songoku said:
If now the point S is moved to the right towards Q, is it still correct to say V at P = V at A? If yes, then it means no potential difference between P and A so why current can flow from P to A?
Maybe it is somwhat confusing that in the circuit diagram PA looks the same as PSQ. They are intended to be different:
PA is supposed to be an ideal conductor (zero resistance, potential the same at both ends and nevertheless a current can flow, the magnitude of which is determined elsewhere. Remember V = IR).
PQ is supposed to be a wire
with resistance, so that there
is a voltage drop.
If The wire length is L
PQ and the resistance is R
PQ, then
RPS / RPQ = LPS / LPQ.
And if the galvanometer shows zero:
VPS / VPQ = LPS / LPQ
the voltage drop is proportional to the length L. In a picture:
So the idea is that if ##\ L_{PS} = 40 ## cm and ##\ L_{PQ}\ = 100 ## cm (are you sure that wasn't in the exercise statement ?) that -- if the galvanometer shows zero -- then ##V_{PS} = 40/100 * 2.00 ## Volt.
This exposes a weakness in the exercise: nothing is said about R. If the idea is to determine the voltage that the solar cell produces by making use of the $$V_{PS} = V_{PQ} * {L_{PS}\over L_{PQ} } $$ and ##\ V_{PS} = V_{AB}\ ## when the galvanometer shows zero, then still ##\ V_{PQ}\ ## has to be known.
Either through a measurement or by allowing the approximation ##\ V_{PQ}\ \approx 2.000 ## Volt. Such an approximation can only be made if ##\ R << R_{PQ}\ ##.
The composer of the exercise should have left out R altogether: its function is current limitation, but the 2.000 V and the 40.0 suggest an accuracy in the milliVolt range, so the voltage dop over R should be less than a mV. If e.g. ##\ R_{PQ} = 10 \Omega\ ## that means ##R < 0.01 \Omega\ ## and therefore useless as current limiter.
I assume B has zero potential (grounded)
You can certainly do that. But it serves little purpose: everything here is about voltage differences. It may even be confusing, because it's not a constant reference point if S moves.
and let say the solar cell has emf of 0.5 V. Is it correct to say potential at A is 0.5 V and at B is zero? If yes, why no current flows even though there is potential difference? Or the thing I write is completely different matter compared to the picture?
Yes: ##\ V_A = 0.5 ## V and B was defined to be zero.
No current flows in AB because ##\ V_{PS} ## is also 0.5 V.
Sorry I don't understand this part. You mean when the potentiometer is not balanced, the current flowing through solar cell will be the same as the current flowing through PS?
No.
I resume:
BvU said:
3) PS potential difference will be greater than AB potential difference, so current will flow from P to A and from B to S.
songoku said:
And I think PS and AB are parallel and in parallel connection the potential difference will be the same but VPS ≠ VAB, what am I missing here?
Ergo they are not parallel ! There is resistance PS between the ends on one side and the solar cell between the ends on the other.
They carry the same current, so at best they are in series.
I think I made a mess of that, sorry
. Strike the last sentences -- must have been the lack of coffee
But then "3) PS potential difference will be greater than AB potential difference, so current will flow from P to A and from B to S." does not make much sense either !
Let me try again -- altough it looks much the same ...
If S is moved away to the right from the balanced point , ## \ V_{PS} \ ## will be greater than the solar cell voltage and current will flow from P to S through the cell, so from A to B.
One thing sure to happen is that the current trough AB will increase the voltage drop over RS, thereby reducing ## \ V_{PS} \ ##
In addition, the voltage drop ## \ V_{AB} \ ## may increase, dependent on the characteristics of the solar cell.
Alone or together these reactions will ensure ##\ V_{PS} = V_{AB}\ ## as becomes a decent parallel circuit
and you pointed out correctly
!I hope this time someone corrects me if I am still mistaking !
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