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(B) Let the potential gradient of the potentiometer wire be $k$.Let the $EMF$ of the two cells be $E_1$ and $E_2$.When the cells assist each other,the

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$3:1$

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(B) Let the potential gradient of the potentiometer wire be $k$.Let the $EMF$ of the two cells be $E_1$ and $E_2$.When the cells assist each other,the resultant $EMF$ is $(E_1 + E_2)$.According to the principle of the potentiometer,$(E_1 + E_2) = k 	imes 120 \ cm$  ---$(i)$When the cells oppose each other,the resultant $EMF$ is $(E_1 - E_2)$.According to the principle of the potentiometer,$(E_1 - E_2) = k 	imes 60 \ cm$  ---(ii)Dividing equation $(i)$ by equation (ii):$\frac{E_1 + E_2}{E_1 - E_2} = \frac{120}{60} = \frac{2}{1}$Applying componendo and dividendo or cross-multiplying:$E_1 + E_2 = 2(E_1 - E_2)$$E_1 + E_2 = 2E_1 - 2E_2$$3E_2 = E_1$Therefore,the ratio $\frac{E_1}{E_2} = \frac{3}{1}$.

Similar Questions

In the potentiometer circuit as shown in the figure,the balance length $l = 60 \ cm$ when switch $S$ is open. When switch $S$ is closed and the value of $R$ is $5 \ \Omega$,the balance length $l' = 50 \ cm$. The internal resistance of the cell $C'$ is : .............. $\Omega$

$A$ potentiometer is used to measure the potential difference between $A$ and $B$. The null point is obtained at $0.9 \ m$. Now,the potential difference between $A$ and $C$ is measured,and the null point is obtained at $0.3 \ m$. The ratio $\frac{E_2}{E_1}$ is $\left(E_1 > E_2\right)$.

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$A$ potentiometer is an ideal device for measuring potential difference because

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