Tuesday, June 25, 2013

Parallel resistors, Part III, final post. I hope.


So, my little brother (if you recall, of the Bruce Lee level electrical/computer kung fu) indicated in his comment on a previous post that:

"Remember, you always have to put your multimeter in series with the CURRENT you want to measure and in parallel with the VOLTAGE you want to measure. If you try to measure the current by putting your meter in parallel with it, then you have just given current another path to flow and diverted it away from your main circuit (meaning you effectively have no idea how much current is flowing through the circuit in question)."

Of course, his use of the word "remember" implies that at some point in the past I learned the fact in question. 

The astute reader may also remember this quote of mine from the first post:

"This does allow for measurement of current but bypasses the switch and closes the circuit.  When I did so, I had a current of about 17-18 microAmps. When I pressed the switch, the current dropped to 0.1 microAmps. "

This made no sense to me at the time.  Things that don't make sense to me annoy me.  Make me stay up at night.  But remember, dear reader, the wisdom of Shakira:  "hips (and multimeters) don't lie".

When you measure current in parallel, you set up a situation as seen below:



In which the meter becomes a resistor in parallel with whatever is in between the two probes.   So it makes intuitive sense to me that the current of electricity--like the current of a river--follows the path of least resistance (but see link). So the ohmic (is that a word?) values of X and the meter dictate what the meter shows.  In the situation from the quote above, the measured current of "17-18 microAmps" was with the switch open, and all the current flowing through the meter (as is appropriate).  If plugged into the diagram above, x = infinity. Once the switch closed, however, then x = 0.4 ohms (measured), and then it all depends on the resistance of the meter as to where the current flows.

If the current drops by 2 orders of magnitude, then the resistance of the meter must be two orders of magnitude higher, or somewhere around 40 ohms.  We will see in a bit that we have some data that indicates it's lower than that, but that still means 90% of the current flows through the switch, not the meter.

So to demonstrate that,  I set up the following experiment:



This runs an LED through a protective resistor and then a potentiometer in series.  The resistor is designed to protect the LED even if the potentiometer is set at zero.  I then measure across the potentiometer and vary the resistance.  The above photo has the resistance of the potentiometer set to almost zero (it measures around 1.4 ohms).  The fact that we have very little current running through the meter indicates that the resistance of the meter is likely higher than that.



Now we have the same setup, but with the resistance of the potentiometer set as high as it will go, something like 45 kohms.  The current running through the multimeter has increased substantially.

Of note the LED did not change intensity at all.  Which indicates that this "sharing" back and forth is a zero sum game, as discussed previously.

Here, then, we have an inverse relationship.  Flow ~ 1/resistance.  What is the total flow through the circuit, then?



To answer that question I had to rewire things. This now bypasses the potentiometer entirely and goes through the switch, where I can put my multimeter in series like I should.  So the total flow is 10.88 milliAmps.

Which means that we now have a means of measuring the resistance of my multimeter on this setting.  If two resistors are equal, then flow through them should be equal. So if I can adjust the potentiometer to give me 5.44 milliAmps precisely, then it should be exactly the same resistance as the meter, and then we can measure the multimeter directly.

Turns out that's difficult to do.

video


The potentiometer is quite nonlinear in the range I need.  Much fiddling was done, over the course of two days, to try and get close to 5.44.  Best I got:



This was the second day, where the measured current across the switch was 10.72 milliAmps.  The measured resistance across the potentiometer at that point was:



Which although a little hard to read, is 15.1 ohms.  Dividing out the percent of current going through each resistor gives 58.39% of the current through the meter, and 41.6% through the potentiometer.  What resistance on the potentiometer will give us exactly 50% flow?  That should equal the resistance of the multimeter.

Can't use a straight ratio because this is an inverse relationship.  It's (1/15.1) x (1/41.6) = (1/50) x (1/x).  Solving for x gives 12.5 ohms.

Not sure if the math is valid but it gave me an answer that makes sense.  And much fun was had!

,m



Resistors in parallel, part 2

Again, I refer the reader to the HO:ME blog, which really should be read in parallel (pardon the pun) with this one.  As we learned last week, two blogs in parallel should reduce your resistance to learning...

Today a variation on the theme from last week, in that we will wire two resistors in parallel, that have different resistances.  We spent some time last week looking at a accidental case of this, in which we wired our "triangle circuit" 



And saw that the flow through the wire should be 10^5 higher through the wire than the resistor.  My little brother saw this and informed me that in a circuits class, this would be viewed as "shorting out" the resistor.  The wire would be assumed to have a resistance of zero and the triangle above would simply be erased.  I actually measured the resistance of the wire (0.4 ohms) and although I am not perfectly certain my multimeter is accurate down in those ranges, it's the data I got, so it's the data I'll use. 

So the setup of today involves a 1K and a 2.2 K resistor, like so:





I have drawn these out using Modified Deshazer Notation, which probably does not coincide with standard notation. 

So let's see what these look like.  Let's first demonstrate the baseline current for each resistor:



So if you can't read that, it's 0.74 mAmps through a 2.2 Kohm resistor on the left, and 1.61 mAmps through a 1Kohm resistor on the left. Baseline voltage is about 1.6. 






And then each one in parallel, just measuring off one of situation A) or B) above.


The pic above is measuring across the 2.2 Kohm resistor arm of the parallel resistors.  Note that the measured amps are identical to the initial, single-resistor-only pic above.

Here is the measurement across the 1 Kohm resistor:


Interesting, no?  It appears that the current across either arm of this setup is identical to the measured amps prior to hooking up the parallel resistor.   I confess I had expected to see a change in the current across each resistor as I added a second one in parallel.  I guess this makes sense, as the voltage drop across each of these resistors is identical.  However, the resistance has changed, so let's measure the total current in the circuit:



The components of the current seen on each arm add to (more or less) the current seen across the entire circuit.  Thus, the more (parallel) resistors you add, the more current you get!

In fact, for certain circuits, this may be an easier way of finding the resistance.  As the current will be the same across each resistor, regardless of how many others will be placed in parallel, it may make sense to calculate the sum of those currents, then use Ohm's law to determine the total resistance of the circuit.  Seems like it could be easier than 1/R = 1/R1 + 1/R2, etc.

So, let's check.  The current is inversely related to the resistance, therefore 1/R = (1/1000) + (1/2200), or 687.5 ohms.  Total current we measured at 2.33 mAmps.  V=IR gives us a calculated voltage of 1.6 volts, which is an accurate reflection of our measured volts.

Good learning!




Friday, June 21, 2013

Hurtin'

Moonlighting, dear readers, is a pleasant diversion from my real job. 

I like the place I work, and of course they pay me like a "real doctor" and not a trainee.  

Occasionally, however, the trainee aspect of my life steps to the front.  With typical disastrous results on my rock-n'-roll lifestyle.  

No progress in the electronics world this week.  

Monday, June 17, 2013

Resistors in parallel, part I

So the next experiment is to place two resistors in parallel.  We start with two resistors of the same strength.

Again, I refer the reader to the HO:ME blog for further discussion of same.  They also refer to some "math" that we will get to in a bit, although probably not at this blog post.

The setup:


This first is just to document the voltage from the single AA battery. 

I then wired the circuit below, and found that as I was wiring the circuit I made a mistake in the way I wired it up.  I had everything connected, below, except for the two dotted lines.  I then had two choices, to connect via path 1 below, or path 2.  I honestly wasn't sure which, so I just did both and measured the resistance:




This is path 1, current of 162 milliAmps:


This is path 2, current of 320 milliAmps:

So right at twice the flow.  Which I thought was wrong.  What happened to the other resistor?  I had two resistors in the circuit, but it seemed that when I ran the path 1 circuit, one of the resistors was effectively cut out.

For comparison, the current just running through a single resistor is 162, same as the path 1 wiring.


So I drew this out with the path 1 section of the circuit tipped over a bit, just to help me visualize, and came up with this funny-looking triangle circuit.   It occurred to me that this triangle loop was a bit deceptive, and what I really have is two resistors in parallel.  One is the actual resistor on the up-and-back part of the triangle, and the other is just a wire.

The resistor itself is 10K ohms.  The wire, 0.4 ohms.  This, in my mind, indicated that there probably is current flowing through the resistor, but it's, oh, 5 or so orders of magnitude lower than that flowing through the wire.  No wonder it looked like I had simply cut that resistor out of the circuit.




This, then, is the diagram for path 2, which is the proper way to hook up the resistors in parallel.


A very simple circuit, which still provided ample opportunity for learning (that is, confusion that required resolution via reading and further experimentation).


Friday, June 14, 2013

Gear up!

Tired of fussing with all the little wires, LEDs, etc.  Saw that the brother had organized his electrical hobby mess into a tackle box.  Found this instead:





That's how you know you are serious.  Or OCD.  Or just found something on sale, that looked like it would work.  Or didn't have any other blog posts ready and realized this post would be the fastest possible thing to do. 

,m

Thursday, June 13, 2013

Varying voltage and zero sum games

Dear reader, off to another experiment.

First, I suppose I should mention my kid sister, who has a Master's degree in Materials Science from some Hole In The Wall University, and after my conversation with her last night on solar cells, suspect she may have even greater electrical Kung Fu than my little brother.  Not sure who will represent her Kung Fu, ya'll can watch and decide.

I confess to minor deviations from the text of the book.  However, isn't that the point?

We are on experiment 3, which you should take the opportunity to review at the HO:ME blog.  I agree wholeheartedly with the author of that blog that it's difficult to read the bands on the resistors.  And whoever thought that particular color counted as purple?  Crazy.  It was bad enough I purchased a magnifying card to help.

No jokes about old eyes, please.

The experiment has you set up a circuit with a set resistor and an LED in series, initially, then with a potentiometer to vary the resistance.

At one point during the experiment the text asks to check the voltage across both the potentiometer and the LED, and look for variation.

Indeed, it does vary.

Here, at the potentiometer:



Note the intensity of the LED increases as the measured voltage across the potentiometer decreases.

Now the voltage drop across the LED itself:



Harder to see here, but the intensity of the LED increases as the measured voltage decreases.

I admit that at first the changes above confused me, especially when I had only measured the voltage at the potentiometer, and not yet at the LED.  I was caught in a mental trap that equated voltage to "available power".  It was not clear to me why I would have increased intensity at the LED--and therefore an obvious increase in "available power"--when I measured a decrease in voltage at the potentiometer.

It was not until I read in the text that the LED and the potentiometer were "sharing" the voltage that things clicked.  This is a zero sum game.

I first heard that concept when working for P. Read Montague (who is much more impressive than the Wiki article implies) at Baylor's Human Neuroimaging Lab.  It means something a little different than sharing; it implies that there is a set and unchangeable amount of stuff, and that if one party takes more it leaves less for others.  See Wikipedia article.

It became clear later that I had not conceptualized the LED as a resistor either.  In my mind, it was some sort of "passive receptor of electrical power that magically transmutes it into light" as opposed to something with inherent resistance.

In addition, I was not looking at the voltage as a "voltage drop", just as a source of power. In this context it's more appropriate to look at the measured voltage on the multimeter as "voltage stolen by the circuit element being measured"--unless you are measuring a battery or solar cell.

Simple stuff, I know.  I'm sure my younger siblings will get a laugh out if this. But good learning, nonetheless!

,m


Tuesday, June 11, 2013

Experiment 2

Again, dear reader, I refer you to the HO:ME blog for the details of experiment 2.  In short (pun not intended) the point of this experiment is to short out a battery.  Feel the heat, baby!

So I shorted out the 1.5 v AA battery.  Nothing to see, really, so I'll post a random picture here to distract you:



Nice, eh?

The next phase was to short out a fuse.  I bought the collection of automotive fuses from RadioShack, as that was the only one that had a 3 amp fuse.  Unfortunately, 1.62 volts/.8 = 2.02 amps, which I would not predict would blow the 3 amp fuse.  So I just turned right to the 9v, presuming that my calculated 12 amps should take care of business.

Hooked it up.  Watched.  Waited.  No fire, no energy, no nothing.  What?

Checked the battery.  Produced less than 7 volts.  Double what?

This was the same battery I had used for the earlier tests, and at one point I had run the battery through a very small resistor and then through the multimeter, showing 2 amps current.  Left it like that for a while.  Didn't think much of it.  Until I picked that stupid battery up and found it was--what else? hot as blazes.  I got lucky, I did.  Hot batteries have been known to cause some problems.

Dreamliner grounded due to battery fire

Turns out that the above lead to some reading on the difference between battery capacity, current, amp-hours (or watt-hours) and why folks use a stack of 1.5 v batteries to power their remote controlled cars instead of a single 9v.

Voltage for a battery is set by the number of cells, etc internal to the battery.  Capacity is how much energy it can store--like the size of the gas tank in the car.  A battery can then produce a certain number of amp-hours, which is number of amps for a certain amount of time.  A 9v rated at 2 amp-hours can produce 2 amps for 1 hour, 0.5 amps for 4 hours, etc.  This is analogous to the range in your car.

However, just like in the car, speed makes a difference.  The faster the car goes, the more inefficient it is (air resistance increases as a square of the velocity).  Same for batteries.  Running batteries at high current depletes it faster than the rated amp-hours would suggest.

A typical 9v is rated at 0.55 amp-hours, or Ah.  A 1.5 volt AA has 2.4 Ah, almost 4x as much.  So a stack of 6 AAs in series that add up to 9v will last much longer, at the cost of increased weight.

My 9v would only be expected to run 2 amps for about 15 minutes.  But that's a high current for a 9V, so probably even less.  Its ability to put out the number of amps required to burn out the fuse was clearly compromised.

So experiment 2 was a bust (or not, depending on how you view the fuse).  But good learning.