Electrical question - value missing?

[PaxVobiscum]

I've been helping my son prepare for a Physics exam and have been going through past and practice papers with him. I have noticed a few mistakes (wrong answers/more than one correct answer) in the practice papers, but I wasn't quite so sure about the one below - sorry for poor quality phone photo.
I couldn't think how to do it (mind you it was 1969 when I passed my Higher Physics!) so maybe I'm missing something. It seems to be missing a value - battery voltage would be the obvious one to me. Is it actually possible to answer it? (It's only worth 1 mark out of 110). TIA for comments.

 

[Crossover]

Current, maybe? V=IxR would be applied that surely? That said, still couldn't be sure how to solve - been a while since A2 Physics!

 

[table38]

Shirley the batteries are wired back-to-back with the negatives together, so the answer is 0V ?

I suppose the sensible answer is, what would the battery voltage be if the voltmeter was reading 4/5ths of the total voltage? So if it were a 1.5V battery, the answer would be 1.2V so I'll go for "B"

Or maybe C if it was a 6V battery.

 

[Nym]

Many problems with that one...

The cells are of an undefined length, and opposing, so the implication is that there is an equal amount of opposing cells, therefore zero voltage and zero current.

You are also correct that V=IR is correct for a DC circuit.

That there are two cells, this would imply that if they where drawn the correct way round this would be a battery voltage of between 1.2 and 1.5 volts, depending on the cell technology.

If this is the case the battery voltage would be between 2.4 and 3 volts, with an overall impedance of 5 ohms, this would result in a current between 480mA and 600mA. This would give a voltage across the 4 ohm resistor of between 1.92V and 2.4V.

Given the constraints of a standard cell, and at AS level the assumption being Lead Acid technology, rounded to 1.5V.

So the potentially valid answers would be B (1 cell), C (4 cells) or E (5 cells).

In reality of taking the question at face value, none of them are correct and it should be 0V.

 

[PaxVobiscum]

Shirley the batteries are wired back-to-back with the negatives together, so the answer is 0V ?

Well spotted, I hadn't noticed that!
0V isn't one of the options though, and that's maybe a bit subtle for this level - I think it's probably just another mistake.

Many problems with that one...

The cells are of an undefined length, and opposing, so the implication is that there is an equal amount of opposing cells, therefore zero voltage and zero current.

You are also correct that V=IR is correct for a DC circuit.

That there are two cells, this would imply that if they where drawn the correct way round this would be a battery voltage

(did you mean "cell voltage" here?)

of between 1.2 and 1.5 volts, depending on the cell technology.

If this is the case the battery voltage would be between 2.4 and 3 volts, with an overall impedance of 5 ohms, this would result in a current between 480mA and 600mA. This would give a voltage across the 4 ohm resistor of between 1.92V and 2.4V.

Given the constraints of a standard cell, and at AS level the assumption being Lithium Ion or Lead Acid technology, rounded to 1.5V.

So the potentially valid answers would be B (1 cell), C (4 cells) or E (5 cells).

In reality of taking the question at face value, none of them are correct and it should be 0V.

Great answer, Nym, thanks - thorough as usual.

I just looked up the answer given at the back of the book and it was C - 4.8V so IF the battery had been drawn correctly, and given the level of this exam, I think it should have given either the battery voltage or the voltage drop across the 1Ω resistor.

 

[Pigeon]

It's a straightforward potential divider, there's no need to know or work out current.

Voltage reading on meter = battery voltage * (4 / (4 + 1)) = four fifths of the battery voltage.

As it stands, though, there is no way to calculate a numerical value. Therefore none of the multi-guess answers are valid. "4.8V" is only "correct" if we are told that the battery voltage is 6V - which we are not.

Also, as others have noted, the battery symbol is incorrectly drawn. Note that this does not mean that they have drawn two cells in inverse series giving a total of 0V. That would only be the case if the two "cell" symbols were connected by a solid line. Two "cell" symbols (which should be the same way round) connected by a dotted line is a "battery" symbol, meaning "n cells in series" where the value of n is not necessarily known; what they have drawn, with the symbols opposite ways round, is just wrong, and conveys no information other than that they have drawn it wrong.

Basically, both the question-setter and the proof-reader are incompetent, and the question as it stands should never have been presented.

 

[ac6000cw]

Basically, both the question-setter and the proof-reader are incompetent, and the question as it stands should never have been presented.

100% agree - especially as the question has two obvious errors (unless of course it's intended to try and get students to point out the mistakes ).