Figure shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor $\text{R}=10.0\ \Omega$ is found to be $58.3 \ cm,$ while that with the unknown resistance $X$ is $68.5 \ cm$. Determine the value of $X$. What might you do if you failed to find a balance point with the given cell of emf $\varepsilon?$
Exercise
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Resistance of the standard resistor, $\text{R}=10.0\ \Omega$
Balance point for this resistance $, l_1 = 58.3 \ cm$
Current in the potentiometer wire $= i$
Hence, potential drop across $R, E_1 = iR$
Resistance of the unknown resistor $= X$
Hence, potential drop across $X, E_2 = iX$
The relation connecting emf and balance point is,
$\frac{\text{E}_1}{\text{E}_2}=\frac{l_1}{l_2}$
$\frac{\text{iR}}{\text{iX}}=\frac{l_1}{l_2}$
$\text{X}=\frac{l_1}{l_2}\times\text{R}$
$=\frac{68.5}{58.3}\times10=11.749\ \Omega$
Therefore, the value of the unknown resistance, $\text{X, is}\ 11.75\ \Omega.$
If we fail to find a balance point with the given cell of emf $, \varepsilon ,$ then the potential drop across $R$ and $X$ must be reduced by putting a resistance in series with it.
Only if the potential drop across $R$ or $X$ is smaller than the potential drop across the potentiometer wire $AB,$ a balance point is obtained.
Balance point for this resistance $, l_1 = 58.3 \ cm$
Current in the potentiometer wire $= i$
Hence, potential drop across $R, E_1 = iR$
Resistance of the unknown resistor $= X$
Hence, potential drop across $X, E_2 = iX$
The relation connecting emf and balance point is,
$\frac{\text{E}_1}{\text{E}_2}=\frac{l_1}{l_2}$
$\frac{\text{iR}}{\text{iX}}=\frac{l_1}{l_2}$
$\text{X}=\frac{l_1}{l_2}\times\text{R}$
$=\frac{68.5}{58.3}\times10=11.749\ \Omega$
Therefore, the value of the unknown resistance, $\text{X, is}\ 11.75\ \Omega.$
If we fail to find a balance point with the given cell of emf $, \varepsilon ,$ then the potential drop across $R$ and $X$ must be reduced by putting a resistance in series with it.
Only if the potential drop across $R$ or $X$ is smaller than the potential drop across the potentiometer wire $AB,$ a balance point is obtained.
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