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Why do we achieve a zero deflection situation in the galvanometer when we slide the jockey on the potentiometer wire ?

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Or what is happening in the secondary circuit that causes zero deflection in the galvanometer ?

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  • $\begingroup$ Let us say point A at zero potential, point B will always be negative. If same as E of the battery, these is no current. $\endgroup$
    – user137289
    Jan 3, 2017 at 11:37

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We obtain a null point when the potential drop across the potentiometer wire (we can vary the potential drop across the wire by sliding the jockey )is the same as that across the cell in the circuit. In your case, I believe, it's assumed that the galvanometer offers no resistance. When the potential drop across the wire equals that across the cell in the secondary circuit, the potential difference between the two would be zero. Ohm's law tells us that the current through a conductor is directly proportional to the potential difference across it. Now if there is no potential difference there will be no flow of current. Hence we obtain a null deflection in the galvanometer.

Note: Since there is no deflection in the galvanometer, there's no current in the secondary circuit (Galvanometer detects the presence of current in a circuit). Since there is no current in the secondary circuit, what we're measuring is the EMF of the cell connected in the secondary circuit.

Hope that helps!

E2 is the cell in the secondary circuit while E1 is the cell in the driving circuitE2 is the cell in the secondary circuit while E1 is the cell in the driving circuit

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When you slide the jockey in the potentiometer wire, you add or eliminate more resistance from/to the external circuit. This changes the potential drop across the points A and B. When the condition occurs that the potential across the points A and B is equal to $E$, the emf of the battery connected to the galvanometer, then the two emfs oppose each other and there is no current.

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  • $\begingroup$ Can the deflection in the galvanometer be zero if we only only resistances in place of the battery of the emf E ? I think we always need Batteries to cancel out the current due to the source battery. Right ? $\endgroup$
    – Mitchell
    Jan 3, 2017 at 12:14
  • $\begingroup$ No, how would that happen? $\endgroup$ Jan 3, 2017 at 12:15
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Suppose the current in the galvanometer wire is 0 for some position of the jockey.

Applying Kirchoff's voltage rule in the lower loop (secondary circuit), $V_1 = IR_1$

For the upper loop (primary circuit), $V_2 = I(R_1 + R_2)$

Dividing them,

$\dfrac{V_1}{V_2} = \dfrac{R_1}{R_1 + R_2} = \dfrac{l_1}{l}$

(As the resistance of a wire of uniform thickness is proportional to its length)

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