Two primary cells of emfs wire $\varepsilon_1$ and $\varepsilon_2(\varepsilon_1>\varepsilon_2)$ are connected to a potentiometer wire AB as shown in fig. If the balancing lengths for the two combinations of the cells are 250cm and 400cm, find the ratio of $\varepsilon_1$ and $\varepsilon_2.$
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In first combination $\varepsilon_1$ and $\varepsilon_2$ are opposing each other while in second combination $\varepsilon_1$ and $\varepsilon_2$ are adding each other,
So,
$\varepsilon_1-\varepsilon_2=\text{Kl}_1$
$\varepsilon_1+\varepsilon_2=\text{kl}_2$
$\frac{\varepsilon_1-\varepsilon_2}{\varepsilon_1+\varepsilon_2}=\frac{\text{l}_1}{\text{l}_1}$
$\Rightarrow\frac{\varepsilon_1-\varepsilon_2}{\varepsilon_1+\varepsilon_2}=\frac{250}{400}$
$\Rightarrow\frac{\varepsilon_1-\varepsilon_2}{\varepsilon_1+\varepsilon_2}=\frac{5}{8}$
$\Rightarrow8\varepsilon_1-8\varepsilon_2=5\varepsilon_1+5\varepsilon_2$
$\Rightarrow3\varepsilon_1=13\varepsilon_2$
$\therefore\frac{\varepsilon_1}{\varepsilon_2}=\frac{13}{3}$
$\therefore\varepsilon_1:\varepsilon_2=13:32$
So,
$\varepsilon_1-\varepsilon_2=\text{Kl}_1$
$\varepsilon_1+\varepsilon_2=\text{kl}_2$
$\frac{\varepsilon_1-\varepsilon_2}{\varepsilon_1+\varepsilon_2}=\frac{\text{l}_1}{\text{l}_1}$
$\Rightarrow\frac{\varepsilon_1-\varepsilon_2}{\varepsilon_1+\varepsilon_2}=\frac{250}{400}$
$\Rightarrow\frac{\varepsilon_1-\varepsilon_2}{\varepsilon_1+\varepsilon_2}=\frac{5}{8}$
$\Rightarrow8\varepsilon_1-8\varepsilon_2=5\varepsilon_1+5\varepsilon_2$
$\Rightarrow3\varepsilon_1=13\varepsilon_2$
$\therefore\frac{\varepsilon_1}{\varepsilon_2}=\frac{13}{3}$
$\therefore\varepsilon_1:\varepsilon_2=13:32$
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