A potentiometer wire is 100 , cm long and a c | vedclass

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(C) Let the emfs of the two cells be $\varepsilon_{1}$ and $\varepsilon_{2}$ (where $\varepsilon_{1} > \varepsilon_{2}$).Let $k$ be the potential grad

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$3:2$

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(C) Let the emfs of the two cells be $\varepsilon_{1}$ and $\varepsilon_{2}$ (where $\varepsilon_{1} > \varepsilon_{2}$).Let $k$ be the potential gradient (potential difference per unit length) of the potentiometer wire.When the cells are connected in series to support each other,the total emf is $\varepsilon_{1} + \varepsilon_{2}$. The balance point is at $l_{1} = 50 \, cm$.So,$\varepsilon_{1} + \varepsilon_{2} = k \cdot l_{1} = 50k$  .....$(i)$When the cells are connected in opposite directions,the total emf is $\varepsilon_{1} - \varepsilon_{2}$. The balance point is at $l_{2} = 10 \, cm$.So,$\varepsilon_{1} - \varepsilon_{2} = k \cdot l_{2} = 10k$  .....$(ii)$Adding equations $(i)$ and $(ii)$:$2\varepsilon_{1} = 60k \Rightarrow \varepsilon_{1} = 30k$Subtracting equation $(ii)$ from $(i)$:$2\varepsilon_{2} = 40k \Rightarrow \varepsilon_{2} = 20k$The ratio of the emfs is $\frac{\varepsilon_{1}}{\varepsilon_{2}} = \frac{30k}{20k} = \frac{3}{2}$.

Similar Questions

In a potentiometer arrangement,a cell of emf $1.5 \ V$ gives a balance point at $150 \ cm$ length of the wire. If the cell is replaced by another cell and the balance point shifts to $210 \ cm$,what is the emf of the second cell (in $V$)?

$A$ cell is connected to a potentiometer,and the balancing length is obtained at $125 \ cm$. When a resistor of $2 \ \Omega$ is connected in parallel with the cell,the balancing length is obtained at $100 \ cm$. What is the internal resistance of the cell in $\Omega$?

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