Wbahn , you say "most" of the drawings you've seen tend to get that right, which probably means that you have seen some that don't get it right.

Not that I recall -- I simply can't rule out having seen one that didn't get it right. I particularly don't recall seeing a published problem that got it wrong, but again can't rule it out. As I've stated on a few occasions, I almost never speak in absolutes unless I can do so with confidence that it IS an absolute -- that's one of the things a career in engineering has thoroughly beaten into me. Notice that, even here, I didn't say that I never speak in absolutes unless I can do so with confidence, because I'm pretty sure I have and will continue to do so from time to time.

Also, I get the impression that you haven't seen many uses of electron current in problems for beginners.

And there you would have a poor impression. I've seen electron current used in problems for over four decades in many different venues -- ham radio, military tech school, industrial tech training, middle school science and and high school votech texts. Not to mention countless web sites and our own AAC electronic text book.

So, what is the relative occurrence of bad battery symbols versus the rate of occurrence of usage of electron current?

In my experience, very few of the former and very many of the latter.

Do we need to take those rates into account as we invoke Occam's razor?

Go right ahead. But don't forget to factor in not having the top of the battery be positive, which you pointed out as being out of place irrespective of the orientation of the battery symbol.

I don't know where the "odd" aspect occurs in this problem, but something's atypical about it.

For people used to conventional current, yes. For people used to electron current, everything looks pretty normal.

 

Apparently you missed post #17

Post #17 wasn't showing until after I submitted my response.

I've deleted the post as being OBE (overtaken by events).

 

Simulation result.

 

Simulation result.

View attachment 169901

Please stop posting solutions (especially wrong solutions) in Homework Help. The intent is to guide the TS along the path of being able to figure out the solution for themselves, not just spoon feed them answers. If you can't abide by those rules, then please don't post in this subforum.

Having said that, does your simulation result make sense?

You might consider, for example, just what it is that your ohmmeter is measuring that is 67 kΩ. Or just how it is that current is flowing left to right in the top two resistors but right to left in the bottom resistor that is in parallel with the other two.

Always, always, ALWAYS ask if your answer makes sense!

 

You might consider, for example, just what it is that your ohmmeter is measuring that is 67 kΩ. Or just how it is that current is flowing left to right in the top two resistors but right to left in the bottom resistor that is in parallel with the other two.

I, like you, wonder what this simulation shows. What does an ohmmeter show when it's connected across a resistor (or combination of resistors) that is energized by a source other than the one contained in the ohmmeter?

I wonder what resistor are you referring to when you say "bottom resistor"? I see a current source in place of the resistor which was labeled R3 in post #1.

 

I, like you, wonder what this simulation shows. What does an ohmmeter show when it's connected across a resistor (or combination of resistors) that is energized by a source other than the one contained in the ohmmeter?

I wonder what resistor are you referring to when you say "bottom resistor"? I see a current source in place of the resistor which was labeled R3 in post #1.

I'm referring to what was R3 in the problem. If this simulation shows a solution to the problem, then the current through and voltage across that current source must be the same as the current through and voltage across R3, meaning that, as far as the rest of the circuit is concerned (under these operating conditions), it is effectively R3.

 

Problem is perfectly solvable using KVL and KCL.

Looks like the professor took a previous exam question where R2 = 240Ω and replaced it with 208Ω.

 

Problem is perfectly solvable using KVL and KCL.

If you solve it assuming conventional current and assuming the top of the battery is positive, do you get a result for R3 that might be considered unusual?

 

If you solve it assuming conventional current and assuming the top of the battery is positive, do you get a result for R3 that might be considered unusual?

No. There is no solution in that case. I simply assume that either the direction of the arrow is incorrect or the creator of the drawing is using electron flow.

 

No. There is no solution in that case. I simply assume that either the direction of the arrow is incorrect or the creator of the drawing is using electron flow.

If you set up the equations ans solve them, don't you get a number which satisfies the equations?

 

If you set up the equations ans solve them, don't you get a number which satisfies the equations?

Not if you adhere to the constraint the passive resistors have positive resistance.

If you can find a bag of -52 Ω resistors, then we can revisit the point.