What is end error in a metre bridge? How is it overcome? The resistances in the two arms of the metre bridge are $\text{R}=5\Omega$ and $S$ respectively. When the resistance $S$ is shunted with an equal resistance, the new balance length found to be $1.5 l_1 ,$ where $l_1$ is the initial balancing length. Calculate the value of $S.$
CBSE 55-1-1 PAPER SET 2019
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The shifting of zero of the scale at different points as well as the stray resistance gives rise to the end error in meter bridge wire. This error arises due to the non$-$uniformity of the meter wire End corrections can be estimated by including known resistances $P_1$ and $Q_1$ in the two ends and finding the null point We have
$\text{R}=5\Omega$ According to the wheat stone Bridge principle: $\frac{\text{R}}{\text{l}_1}=\frac{\text{S}}{100-\text{l}_1}$
$\frac{5}{\text{l}_1}=\frac{\text{S}}{100-\text{l}_1}$ equation $(1)$ After shunting means we are connecting resistance in parallel $\text{S}\rightarrow\frac{\text{S}}{2}$
$\frac{5}{1.5\text{l}_1}=\frac{\text{S}}{2(100-1.5\text{l}_1)}$ equation $(2)$ Equation $(1)$ can be written as: $500 - 5l_1 = Sl_1$ equation $(3)$ And, equation $(2)$ can be written as $10(100 - 1.5l_1) = 1.5Sl_1$ equation $(4)$ From equation $(3)$ and $(4) \frac{500-5\text{l}_1}{\text{l}_1}=\frac{1000-15\text{l}_1}{1.5\text{l}_1}$
$750-7.5\text{l}_1=1000-15\text{l}_1$
$ -250=-7.5\text{l}_1$
$\text{l}_1=\frac{100}{3}$
$\text{S}=\frac{500-\frac{5\times100}{3}}{\frac{100}{3}}=\frac{500-\frac{500}{3}}{\frac{100}{3}}=\frac{1000}{3}\times\frac{3}{100}$
$\text{S}=10\Omega$
$\text{R}=5\Omega$ According to the wheat stone Bridge principle: $\frac{\text{R}}{\text{l}_1}=\frac{\text{S}}{100-\text{l}_1}$
$\frac{5}{\text{l}_1}=\frac{\text{S}}{100-\text{l}_1}$ equation $(1)$ After shunting means we are connecting resistance in parallel $\text{S}\rightarrow\frac{\text{S}}{2}$
$\frac{5}{1.5\text{l}_1}=\frac{\text{S}}{2(100-1.5\text{l}_1)}$ equation $(2)$ Equation $(1)$ can be written as: $500 - 5l_1 = Sl_1$ equation $(3)$ And, equation $(2)$ can be written as $10(100 - 1.5l_1) = 1.5Sl_1$ equation $(4)$ From equation $(3)$ and $(4) \frac{500-5\text{l}_1}{\text{l}_1}=\frac{1000-15\text{l}_1}{1.5\text{l}_1}$
$750-7.5\text{l}_1=1000-15\text{l}_1$
$ -250=-7.5\text{l}_1$
$\text{l}_1=\frac{100}{3}$
$\text{S}=\frac{500-\frac{5\times100}{3}}{\frac{100}{3}}=\frac{500-\frac{500}{3}}{\frac{100}{3}}=\frac{1000}{3}\times\frac{3}{100}$
$\text{S}=10\Omega$
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