Calculating resistances via a potentiometer

[Shivang kohlii]

Homework Statement

Figure shows a potentiometer circuit for comparison of two resistances , the balance point with standard resistor R = 10 ohm , is found to be 58.3 , while that with unknown resistance X is 68.5 cm , determine the value of X .
b) what might you do if you failed to find the balance point with the given cell end e?

Homework Equations

R/X = L1/L2
K = potential gradient across AB
I1 = current when R is connected
I2 = current when X is connected
3.attempt at a solution
I understand that the potential difference across AP = EF
I.e. k×58.3= I1 × R , where I is the current when R is connected to circuit
And I1×R = E - (2)
Also when X is connected and R is removed ,
k× 68.5 = I2 × X
Also I2 × X = E ... - (1)
But then that means that from 1 and 2 , k × 58.3 = k× 68.5 ? That's obviously wrong!
I don't get it ... I guess somehow I1 × R = E1 and I2× X = E2 with E1 is not equal to E2... If I do that answer comes rightly ...
But shouldn't E1 = E2 as ,
potential difference across EF = potential difference across LM = E always ??

Thank you!

 

[BvU]

from 1 and 2 , k × 58.3 = k× 68.5 ? That's obviously wrong!

Pity you use E in the picture and in the equations -- with different meaning.
Only possible conclusion, my dear Watson, is that E in case of I₁ is not equal to the E in case I₂. In other words: $\varepsilon$ doesn't come from an ideal voltage source.

But shouldn't E1 = E2 as ,
potential difference across EF = potential difference across LM = E always ??

Apparently not !

 

[Shivang kohlii]

Pity you use E in the picture and in the equations -- with different meaning.
Only possible conclusion, my dear Watson, is that E in case of I₁ is not equal to the E in case I₂. In other words: $\varepsilon$ doesn't come from an ideal voltage source.
Apparently not !

I didn't ask for your pity mister , only solution.. also don't presume I didn't consider the non ideal voltage source with some internal resistance , I didn't add it to my answer as the question doesn't mention any internal resistance .. which is odd for the 12th level book!

 

[BvU]

I didn't ask for your pity mister

I didn't offer it, on the contrary: it was an ironic reproach, intended to gently steer you in a direction that avoids future errors caused by this kind of sloppiness.

What I offered was the only explanation I could come up with. I have seen this circuit before in PF, you're welcome to search for it.
12th level sounds like a computer game. Guess it's an age indication in your culture, like around 18 ?

 

[DaveE]

Sorry, you have to be using that textbook. The schematic given is non-standard (according to what I'm used to, anyway) and vague when combined with the problem statement. For example, what currents are being measured? "unknown resistance X is 68.5 cm" requires further explanation, I don't know what this means.
Problem solutions are frequently an exercise in describing the question carefully so that the answer becomes more readily apparent. I would suggest redrawing/rephrasing this more carefully either for your own understanding, or for ours. You will get better assistance that way.