Voltage in a simple circuit (MITx 6.002, Circuits and Electronics)

[zenterix]

Homework Statement

Joe was debugging part of an experimental apparatus, probing around with his voltmeter. Part of the apparatus had two obvious resistors in series with an unknown element, as shown in the diagram below:

Relevant Equations

The unknown element is hard to reach, so Joe put the negative (black) probe of his voltmeter at the interconnection of the two obvious resistors and then put the positive (red) probe at the other end of each resistor, measuring $v_1=1.4V$ and $V_2=0.89999999V$.

What is the voltage $v_3$ measured across the unknown element?

It seems that in this example Joe measured the difference $V_{1,+}-V_{1,-}=1.4V$ and then $V_{2,+}-V_{2,-}=0.8999999V$.

Shouldn't one of these be negative? That is if current is flowing clockwise, then shouldn't we have
$V_{2,+}-V_{2,-}=-0.8999999V$? And if the current is counterclockwise, then shouldn't we have $V_{1,+}-V_{1,-}=-1.4V$?

The answer to the problem is apparently $0.5V$. I assume this comes from $1.4V-0.8999999V$.

The answer I originally came up with was $-2.3V$, since I assumed that the voltage across the resistors were both of same sign, so I added them.

I think I have a good grasp of all the concepts here, but either 1) I don't or 2) the problem is somehow stated in a way that is ambiguous or incorrect.

I'm looking for someone to tell me which one it is.

 

[berkeman]

How could both of those voltages measured with the DVM be positive? That makes no physical sense, IMO.

 

[zenterix]

How could both of those voltages measured with the DVM be positive? That makes no physical sense, IMO.

So would you say the problem statement is incorrect?

 

[kuruman]

So would you say the problem statement is incorrect?

If the two resistors are obviously in series and there is a measured voltage across each one of them, the current in each resistor is the same and flowing in the same direction. This means that the point between the two resistors must be at an intermediate potential between the ends.

I would say that the problem statement is not self-consistent. Something has to give.

 

[berkeman]

So would you say the problem statement is incorrect?

As you wrote it, yes, I think it's non-physical (violates the continuity equation for charge flow). Are you sure you copied it word-for-word? Can you upload a screenshot of the question? (use the "Attach files" link below the Edit window)

 

[zenterix]

As you wrote it, yes, I think it's non-physical (violates the continuity equation for charge flow). Are you sure you copied it word-for-word? Can you upload a screenshot of the question? (use the "Attach files" link below the Edit window)

 

[berkeman]

Yeah, that's a problematic question. Two red flags: 1) The guy with the DVM is named "Joe", and 2) The problem is worth zero points. Watch out for those red flags in your future schoolwork...

 

[kuruman]

Additional red flag: $v_2=0.8999999999999999 ~$V. It seems that Joe's voltmeter is very precise but, as far as accuracy is concerned, it sucks . One thing I don't understand is why 0.5 V is given as the correct answer.

 

[Steve4Physics]

Additional red flag: $v_2=0.8999999999999999 ~$V. It seems that Joe's voltmeter is very precise

And very wide!

A poor question, as already noted.

Possibly the intention of the question is this:

The upper and lower resistors ($R_1$ and $R_2$) and the unknown component ($X$) are the only components in the loop.

However, although the question says these components are in ‘in series’, I’m guessing this is a silly mistake.

The ‘loop’ shown could be part of a larger circuit which isn’t shown. Arguably, this is implied in the Homework Statement by the phrase 'part of the apparatus'. For example there could be connections to other parts of the apparatus between $R_1$ and $R_2$, between $R_1$ and $X$ and between $R_2$ and $X$. So, the 3 components don’t necessarily carry the same current.

On this basis, the question is trivial - which maybe is why it carries no marks

For convenience, take the junction between $R_1$ and $R_2$ as 0V.

The potential at the top of $R_1$, and hence at the top of $X$ is $1.4V$.

The potential at the bottom of $R_2$, and hence at the bottom of $X$ is $0.9V$.

The p.d. (ignoring any sign issues) across $X$ is therefore $V_3 = 1.4V - 0.9V = 0.5V$ as required.

 

[berkeman]

And very wide!

(Joe was using it in Landscape mode...)

 

[DaveE]

View attachment 328954

Not the first poorly written HW problem. It surely won't be the last.

I would interpret it as you did where $v_1=1.4v$ should be $|v_1|=1.4v$, etc.

Also, kind of nitpicking, $0.99999 \neq 1$, but $0.99999... = 1$.