thus potential for magic to happen.
 

Thread Starter

Hi_Im_Andy

Joined May 31, 2017
3
Wow! What a wonderfully detailed explanation. I was not expecting this all, especially so much later on. I really really appreciate you taking the time to explain this to me. I must say that I did finally get a decent grasp of the concept some time later finally but this is a great explanation to help me further understand the grey areas that I had. Seriously thank you again for this. Sometimes the little things just hold me back from progressing and it can be so frustrating lol. This is great and I hope it helps anyone else who comes along as well!
 
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