Help with a assignment question - Page 1
Forum section: electronics / beginners

author: minipower
author_rank: Contributor
timestamp: November 15, 2015, 11:15:09 pm
content: I had this fill in the blank Combination Circuit question on a class Lab assignment and I am having a hard time working it out.I got a different answer than the instructor and both the instructor and I got a different answer then Multisim.Do you guys have the time to help me understand what I am doing wrong.20VR1 470R2 1000R3 2000R4 10000R5 4700-------------------------\/\/\/\----------------------------------------------------------------------------l R1 l l ll l l ll l l l--------- \ \ \--- 20V R2 / R3 / R4 /-------- \ \ \--- / / /l l l ll l l ll R5 l l l--------------------------\/\/\/\--------------------------------------------------------------------------I gotR1 + 1/R2 + 1/R3 + 1/R4 + R5 = 5795Rt=5795 ohm or 5.8K ohm20V / 5795 = .00345A or 3.5mA20v x .00345A = .069W or 69mW20V.00345A5795ohm.069WR E I PR1 470 1.645 .00345 .0056R2 1000 18.35 .01835 .3367R4 2000 18.35 .00917 .1682R4 10000 18.35 .00183 .0335R5 4700 16.21 .00345 .0559Rt 5795 20 .00345 .069I got my R2 R3 R4 Voltage by subtracting the Voltage Drop of R1 = 1.645V...... So 20V - 1.645V = 18.35VInstructor gotR E I PR1 470 1.7 .00345 .0058R2 1000 1.8 .0018 .0032R4 2000 1.8 .0009 .0016R4 10000 1.8 .00018 .0003R5 4700 16.5 .00345 .0569Rt 5795 20 .00345 .069The Instructor got his R2 R3 R4 Voltage by adding the Voltage Drop of R1 + R5 = 18.2V...... So 20V - 18.2V = 1.8VMutisim showedR E I PR1 470 1.62 .00345 .0056R2 1000 18.4 .00216 .00456R4 2000 18.4 .00108 .00233R4 10000 18.4 .000216 .000456R5 4700 16.21 .00345 .056Rt 5795 20 .00345 .069Looks like Multisim got its R2 R3 R4 Voltage by subtracting the Voltage Drop of R1 = 1.645V...... So 20V - 1.645V = 18.35V and rounding up to 18.4VI probably need to make different post to understand this part.The Multisim 2.16V Voltage Drop on the Parallel R2 R3 R4.. Which might account for the different R2 R3 R4 Amps I got compared to Multisim

author: helius
author_rank: Super Contributor
timestamp: November 15, 2015, 11:43:12 pm
content: QuoteThis doesn't make any sense. What about R5? If R2 || R3 || R4 were replaced by the equivalent Rx, then E(R1)+E(Rx)+E(R5)=20.

author: ataradov
author_rank: Super Contributor
timestamp: November 15, 2015, 11:45:34 pm
content: Instructor is correct and quick check in LTSpice confirms it.I've never used multisim, but it is possible that you have selected non-ideal components (voltage source with internal series resistance, for example).

author: helius
author_rank: Super Contributor
timestamp: November 15, 2015, 11:48:52 pm
content: Quote from: blueskull on November 15, 2015, 11:44:26 pmThe circuit shown does not contain a ground.

author: helius
author_rank: Super Contributor
timestamp: November 15, 2015, 11:55:51 pm
content: Quote from: blueskull on November 15, 2015, 11:53:25 pmGood to know, thanks!One question: this makes the analysis quite confusing because the components R1, R2, etc are not nodes. If it refers those to adjacent nodes than confusion between high/low side is inevitable.

author: TimFox
author_rank: Super Contributor
timestamp: November 16, 2015, 12:27:47 am
content: In Spice, all nodes are numbered, but node 0 is reserved for the ground reference.Each resistor must have a name starting with "R" and must connect between two nodes.Spice voltages can be defined between two nodes or between a node and ground, following Spice syntax.

author: IanB
author_rank: Super Contributor
timestamp: November 16, 2015, 12:35:20 am
content: Also, in Spice you can designate any node as the "ground" node (i.e. you can number any node as 0--it's just a label).So for this kind of problem it is expected and logical to indicate for each resistor the voltage difference between its terminals, the current through it, and the power dissipated within it. The (arbitrary) choice of a "ground" node does not affect the result. Neither can it be assumed. If you indicate voltages with reference to some particular node, you must state which node that is.

author: IanB
author_rank: Super Contributor
timestamp: November 16, 2015, 12:43:37 am
content: Quote from: minipower on November 15, 2015, 11:15:09 pmIf you write this algebraically, you have:Rt = R1 + (R2||R3||R4) + R5where || means "in parallel with".It = V / Rt = 20 / RtI(R1) = ItI(R5) = ItV(R1) = It * R1V(R5) = It * R5V(R2) = V(R3) = V(R4) = 20 - V(R1) - V(R5)I(R2) = V(R2) / R2I(R3) = V(R3) / R3I(R4) = V(R4) / R4There are other ways to break down this problem, but they will all give the same answer.

author: minipower
author_rank: Contributor
timestamp: November 16, 2015, 04:45:35 am
content: Thank you for all the reply'sI am still not understanding why the circuit is solved as IanB writes itV(R2) = V(R3) = V(R4) = 20 - V(R1) - V(R5)Instead ofV(R2) = V(R3) = V(R4) = 20 - V(R1)I am looking at it as solving it from a start to finish " linearly " Since electricity flows.

author: IanB
author_rank: Super Contributor
timestamp: November 16, 2015, 05:51:49 am
content: Quote from: minipower on November 16, 2015, 04:45:35 amVoltage in a circuit is always expressed as a difference between two points. When I say V(R2) I mean the voltageacrossR2, the difference in voltage between one end of R2 and the other.This is important to make the calculations work. For instance, the currentthroughR2 is a function of the voltageacrossR2 and the resistance of R2. The equationI(R2) = V(R2) / R2is only true when V(R2) is the difference in voltage between each end of R2.

author: tec5c
author_rank: Frequent Contributor
timestamp: November 16, 2015, 06:15:41 am
content: What are you trying to find? You just have a circuit with a 20v supply, 5 resistors (with unknown values?). You don't mention anything about what the question asks you to do."Fill in the blanks" - The resistor values? Find the current for each branch and total current? Find the Thevenin/Norton equivalent?

author: ataradov
author_rank: Super Contributor
timestamp: November 16, 2015, 06:21:27 am
content: Quote from: tec5c on November 16, 2015, 06:15:41 amAll the values are listed right before the schematics. And OP's question is who's calculations are correct (his or teacher's).

author: tec5c
author_rank: Frequent Contributor
timestamp: November 16, 2015, 07:21:22 am
content: Quote from: minipower on November 15, 2015, 11:15:09 pmEverything is correct so far.Quote from: minipower on November 15, 2015, 11:15:09 pmLook at the current values in bold. How can these make sense? You have more current going through R2 and R3 than the total current for the circuit. This should have been your first red flag.What is the class that this is for? High School? University?

author: minipower
author_rank: Contributor
timestamp: November 17, 2015, 06:07:15 am
content: Thank you for pointing that out tec5c. I didn't realize that till you said something.I subtracted R1 voltage drop from Vt20 and got 18.35V. Then since Voltage stays the same in parallel circuits I took 18.35V\1000 which got me 0.01835I. I divided the 18.35V by each of R2 R3 R4 to get their amps.That is what I did wrong. I am not understanding why it should be solved by add the voltage drop of R1+R5=V of R2 R3 R4.I'm missing something.I am finally taking Electronic classes at the community college after years of wanting to, but not having the time too. Over the years I have tried to work out circuits by myself, by trying to repair broken electronics. That my be what is getting in my way and mixing me up.Edit: I apologize. Thinking back I remember I did realize the current problem. When I was cross checking my answers and seeing the parallel banks current didn't add up. That was when I asked for help to understand where and why I went wrong. And when I plugged it into Multisim and saw mine and Multisims different current numbers.

author: tec5c
author_rank: Frequent Contributor
timestamp: November 17, 2015, 06:45:43 am
content: Quote from: minipower on November 17, 2015, 06:07:15 amIt sounds like you forgot to account for the loss of current when calculating the current through R3 and R4. So for R3 you need to take the loss of current through R2 into consideration. Similarly for R4, you need to consider IR3 and IR2.Sorry if that doesn't make much sense. I am in a hurry, but can help more later on.

author: IanB
author_rank: Super Contributor
timestamp: November 17, 2015, 06:51:08 am
content: Quote from: minipower on November 17, 2015, 06:07:15 amWhat you are missing is that all voltages are voltage drops (see my previous answer to your question above). There is no such thing in a circuit as a "voltage" at a point. There is only a difference in voltage between two points.Imagine that you have a voltmeter (a multimeter). It has two probes, a red one and a black one. To measure a voltage you have to touchbothprobes to the circuit. If you touch just the red probe nothing will happen. When you touch both probes to the circuit the meter displays the voltagedifferencebetween the two probes. This is the only voltage that can ever exist, the difference between two points.Now in your circuit, the voltagedifferenceacross the battery terminals is 20 V. In the circuit you have three resistances in series: R1, (R2||R3||R4), and R5. The voltagedifferencesacross R1, across (R2||R3||R4), and across R5 must add up to the voltage difference across the battery terminals. Put another way:V(battery) = V(R1) + V(R234) + V(R5)This is the basic voltage law of circuits.Now, by algebra, we must have that:V(battery) - V(R1) - V(R5) = V(R234)If we were to write this:S = A + B + Cthen you would agree (I hope) thatS - A - C = BIt is just the same with the voltages.

author: AustinEngy
author_rank: Regular Contributor
timestamp: November 17, 2015, 02:30:40 pm
content: I did some calculations on paper and got the same answers as your simulation program got, but cannot for the life of me figure out how your professor got the answers he did. I just turned the circuit into two different voltage dividers by combining the parallel combination into an equivalent resistance and found node voltages with reference to 0V. Then it simply requires an application of ohms law to figure out the current through each resistor.Looking at your professors answers, they really don't make much sense, since the parallel combination's currents should add up to the current found in R1 and R5.Hope this helps!

author: IanB
author_rank: Super Contributor
timestamp: November 17, 2015, 04:40:43 pm
content: I think your instructor has an error. Here is a worked solution:> V=2020> R1=470470> R2=10001000> R3=20002000> R4=1000010000> R5=47004700> J2=1/R20.001> J3=1/R30.0005> J4=1/R40.0001> J234=J2+J3+J40.0016> R234=1/J234625> Rt=R1+R234+R55795> It=V/Rt0.00345125107852> I2=It * J2/J2340.00215703192407> I3=It * J3/J2340.00107851596204> I4=It * J4/J2340.000215703192407> I2+I3+I40.00345125107852> V1=V * R1/Rt1.6220880069> V234=V * R234/Rt2.15703192407> V5=V * R5/Rt16.220880069> V1+V234+V520>

author: bson
author_rank: Unknown
timestamp: November 18, 2015, 03:44:54 am
content: Quote from: minipower on November 15, 2015, 11:15:09 pmShould beR1 + 1/(1/R2 + 1/R3 + 1/R4) + R5

author: tec5c
author_rank: Frequent Contributor
timestamp: November 18, 2015, 03:50:52 am
content: Quote from: bson on November 18, 2015, 03:44:54 amHe still got the correct value though.

author: minipower
author_rank: Contributor
timestamp: November 21, 2015, 12:45:24 am
content: Thank you guys for helping me get this down...You Got me IanB. I am getting over the misconception of Voltage being a force. At first I was trying to work through the Classes example circuits by thinking Say.. the voltage drop from R1 would force a change to the values of R2 R3 R4 and then that voltage drop would force a change at R5 " if that makes any sense? " That error most definitely came from trying to learn this on my own. With electronic devices rated in Volts and then ohm's law " current is directly proportional to voltage and inversely proportional to resistance " helped enforce the error that voltage was a tangible force that could cause a change.. I'm still working to pull that out of my brain lolThanks bson for correcting my mistake there. I'm glad you guys are good enough to understand what I was trying to say..IanB do you know of another formula for> I2=It * J2/J2340.00215703192407> I3=It * J3/J2340.00107851596204> I4=It * J4/J2340.000215703192407> I2+I3+I40.00345125107852It is a little confusing to me. Switch to an inverse of and from the inverse of in the same formula will have me hitting a speed bump everytime.

author: IanB
author_rank: Super Contributor
timestamp: November 21, 2015, 01:02:06 am
content: Quote from: minipower on November 21, 2015, 12:45:24 amBut every force always has two sides to it. Just like Newton's third law: "every force has an equal and opposite reaction". No force ever acts alone. If a magnet is pulling on a piece of metal, is the metal attracted to the magnet, or is the magnet attracted to the metal? There are two sides, the magnet side and the metal side, and between them there is a force. As it is with magnets, thus it is with voltage. A resistor has two ends, and if between each end there is a voltage then a current flows.Just consider Ohm's law like this and everything will be correct: "current is directly proportional to voltage difference and inversely proportional to resistance".QuoteJ here is conductance. More conductance gives more current, less conductance gives less current. We can re-state Ohm's law like this: "current is directly proportional to voltage and also directly proportional to conductance".There is a symmetry here. Voltage in a series circuit divides according to the resistances, while current in a parallel circuit divides according to the conductances.Another formula for I2 is of course like this:I2 = V2 / R2Just apply Ohm's law to each resistor.

author: minipower
author_rank: Contributor
timestamp: November 30, 2015, 02:54:02 am
content: Hello guys.. I hope you all enjoyed your holiday, and if you do not celebrate the Thanksgiving tradition. I hope you enjoyed watching us stuff ourselves past any definition of sane limits. hahaFirstly I want to apologize if I make anyone pull there hair out. lolIanB.. I was able to get your> I2=It * J2/J2340.00215703192407numbers by putting in the calculator It 0.00345125107852 * .001 The inverse of J2 / .0016 The inverse of J234 = 0.00215703192407.. Do you think that would work right if I calculate it like that.This statement had me scratching my head.Another formula for I2 is of course like this:I2 = V2 / R2Just apply Ohm's law to each resistor.I was all like... But that's what I did and I got it wrong..lolBy reading your guy's reply's and studying IanB's work in Post #21. I am being able to make since of it and understand where I was going wrong.I took the the 18.35V I got from subtracting V1=1.65V from Vt=20V and then using Ohm's law I=V*R. Instead of adding up R234 = 625*It=V234. My instructor got to the same answer by adding V1 and V5 and subtracting to get V234.Most likely My instructor got the different numbers as IanB did, from not using the whole decimal places and or rounding up.

author: minipower
author_rank: Contributor
timestamp: December 06, 2015, 09:04:54 pm
content: I went back and read the definition for Conductance. "Conductance is the reciprocal of resistance" So yes, that formula would always work.I am continuing to learn the new terms and definitions. lol I won't continue to mix up Voltage with Power.. Captain Hindsight told me I am a tard for mixing them up.Again... thank you guys.