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Standard XII

Physics

EMF and EMF Devices

In the given ...

Question

In the given circuit of potentiometer, the potential difference $E$ across $A B$ $(10 m$ length $)$ is larger than $E_{1}$ and $E_{2}$ as well. For key $K_{1}$ $($ closed $)$ , the jockey is adjusted to touch the wire at point $J_{1}$ , so, that there is no deflection in the galvanometer. Now, the first battery $(E_{1})$ is replaced by the second battery $(E_{2})$ for working by making $K_{1}$ open and $K_{2}$ closed. The galvanometer then gives null deflection at $J_{2}$ . The value of $\frac{E_{1}}{E_{2}}$ is $\frac{a}{b}$ , where, $a =$ _____.

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Solution

We know that,

$\frac{E_{1}}{E_{2}} = \frac{l_{1}}{l_{2}}$

$\Rightarrow \frac{E_{1}}{E_{2}} = \frac{3 \times 100 + (100 - 20)}{7 \times 100 + 60} = \frac{1}{2} = \frac{a}{b}$

$\Rightarrow a = 1$

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We know that, E1E2=l1l2 ⇒E1E2=3×100+(100−20)7×100+60=12=ab ⇒a=1

We know that,

$\frac{E_{1}}{E_{2}} = \frac{l_{1}}{l_{2}}$

$\Rightarrow \frac{E_{1}}{E_{2}} = \frac{3 \times 100 + (100 - 20)}{7 \times 100 + 60} = \frac{1}{2} = \frac{a}{b}$

$\Rightarrow a = 1$