# A trivial circuit analysis question

1. Dec 30, 2013

### CDTOE

1. The problem statement, all variables and given/known data

I had the question above in an exam. Although it's very trivial to analyze, I'm wondering what's wrong with part (e) of the question. It could be that I don't understand the hidden meaning behind this question, but I see it as obvious as that the current through RL will not be reduced, contrary to what the part (e) wants to prove, because it's independent of any resistance we add in parallel.

I would like my solution of part (e), in particular, to be looked at and be seen if there's anything wrong. the 'V' component is actually a DC supply not a capacitor as it looks like.

2. Relevant equations

none.

3. The attempt at a solution

2. Dec 30, 2013

### phinds

Makes no sense. If "RL" is indeed a pure resistor, then you cannot possibly change the current through it if you don't change the voltage across it. Ohm would be offended.

3. Dec 30, 2013

### Staff: Mentor

The question is nonsense. The current through a resistor is I = V/R, always. If V is to remain unchanged as stated then I must be unchanged, too. Further, the symbol employed for the voltage source is that of a capacitor, not a battery. Methinks something is rather fishy here...

4. Dec 30, 2013

### CDTOE

The symbol is for a capacitor, but it's just a misprint from the instructor's side who also emphasized that it's a DC voltage source and the load is pure resistive. I actually didn't pay attention to the wording of the question during the exam, so I didn't figure out what's wrong with it until I was driving my car back to home.

Makes no sense indeed!

5. Dec 30, 2013

### Staff: Mentor

I would say it's intended that RL represents the load, i.e., the resistance between the battery posts. So adding extra series resistance to make RL greater will achieve what the examiner is asking for.

There was probably a prac class where this exact phenomenon was demonstrated some days before you were given this test paper?

6. Dec 31, 2013

### CDTOE

If I add a resistor in series with the load, I would still be asked to have a constant voltage around RL only, which wouldn't be achievable in this case because of the voltage drop on the added resistor. The questions explicitly asks to prove that the voltage on RL hasn't changed after modification.

If the intent is to measure the overall voltage between the leads of RL and the added resistor (R), then yes, we will have a constant voltage with reduced current, but this wouldn't be RL anymore; it would be RL + R which is not asked for.

Let's not go big in a test of words and symbols and suggest that his intent is that the series resistor and the load will merge into one big RL different from RL given in the schematic. This could be his excuse, I'm afraid, after he finds his mistake in writing a trivial question.

:uhh:

Last edited: Dec 31, 2013
7. Dec 31, 2013

### CWatters

Do post his solution :-)

8. Dec 31, 2013

### CDTOE

He doesn't give away a model answer sheet. This question came in the final exam, which is very easy overall, for a course that's titled Electrical Engineering Technologies. It covers a wide range of topics on electrical and electronics measurement systems, devices, components, conductors/insulators, counters, sensors, transducers, uncertainty analysis, temperature measurements, electrical safety and so on. It's supposed to be an introductory course after you take the first course in electrical circuits analysis techniques.

The instructor has PhD in EE with him being specialized in some field in medical instruments engineering/medical imaging. I would insist that he has forgotten some fundamental techniques and knowledge in circuits and electronic components, both in analog and digital domain. This insistence, from my side, is based on a complete semester of lecturing and we barely completed half of the topics that we were supposed to cover. However, he's at ease when writing exams as they're very easy and barely need any preparation. Even though, this kind of mistake is unforgivable.

Last edited: Dec 31, 2013