How can this be? (picture) - Page 1
Forum section: electronics / beginners

author: billbyrd1945
author_rank: Regular Contributor
timestamp: February 14, 2019, 03:14:27 pm
content: In an effort to learn circuit analysis, I built up this circuit. The math for parallel and/or series circuits agrees with the measured current. I should've stopped there. But I decided to go a step further and measure voltage coming from each resistor using a common ground. Please notice R2 and R3. How can these two resistors pass voltages so different from the others in the circuit? And if R3 is somehow defective, how can R4 inherit the correct voltage. Adding all voltages yields a number greater than the 7.0 set voltage from my power supply. I appreciate any help.

author: Tom45
author_rank: Frequent Contributor
timestamp: February 14, 2019, 04:05:20 pm
content: Resistors pass currents, not voltages. Measure the voltages across each resistor, rather than from one side of a resistor to some other point in your circuit.For example, V1 is voltage across R1, V2 is across R2, etc. You should find that:V1 + V2 + V5 = 7 volts and V3 + V4 + V5 = 7 volts.Read up on voltage dividers and then try some simple experiments with 2 resistors in series connected to your 7 volt supply. Once you understand that, you should be able to explain what you are seeing in your example circuit.

author: ebastler
author_rank: Super Contributor
timestamp: February 14, 2019, 04:18:06 pm
content: When you give this another look, following Tom45's guidance, be sure that you don't confuse yourself with the measurement you have indicated for R5. Your red probe connection for R5 must have been on the bottom side of the resistor, not on the top as drawn.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: February 14, 2019, 05:08:06 pm
content: Especially when learning electronics, start by learning how to draw schematics properly.General rules are very simple:Signals flow from left to right.Voltages go from top to bottom.So:draw your battery / voltage source on the left of your screen.Then draw 2 horizontal lines. One on the top, that is your "positive voltage rail". and the other on the bottom. That is your "ground"(These names are arbitrary! but keeping to this convention helps with understanding.Then connect your voltage source to the 2 horizontal lines.Then draw your resistors between the 2 voltage rails.

author: billbyrd1945
author_rank: Regular Contributor
timestamp: February 14, 2019, 05:09:22 pm
content: Thanks to all of you. You've given me something to work with. I'll probably post a follow-up image.

author: billbyrd1945
author_rank: Regular Contributor
timestamp: February 14, 2019, 06:25:19 pm
content: Would this be the correct format for the schematic? I wasn't sure how to handle the split.

author: ebastler
author_rank: Super Contributor
timestamp: February 14, 2019, 06:52:06 pm
content: Here's how I would draw it. Doctorandus summed up the typical conventions nicely: Source on the left, signals propagating left to right, voltages dropping top to bottom. But these are just conventions to help yourself and others get your bearings, and they often can't be followed to the letter in more complex designs. There is no absolute "right" or "wrong" here.

author: Rick Law
author_rank: Super Contributor
timestamp: February 14, 2019, 07:39:41 pm
content: Quote from: billbyrd1945 on February 14, 2019, 06:25:19 pmYour initial diagram made it easy to be confused. This new one is much improved. I don't know if this new drawing will pass expert's opinion (since I am just another hobbyist with limited experience), but I think it sure would help you think about the problem.If you write down the same voltage you measured, you can now following it easier.For example, in youroriginal diagram, the red dot left of R1 and the red dot to the right of R4 are actually the same point electrically speaking - since they are directly connected by (ideally 0-ohm) wire. It would still be the same point if you put the dot right under R5.I bet you you've a mistake there: The dot you haveabove R5is actuallybelow R5. Above R5, you are directly connected to the black dot ground. So, it should read 0V. Below R5, that is the same electrical point as left-of-R1 and right-of R4 described above, so they are all at 6.34V.That said, lets go around your diagram and see if I can help you understand how that flows (using your original diagram and not the redrawn one since your original has voltage written down) :Right-of-R2 is connected to +7v, so at +7V it is.Left-of-R2 is 6.38V -> R2 dropped (7V-6.38V) = 0.62V*** This tells you your current running through R2 isI=V/R=0.62V/148.5ohm=0.004175AmpRight of R1 is 6.34V, Left of R1 is 6.38V, so R1 dropped 6.38V-6.34V=0.04V.*** current thought R1 isI=V/R=0.04V/10ohm = 0.004AmpThe current that flows through R1 and R2 should be the same (kirchoff's law) - even without known the law, common sense would say, the current can't disappear, they must be the same since they are inline with each other. Think of that as two connected garden hose R1 and R2. What goes through garden hose R1 and garden host R2 should be the same unless you have a leak.Knowing they should be the same, but they are not - you have 0.004 on one and 0.004175 on the other. That is your system's error: perhaps measurement error, perhaps your resisters has resistance different than labeled..000174/0.004=0.043750 => your error there is is 4.375 percent. Well within the typical expectation.Notice I did not say current through R5 would be the same at about 0.004Amp - they are not. Current from the right side with R3 and R4 also go through R5. So, whatever is going through R3, it is the same as through R4. That two "hose" joins up with the R1-R2 hose. So, R3-R3 current, plus current through R1-R2 would be the current going through R5.Another thing you can surmise quickly is, R3-R4 side should have less current than R1-R2 side. There is more resistance on the R3-R4 side.I hope this helps you see how current is flowing in your system, and how voltage is dropping in your system. Take the time to evaluate the R3-R4 side, see how much current actually is going through. Doing that exercise will have you in the future.Rick

author: billbyrd1945
author_rank: Regular Contributor
timestamp: February 14, 2019, 07:47:31 pm
content: Here are the results measured as suggested. Now I will read the Rick Law response.

author: billbyrd1945
author_rank: Regular Contributor
timestamp: February 14, 2019, 08:02:22 pm
content: Thanks Rick! I actually think I understand everything you said. With regard to the misplaced red test point on R5, I think I actually got a negative value when measured but went ahead and recorded it as positive, so I knew I was probably confused about the polarity. Back to the drawing board.

author: Rick Law
author_rank: Super Contributor
timestamp: February 14, 2019, 08:38:22 pm
content: Quote from: billbyrd1945 on February 14, 2019, 08:02:22 pmGlad it helps - I had a few typos there but that apparently didn't hurt too much.Re: that misplaced red-dotSince you had the number you should (ie: that point is indeed below and not above R5), yeah, you did reverse the probe but that is no big deal. Now had that point really been above R5 and you got a non-zero number, that would be an error. If that number is big, it is telling you something and you should figure it out. That something is: either your meter is ill, or you can improve your technique, or your circuit is shorting out with something that it shouldn't be touching.I have the habit of not applying enough pressure on the probe-tips. I am too concern that the probe may slip, touch something else and shorting things out.When you see some off-the-wall reading like negative when you should see something else, you can use that kinds of opportunity to see if there is any way of improving your measurement techniques. (Or an excuse to get something nicer without feeling guilty).

author: Tom45
author_rank: Frequent Contributor
timestamp: February 14, 2019, 11:58:42 pm
content: Quote from: billbyrd1945 on February 14, 2019, 07:47:31 pmWith the redrawn circuit it is easier to see what is happening. There are effectively 2 resistors in series:R5 in series with ( (R3 + R4) in parallel with (R1 + R2) ). That is the voltage divider that I referred to earlier.158.5 || 296.5 is 103.29 ohms.The total resistance in the circuit is 1103.29 which means 0.00634 amps of current is flowing through R5 (7 volts / 0.00634 amps).Part of that 0.0634 amps comes from the R3 + R4 branch, and the remainder comes from the R1 + R2 branch.I have 45 in my user ID because I was born in 1945. Is that the case for you too?Tom

author: billbyrd1945
author_rank: Regular Contributor
timestamp: February 15, 2019, 12:52:16 am
content: "I have 45 in my user ID because I was born in 1945. Is that the case for you too?"That's a big Roger. December 7, 1945, and just now getting around to a lifelong desire to learn electronics. And I really appreciate all the help.

author: Tom45
author_rank: Frequent Contributor
timestamp: February 15, 2019, 03:17:35 am
content: Quote from: billbyrd1945 on February 15, 2019, 12:52:16 amSo, you are a youngster. I am a day older than you are.We are both too old to be baby boomers by a few weeks.Tom

author: billbyrd1945
author_rank: Regular Contributor
timestamp: February 15, 2019, 05:24:11 pm
content: Too Cool!

author: billbyrd1945
author_rank: Regular Contributor
timestamp: February 16, 2019, 02:59:49 am
content: Riddle me this: My adjustable power supply, and, my Fluke 15B+ both indicate a draw of 0.005A while the math is 0.00634A. And, in the math I used measured values rather than color bands for the resistors. I realize we're talking third decimal point to the right, but isn't something wrong here? It doesn't bother me to admit that my $60 PS and my Chinese Fluke are not top of the line. And, if that's the cause of the discrepancy, I'm okay with it. I just need to know. Anyone?

author: mikerj
author_rank: Super Contributor
timestamp: February 16, 2019, 11:47:50 am
content: The Fluke 15B+ has a resistance accuracy of 0.5% + 2 or 3 digits depending on the range. Did you also measure the actual supply voltage (again 0.5% +3 digits) and use that in your math?

author: billbyrd1945
author_rank: Regular Contributor
timestamp: February 16, 2019, 12:47:49 pm
content: I was released from high school on a plea bargain. Can you show me the math you just suggested. I know how to figure 5% of the displayed power supply values. You lost me on the "+ 2 to 3 digits". The Fluke is on auto-range and displays the same values as the PS. (Then I have a follow-up question regarding series vs parallel.)0.005A + 0.00025(5%) = 0.00525A vs 0.00634A

author: Dave
author_rank: Super Contributor
timestamp: February 16, 2019, 01:07:20 pm
content: Are you taking into account that the Fluke itself has internal resistance when measuring current?

author: ArthurDent
author_rank: Super Contributor
timestamp: February 16, 2019, 04:10:58 pm
content: Any measurement you make on the resistors in your circuit using your multimeter has an effect on the circuit so knowing how the multimeter loads the circuit can determine how to use it to get the most accurate readings.First, the basic accuracy of your Fluke meter is about 0.5% on DCV and 1.5% or 3 times worse on DCMA So if you can just take voltage measurements your readings will be more accurate.If you’re measuring the voltage drop across a resistor the ‘burden’ a voltmeter puts on the resistor you’re measuring will form parallel resistors (say R5, 1K and the meter input resistance of 10M ohm) so the 1K resistor is now the equivalence of a 999 ohm resistor giving a slightly lower voltage drop across that resistor and slightly higher voltage drop across the other resistors in the circuit. Because this ratio is about 1000:1, the high impedance voltmeter will not have much of an influence on the reading accuracy.Now if you wire your multimeter set to Ma in series with R5, 1K, the equivalent value could be 1020 ohms because the burden or series resistance on the Ma range could be 20 ohms. On uA ranges the burden could be 1000 ohms making R5 2000 ohms and this would influence the reading to a far greater extent. This added to the DC current accuracy being 3x worse than the DCV specs means measuring values this way will give far less accurate readings and the readings will not agree with the math.So in this case with the circuit values you have it is best to measure voltage drops and use Ohm’s law to calculate the voltage and current values. If you had a much higher voltage and all the resistors were in M ohms and not ohms, then measuring the voltage drops with a 10M input impedance would cause fairly large errors and using the Ma meter in series wouldn’t have as large an effect. Some better meters have an input impedance of 10G ohms that make them useful for voltage measurement in very high impedance circuits.This video does an excellent job of explaining the current measuring problem in detail.

author: Tom45
author_rank: Frequent Contributor
timestamp: February 16, 2019, 04:40:30 pm
content: Quote from: billbyrd1945 on February 16, 2019, 12:47:49 pmThe +- 2 or 3 digits is really 2 or 3 counts.As you know, the 0.005 reading could be off by 5% either way. The 0.00025 you stated.The +- counts says that the 0.005 reading could be as much as 0.008 or as little as 0.002. Pretty dreadful! But that is because the meter scale is amps. If you have a milliamp scale, it could indicate 5.000. Then the +- counts would give a range of 4.997 to 5.003 milliamps. A lot better.OK, I just looked up the Fluke 15B and it does have a milliamp scale. Also a microamp scale but that wouldn't be appropriate here. See what the current reading is on the milliamp scale. Note that there is a separate input for the milli and micro amp scales.Tom

author: billbyrd1945
author_rank: Regular Contributor
timestamp: February 16, 2019, 08:15:58 pm
content: (Response to Tom's last) Oh, I see what you mean. I have another meter with more counts. I'll check it with that. But-- I actually only used the multimeter (one time) to see how it compared to the PS display. After learning that the PS doubles as an ammeter, I just check to see what it shows. And it isthatfigger (a little Churchill) that's too low. Might there be a trim pot in the PS for calibrating it to agree with the math? (And, I do understand that meters used in checking skew the very values they're measuring.)

author: billbyrd1945
author_rank: Regular Contributor
timestamp: February 16, 2019, 09:53:40 pm
content: Thanks Tom! Learned a lot. Essentially splains everything! Build with math, then check with meters. If they agree, you're probably off a little bit.

author: billbyrd1945
author_rank: Regular Contributor
timestamp: February 16, 2019, 11:12:43 pm
content: It seems obvious in this simple circuit that R5 will be used as a whole number added to the resistance equivalents of R1-R5. In more cluttered circuits (for noobs like me) it isn't always so obvious. I was trying to come up with a rule. At first, I thought "Any component followed or proceeded by a node, would be in parallel". But then... that's probably not correct. Any suggestions?

author: ebastler
author_rank: Super Contributor
timestamp: February 17, 2019, 06:46:15 am
content: I may be misunderstanding you; but I think you have your terminology mixed up, or have misread the circuit diagram. In your diagram,R1 and R2 are connected in series, and R3 and R4 are in series,the group (R1+R2) is inparallelwith the group (R3+R4),R5 is inserieswith the group ((R1+R2) || (R3+R4)).Whenever two resistors are wired in series, theirresistancesadd up:Rtotal= Ra+Rb.Whenever two resistors are wired in parallel, theirconductivities(i.e. the inverse of the resistances) add up to the total conductivity:1/Rtotal= 1/Ra+ 1/Rb.May I again recommend drawing the circuit as shown in reply #6 above? As I said back then, there is no "right" or "wrong" here. But, at least for me, that way of arranging the components makes the parallel vs. series connections more obvious.

author: billbyrd1945
author_rank: Regular Contributor
timestamp: February 17, 2019, 12:52:59 pm
content: To ebastler: No, I understand that R1 and R2 are in series within a parallel branch and therefore added together and then inverted (2 steps). Ditto for R3 and R4. And I intuitively know that R5 would be a series resistor outside the parallel branches. But how would I explain to my 12 year old granddaughter how to rule out R5 as a parallel circuit resistor? How would she know to use it's value rather than its inverse? Thank you

author: ebastler
author_rank: Super Contributor
timestamp: February 17, 2019, 12:59:13 pm
content: Quote from: billbyrd1945 on February 17, 2019, 12:52:59 pmWell, all of the current that wants to flow from one battery terminal to the other needs to flow through R5. There is no alternative (parallel) path to R5 which the current could take. The current first has to find its way through the resistor network made of R1 to R4, and then (afterwards, sequentially, serially) flows though R5. So R5 is in series with the other resistor network.

author: Brumby
author_rank: Unknown
timestamp: February 17, 2019, 01:06:35 pm
content: Quote from: billbyrd1945 on February 17, 2019, 12:52:59 pmRE-DRAW THE CIRCUIT - in the 'conventional' format: Power rails at top and bottom - circuit in between, working left to righteblasterhas done this for you:

author: ebastler
author_rank: Super Contributor
timestamp: February 17, 2019, 01:08:22 pm
content: To add to my prior post:I still like the "plumbing" analogy to explain such simple circuits. Think of your circuit as a system of water pipes. "Current" is the flow rate of the water passing through a pipe (gallons per minute, liters per second or whatever). "Voltage" is the water pressure you apply (e.g. via an elevated reservoir and a pump which brings the water back up to the reservoir). "Resistors" are narrow sections of tubing which constrain the flow.If your force the water to flow through two narrow sections in series, their resistances will add up -- for a given applied pressure, less water will flow through that circuit. If you provide two parallel paths to the water, the water can chose either way. You will get more water flow through those two paths combined than through each individual path. In other words, each parallel path can "conduct" water -- their conductivities (inverse resistances) add up.

author: billbyrd1945
author_rank: Regular Contributor
timestamp: February 17, 2019, 02:00:50 pm
content: Okay. I'm thinking that a rule would be: When using conventional flow, the series resistor's output cannot be divided on its way back to the power source (or the next component). And such a resistor (or component having resistance) would be calculated in the analysis as a whole (non-inverted) value. In this case, the value would be 1000, not 1/1000.

author: ebastler
author_rank: Super Contributor
timestamp: February 17, 2019, 02:09:52 pm
content: Quote from: billbyrd1945 on February 17, 2019, 02:00:50 pmNo, sorry, you still got this wrong. It does not matter at all what comes after the series resistor's output. There could be another complex network following the output of R5 in your example (including a division into multiple paths); and R5 would still bein serieswith both, the complex network in front of its input, and the complex network behind its output. The only thing that matters is that there is no "bypass" in parallel with R5, where the current could flow without going through R5.Sometimes we develop a mental block, and you seem to be in this situation -- maybe overthinking things. Perhaps you should ask that 12-year-old granddaughter of yours to explain it in her words.

author: Brumby
author_rank: Unknown
timestamp: February 17, 2019, 02:28:59 pm
content: I think the OP needs to stick to resistance - forget conductance altogether. (Yes, it can make the math easier, but it is confusing thehellout of understanding the concept!)Resistors in series: Rtot= Ra+ RbResistors in parallel: Rtot= 1/( 1/Ra+ 1/Rb)Please - just stick to these equations.

author: Brumby
author_rank: Unknown
timestamp: February 17, 2019, 02:39:30 pm
content: Process:(A) Solve for R1 and R3 in series(B) Solve for R2 and R4 in series(C) Solve for (A) and (B) in parallel(D) Solve for (C) and R5 in seriesThe value for (D) is the equivalent resistance for the circuit - and you can get the current flowing, using Ohm's Law.If you then want to find out what current is flowing within the network:Using the current calculated above, you can find the voltage across R5 (Ohm's Law again)You can then work out the voltage that is across (A) and (B) and, once more, using Ohm's Law, work out the current passing through each. Note: These 2 current values should add up to the total current.You can then work out the voltages across each resistor: R1, R2, R3 and R4.

author: billbyrd1945
author_rank: Regular Contributor
timestamp: February 17, 2019, 04:45:17 pm
content: Okay. Believe it or not, you provided the rule with this "The only thing that matters is that there is no "bypass" in parallel with R5, where the current could flowwithoutgoing through R5.Thank you. I'm done. Thanks to everyone. Sorry if I seem a little hard headed.