Cannot calculate load current in this simple diagram!

I am being fool. Various doubtful answers made me more fool. Please explain the voltage and current (load). May be it needs few theorems to explain

That circuit will blow up since there are two voltage sources in parallel. The higher voltage source will generate an infinite current through the lower voltage source (in theory).

The circuit as drawn is ridiculous.
Unless the voltage sources have some internal resistance which is not shown in the drawing, the set-up is impossible in both theory and practice.

JimB

Its one of those questions we had to do in college... 15 v across the resistor....

Ian Rogers said:

Its one of those questions we had to do in college... 15 v across the resistor....

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Did you graduate?

LOL.... The questions are in a stupid format... You must have seen them... The voltage sources are just literal they could be coming from anywhere... The point is, some lecturers start making these questions up!!! There will be one or two examples in a text book, but as the students have worked through them, the lecturers start to make their own up... And sometimes ...Bites them in the bum!!!

Did you graduate?

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Oh!! Sorry my post didn't come across as I intended....

Ian Rogers said:

There will be one or two examples in a text book, but as the students have worked through them, the lecturers start to make their own up... And sometimes ...Bites them in the bum!!!

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Presumably those lecturers knew nothing about electronics?, as people setting and marking exams often don't.

Going back to my RAE (Radio Amateurs Exam) days there were a number of examples from previous years that were utter nonsense, and you essentially had to 'guess' what they wanted the result to be - even though it made no sense.

But slightly more recently, though still a good while back, I used to take part in the "Sharp Electronics Engineer of the Year" - I won it twice (and still held the title when the competition was stopped). But in the final round one year, which were held at posh country hotels, we all came out of the exam room moaning like mad about one of the questions - it was multiple choice, and none of the answers were anywhere near right, no matter how you looked at it (it was also yet another completely nonsense question).

So we all complained bitterly to the Sharp staff present (whom we all knew well), and they all pointed at one of their members and said "he set the questions!!!" - so we all descended on him (Jim Stone if I remember correctly?) and gave him some pain about it, his excuse was that he had copied all the questions from existing sources, and hadn't bothered checking them.

But it's nice when you can get face to face with someone responsible!!

The solution and correct response would be:



Wasn't that simple?

While I have memories of such nonsense circuits in school in over 40 years of practical applications I never saw one. Go figure?

Ron

OK, then which/what type of 'answer' a student have to write for this type of 'nonsense' question in exam? Should they complain instead or...?

Nepaliman said:

OK, then which/what type of 'answer' a student have to write for this type of 'nonsense' question in exam? Should they complain instead or...?

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You should complain, and write on the paper that it's a nonsense question that has no possible answer.

I think it's more appropriate to state the issues with the circuit, i.e. what is wrong with it and why that is a problem, as some of the other posters have done above. This shows that you have an understanding of the circuit. The question doesn't require a complaint; just give an appropriate answer.

Nepaliman said:

OK, then which/what type of 'answer' a student have to write for this type of 'nonsense' question in exam? Should they complain instead or...?

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The student should politely complain that the circuit as drawn is incorrect and couldn't function. That or use my answer above. You don't want to bruise the instructors / professors ego.

Ron

Having seen many of this particular "unanswerable" question over the years, for some reason while looking at this one, it occurred to me that a somewhat defensible answer could be obtained like this:

Assume that the 15V and 10V batteries are each made up of identical 1V (for example) cells, each having an internal resistance of R ohms. We then have a 10 volt battery with 10R ohms internal resistance, in parallel with a 15 volt battery with 15R ohms internal resistance. Convert the 10V battery to a current source of 10/10R = 1/R amps, and the 15 volt battery to a current source of 15/15R = 1/R amps. Now we have 2 current sources of 1/R amps each in parallel, giving a single source of 2/R amps in parallel with 15R || 10R = 6R ohms (that's 15R ohms in parallel with 10R ohms). Converting back to a Thevenin voltage source, we have an equivalent battery of (2/R) * 6R = 12 volts with an internal resistance of 6R ohms.

The 5 ohm load will give a load current of 12/(5 + 6R) amps. As the R ohm internal resistance of the postulated 1 volt cells making up the batteries approaches zero, the load current will be 12/5 amps, with an increasingly large circulating current flowing around the loop formed by the two batteries.

There is only a fragment of the actual question included in the picture in post 1.

As such, we don't know if there was any additional information as part of the problem.

For instance, there may have been some statement like:
Assume that all voltage sources have an internal resistance of 1 ohm.

ChrisP58 said:

There is only a fragment of the actual question included in the picture in post 1.

As such, we don't know if there was any additional information as part of the problem.

For instance, there may have been some statement like:
Assume that all voltage sources have an internal resistance of 1 ohm.

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What is this? - make up possible excuses and rewrite the question?

We've all seen such stupid questions in the past, and will continue to do so in the future - it's almost entirely certain that there's no extra information that would provide a feeble excuse for the useless question.