The potentiometer wire AB is 600 , cm long. A | vedclass

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(B) The potential difference across the potentiometer wire $AB$ is given by the voltage divider rule: $V_{AB} = E 	imes \frac{R_{AB}}{R_{AB} + r} = E

Vedclass · Accepted Answer

$320$

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(B) The potential difference across the potentiometer wire $AB$ is given by the voltage divider rule: $V_{AB} = E 	imes \frac{R_{AB}}{R_{AB} + r} = E 	imes \frac{15r}{15r + r} = E 	imes \frac{15r}{16r} = \frac{15}{16}E$.The potential gradient $k$ along the wire is $k = \frac{V_{AB}}{L} = \frac{15E}{16 	imes 600} = \frac{E}{640} \, V/cm$.For zero deflection in the galvanometer,the potential drop across the length $\ell_1$ must equal the $EMF$ of the secondary cell,which is $E/2$.Therefore,$k 	imes \ell_1 = \frac{E}{2}$.Substituting the value of $k$: $\frac{E}{640} 	imes \ell_1 = \frac{E}{2}$.Solving for $\ell_1$: $\ell_1 = \frac{640}{2} = 320 \, cm$.

The potentiometer wire $AB$ is $600 \, cm$ long. At what distance from $A$ should the jockey $J$ touch the wire to get zero deflection in the galvanometer?

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