In a meter bridge (Figure), null point is found at a distance of 33.7 cm from A. If now a reststance

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3.Current Electricity

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In a meter bridge (Figure), the null point is found at a distance of $33.7 \;cm$ from A. If now a reststance of $12\; \Omega$ is connected in parallel with $S$, the null potnt occurs at $51.9 \;cm$. Determine the values of $R$ and $S$.

Option A

Option B

Option C

Option D

Solution

Solution From the first balance point, we get

$\frac{R}{S}=\frac{33.7}{66.3}\dots(i)$

After $S$ is connected in parallel with a resistance of $12 \,\Omega,$ the resistance across the gap changes from $S$ to $S_{e q},$ where

$S_{e q}=\frac{12 S}{S+12}$

and hence the new balance condition now gives

$\frac{51.9}{48.1}=\frac{R}{S_{e q}}=\frac{R(S+12)}{12 S}\dots(ii)$

Substituting the value of $R / S$ from Eq. $(i)$ we get

$\frac{51.9}{48.1}=\frac{S+12}{12} \cdot \frac{33.7}{66.3}$

which gives $s=13.5\, \Omega .$

Using the value of $R / S$ above.

we get $R=6.86\, \Omega$

Standard 12

Physics

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