The circuit shown here is used to compare the emf of two cells $E_{1}\ $and$E_{2}(E_{1} E_{2})$. The null point is at $C$ when the galvanometer is connected to $E_{2}$. when the galvanometer is connected to $E_{2}$, the null point will be [image:3_image51.emf]

Question

The circuit shown here is used to compare the emf of two cells $E_{1}\ $and$E_{2}(E_{1} > E_{2})$. The null point is at $C$ when the galvanometer is connected to $E_{2}$. when the galvanometer is connected to $E_{2}$, the null point will be [image:3_image51.emf]

Dr. Abubacker · Accepted Answer

(a) $E_{1}E_{2} = l_{1}/l_{2}$. As, $E_{1}/E_{2}$, therefore $l_{1} > l_{2}$. Therefore the null point for thr cell of emf $E_{2}$must be at shorter length than that of cell $E_{1}.$Thus null point on potentiometer wire should shift towards left of $C$.

Q. The circuit shown here is used to compare the emf of two cells $E_{1}\ $and$E_{2}(E_{1} > E_{2})$. The null point is at $C$ when the galvanometer is connected to $E_{2}$. when the galvanometer is connected to $E_{2}$, the null point will be

Explanation

Reviewed by Dr. Abubacker

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