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Based on above figure, balanced point is point N. If the jockey is put at the left or right of point N, there will be current flowing through the galvanometer. On my note, when jockey is at point L, V

_{AL}< E (where E is potential difference of cell at lower part of the circuit). When the jockey is at point R, V

_{AR}> E.

I want to ask several questions:

1.

when the jockey is at L, there will be current flowing through galvanometer (let's call this

*P)*and the direction is clockwise (in my opinion). So the current flowing through AL will be the sum of

*P*and the current produced by upper cell X, directed counterclockwise (let's call this

*Q*). Am I right?

2.

when the jockey is at N, there is no current through galvanometer. The current flowing through AN will be equal to

*Q*. When the jockey is at R, there will be current flowing through galvanometer in counterclockwise direction with value less than

*P*and again the current passing through AR will be the sum of both current from both cells. Am I right?

3. Why:

a. at point L, V

_{AL}< E

b. at point R, V

_{AR}> E

c. at point N, V

_{AN}= E

Thanks