Help solving a homework problem... - Page 1
Forum section: electronics / beginners

author: cvriv
author_rank: Frequent Contributor
timestamp: March 27, 2020, 09:01:56 am
content: I'm studying for a test and decided to do some extra problems just for the hell of it, so I started with the easy ones and I can't, for the life of me, solve this one problem. I think I'm having an algebraic issue more than a misunderstanding of what I need to do. I pretty much solved through most of it but can't seem to isolate the current to solve for it. Please have a look to see what it is that im missing here. I'm 4 hours in trying to solve this. 😆

author: aneevuser
author_rank: Frequent Contributor
timestamp: March 27, 2020, 02:25:02 pm
content: It will probably help if you draw and label a diagram, showing all currents and voltages.Note that you know neither I nor R, the external load resistance, so you will need a relationship between these to eliminate one or the other. It looks like you will end up with a quadratic in I.

author: Wimberleytech
author_rank: Super Contributor
timestamp: March 27, 2020, 02:36:18 pm
content: You just solve the quadratic and you are done!Ptot = IVterm + I2RintI2Rint + IVterm - Ptot = 0I2+ IVterm/Rint - Ptot/Rint = 0I2+ 90I - 150 = 0

author: wraper
author_rank: Unknown
timestamp: March 27, 2020, 02:51:30 pm
content: Question is worded in such a weird way that battery internal resistance appear to be irrelevant.Like battery has 9V on terminals under load and all 15W goes into load.It's particularly said that battery supplies 15W of power, not that the whole system dissipates such power. So I=P/V -> 15/9 = 1.67A, and power dissipated by resistor = power in the question = 15W, LOL.

author: MK14
author_rank: Super Contributor
timestamp: March 27, 2020, 03:20:42 pm
content: Quote from: wraper on March 27, 2020, 02:51:30 pmNo.Details withheld, as I want them to solve the homework, themselves.OP: Would this be from (cough cough cough) Chapter 18.

author: iMo
author_rank: Super Contributor
timestamp: March 27, 2020, 03:23:50 pm
content: Quote from: wraper on March 27, 2020, 02:51:30 pmThe question is whether the 15W is delivered by the "battery", or the 15W is available at the battery terminals.. I would think the 15W comes from the chemistry, thus 0.1ohm internal resistance plays a role.. Otherwise it would be too easy exercise..

author: Zero999
author_rank: Super Contributor
timestamp: March 27, 2020, 03:28:58 pm
content: Quote from: wraper on March 27, 2020, 02:51:30 pmYes, that's how I read it. Always draw a schematic, before answering questions like this. Hopefully the examiner will be able to follow your reasoning.

author: MK14
author_rank: Super Contributor
timestamp: March 27, 2020, 03:34:21 pm
content: Quote from: Zero999 on March 27, 2020, 03:28:58 pmI "cheated" and had a little peak at the answer (or partial answer). It's not my homework question, and I could have solved it, if necessary.I agree, it is arguably ambiguous, if the 15 watts, was for the output terminals of the battery, or the total overall power, including the batteries internal resistance.But the "answer peek" by me, seems to imply that it is the latter.EDIT: I can't be 100% sure, it is the same question/book.

author: aneevuser
author_rank: Frequent Contributor
timestamp: March 27, 2020, 04:03:54 pm
content: Unlike others, I see no ambiguity here: the expression "total power" seems to make it clear that it includes power dissipated in the internal resistance.

author: MK14
author_rank: Super Contributor
timestamp: March 27, 2020, 04:23:00 pm
content: Quote from: aneevuser on March 27, 2020, 04:03:54 pmThat's also what I thought, when I first looked at this thread.But, I can understand if others think/see it as being ambiguous.It is tricky to describe things like that, without causing at least a bit of confusion.

author: Zero999
author_rank: Super Contributor
timestamp: March 27, 2020, 04:52:47 pm
content: Quote from: aneevuser on March 27, 2020, 04:03:54 pmThe battery supplies a total power of 15W with a terminal voltage of 9V. That means the battery supplies 15W, to a load, whilst the voltage across its terminals is 9V. It seems pretty clear to me.Look at the data sheets for transformers and you'll find that the secondary voltage is always when it's fully loaded to the full power rating. A 15VA 9V transformer will be specified at delivering 15VA to the load, with a secondary voltage of 9V. The regulation specification will tell you the secondary voltage will be closer to 11V, when unloaded.I gave the right answer. The answer in the book is wrong. They should provide clear schematics, rather than vague phrases, in exams.

author: MK14
author_rank: Super Contributor
timestamp: March 27, 2020, 05:03:45 pm
content: Quote from: Zero999 on March 27, 2020, 04:52:47 pmIf you look at a transformer datasheet.Here is one:https://docs.rs-online.com/7a07/0900766b815bd318.pdfIt says words to the effect that the output power is 80 watts (2 x 40VA), BUT it also says the power losses are:QuoteSo the TOTAL POWER (consumed) is 80 + 0.49 + 13.8 = 94.29 Watts (assuming a suitable 80 watt load is used)Total Output power is 80 watts, which is what you are referring to.

author: tunk
author_rank: Super Contributor
timestamp: March 27, 2020, 05:08:11 pm
content: I guess it depends on who the question is aimed at.If it's university level, then I would expect thestudent to understand that it includes the internalbattery loss.

author: Wimberleytech
author_rank: Super Contributor
timestamp: March 27, 2020, 05:20:13 pm
content: Quote from: aneevuser on March 27, 2020, 04:03:54 pmI agree. Otherwise, it is a trivial problem.

author: wraper
author_rank: Unknown
timestamp: March 27, 2020, 05:44:27 pm
content: Quote from: tunk on March 27, 2020, 05:08:11 pmOr it may include professor with 'gotcha' questions for students who don't read with scrupulous attention. Trivial answer does not mean it's incorrect, and I've seen 'smart' idiots who make their life harder by overthinking. In this case what was supposedly meant while creating the question and what's actually written in the question is different. And correct would be to answer to what was actually written, not what was 'supposedly' implied.If battery supplied something, it's already outside the battery. If intention was to write total power dissipated, including the battery itself, then question should have been written differently.QuoteTry this argument in court when interpreting the law or contract written by lawyers.

author: MK14
author_rank: Super Contributor
timestamp: March 27, 2020, 05:58:42 pm
content: Quote from: wraper on March 27, 2020, 05:44:27 pmBasically, this:

author: Zero999
author_rank: Super Contributor
timestamp: March 27, 2020, 06:27:15 pm
content: Quote from: MK14 on March 27, 2020, 05:03:45 pmYes, that's true, but if you search for transformers in RS Components, Digi-Key, Farnell etc. you'll find they're listed by the secondary power rating, not by power consumption.The question states that the battery supplies a total power of 15W. If some power is lost inside the battery, then it isn't supplied, but lost inside the battery. It would be like saying the power company supplies my house with 10kW/h per day, when 2kW/h is lost in the cables and transformers and I only get 8kW/h.

author: wraper
author_rank: Unknown
timestamp: March 27, 2020, 06:31:08 pm
content: Quote from: MK14 on March 27, 2020, 05:58:42 pmIMHO more like this:

author: cvriv
author_rank: Frequent Contributor
timestamp: March 27, 2020, 06:45:06 pm
content: Holy replies Batman!Quadratic equation..... Wow. Nowbim feel stupid. I knew I was right there but just couldnt see it. Probably because my terms are labeled variables with no values. I dont know. I got this. Thanks.Yea this is from chapter 18! Lol. I was doing the problems because I love solving problems like this. It's almost like solving puzzles, but this one problem.... Lol.I understand everything thats going on with the problem. I knew I was having an algebraic issue.Thanks everyone.

author: MK14
author_rank: Super Contributor
timestamp: March 27, 2020, 06:52:32 pm
content: Quote from: Zero999 on March 27, 2020, 06:27:15 pmWhen you talk about transformers or batteries and things. Datasheets/specifications, are indeed usually talking about the available output power, and ignoring losses. I agree.So for a 100 Horse Power car engine, it usually means it creates 100 HP, for the driver to use.But in reality, it also generates more HP, which is used to overcome the friction/losses in the engine, and power the ancillary equipment. Such as the alternator and fan (belt), etc.So a real life 120 HP engine (full generated power), would probably be described as being a 100 HP engine.Similarly, when something says a battery is 9 Volts (at 1 amp load). So it is a 9V battery.What they really mean (but is rarely stated), it is a 9.5V (guestimate, half ohm internal resistance) battery internally/chemically, and 0.5 volts is lost by the time the 9V reaches the battery terminals.But there is bias, depending on where/how the question comes from.If I ask, how much you earn, you may say £500 per week. Because there is £500, paid into your bank account, each week.But if the tax man, on a tax form asks how much you earn, you might say £953.23 per week (Before Tax, NI and any other deductions).Because the question differentiates, between TOTAL POWER and "Power delivered to the Load resistor". It seems to be phrased, so that TOTAL POWER, is the absolute total power the battery generates, BEFORE losses. As I see it.

author: wraper
author_rank: Unknown
timestamp: March 27, 2020, 06:58:30 pm
content: Quote from: MK14 on March 27, 2020, 06:52:32 pmYou forgot one word: A battery...suppliesof total power. My argument is that what it lost internally, it did not supply.

author: MK14
author_rank: Super Contributor
timestamp: March 27, 2020, 07:04:16 pm
content: Quote from: wraper on March 27, 2020, 06:58:30 pmI agree. (But would prefer to say it is ambiguous, or can be ambiguous to some people).Also, many questions can be torn to pieces, if you really want to.What time is it ?Alternative interpretations:What time zone or time where ?What units of time do you want ?Etc.

author: tunk
author_rank: Super Contributor
timestamp: March 27, 2020, 07:09:41 pm
content: I guess this basically is a question about understanding thewhole system, i.e. how the power supply interacts with the load.

author: cvriv
author_rank: Frequent Contributor
timestamp: March 27, 2020, 07:14:17 pm
content: Quote from: tunk on March 27, 2020, 07:09:41 pmExactly. That I understand. Just got stuck on the algebra. Couldn't see the quadratic starring me in the face.

author: Zero999
author_rank: Super Contributor
timestamp: March 27, 2020, 07:23:11 pm
content: Quote from: wraper on March 27, 2020, 06:58:30 pmExactly. I can't see what else the question could possibly mean.

author: aneevuser
author_rank: Frequent Contributor
timestamp: March 27, 2020, 08:26:26 pm
content: Quote from: Wimberleytech on March 27, 2020, 05:20:13 pmRight, that is indeed the key point, and I'm wondering if those people arguing otherwise have actually worked the problem with both interpretations.Quite apart from that, the question requires that you "Determine ... the power delivered to the load resistor". If the 15W does not include internal losses, then there's nothing to "determine" here.

author: wraper
author_rank: Unknown
timestamp: March 27, 2020, 08:39:25 pm
content: Quote from: aneevuser on March 27, 2020, 08:26:26 pmFWIW in some cases the catch might be to fully understand the the thing, rather than stupidly proceeding to calculate something on harder side.

author: Zero999
author_rank: Super Contributor
timestamp: March 28, 2020, 08:25:22 am
content: Quote from: wraper on March 27, 2020, 08:39:25 pmIt says "Determine (a) the currentIand (b) the power delivered to the load resistor." so I would have probably thought it was unusual, since it already said the battery supplies W.It also says a "a 9.00V terminal voltage": is this supposed to the open or voltage with the load connected?It's very poorly worded. They should always provide schematics for these sorts of questions.

author: iMo
author_rank: Super Contributor
timestamp: March 28, 2020, 01:45:36 pm
content: Quote from: cvriv on March 27, 2020, 09:01:56 amYour equation put into the excel..

author: cvriv
author_rank: Frequent Contributor
timestamp: March 28, 2020, 01:46:09 pm
content: I can't believe this turned into such a thing. LOL. Just want to know where I was going wrong with me algebra. I didn't think the question was worded poorly at all. This is a question from a Physics book, not an engineering book. I wonder how all the other physics students are doing with this one if they even tried it. I think I might be the only EE student in my class.

author: cvriv
author_rank: Frequent Contributor
timestamp: March 28, 2020, 01:48:20 pm
content: Quote from: imo on March 28, 2020, 01:45:36 pmThanks, but im not going to lookWant to finish the math myself. I want that "feeling" I get when I finish stuff on my own.

author: iMo
author_rank: Super Contributor
timestamp: March 28, 2020, 01:54:49 pm
content: Your equation is ok, imho.Yours:Ptot = I*(Vterm + I*Rint)rewrite it:I*I*Rint + I*Vterm - Ptot = 0 (a simple quadratic eq.)P_Rint + P_Rload - P_tot = 00.1*I*I + 9*I - 15 = 0PS: Power into the load:P_Rload = I*Vterm = I*9

author: Psi
author_rank: Super Contributor
timestamp: March 28, 2020, 02:09:34 pm
content: In general, i think education questions designed on purpose to confuse people are counterproductive.If the goal is to educated people then questions should be as easy to understand as possible. With no ambiguity about what is being asked.Now.. if the course being taught is "attention to detail and problem solving skills" then that's a different matter.Note: i'm not saying the OP question is confusing on purpose.It just reminded me how much i hate confusing questions when trying to learn a subject. Especially when the question is confusing in a way that has nothing to do with the subject being taught. Like English word games in a math question.

author: cvriv
author_rank: Frequent Contributor
timestamp: March 28, 2020, 02:30:11 pm
content: Quote from: imo on March 28, 2020, 01:45:36 pmI got 1.29A. You sure you plugged your values into the quadratic equation correctly? I got 14.8W of power for whatever load is connected.

author: iMo
author_rank: Super Contributor
timestamp: March 28, 2020, 02:39:09 pm
content: Read my eq above..Quote from: cvriv on March 28, 2020, 02:30:11 pmThere is an error in your calculation.. your mess starts at line 4..

author: cvriv
author_rank: Frequent Contributor
timestamp: March 28, 2020, 03:27:15 pm
content: Quote from: imo on March 28, 2020, 02:39:09 pmI'm confused now. If P_load = I(V_term) = 11.6W, then then P_int = 15W - 11.6W = 3.39W. R_int = P_int / I(I) = 3.39W / 1.29A(1.29A) = 2.03Ohms. That's not 0.1Ohms.Whats happening here?

author: iMo
author_rank: Super Contributor
timestamp: March 28, 2020, 03:29:58 pm
content: Quote from: cvriv on March 28, 2020, 03:27:15 pmRead all my above posts.

author: cvriv
author_rank: Frequent Contributor
timestamp: March 28, 2020, 03:52:09 pm
content: I don't understand whats going on here. I know I'm doing the quadratic right. Can you please check it for me.EDIT: I think I may have swapped my terms somehow...... I have to back track.

author: iMo
author_rank: Super Contributor
timestamp: March 28, 2020, 03:57:14 pm
content: Quote from: cvriv on March 28, 2020, 03:52:09 pmYes, still the same mess

author: Wimberleytech
author_rank: Super Contributor
timestamp: March 28, 2020, 04:01:29 pm
content: Quote from: cvriv on March 28, 2020, 03:52:09 pm

author: cvriv
author_rank: Frequent Contributor
timestamp: March 29, 2020, 02:59:42 am
content: I found my mistake. I switched the square as I continued to write out my work. Thanks for your help!