A $50\,\Omega $ resistance is connected to a battery of $5\,V$. A galvanometer of resistance $100\, \Omega $ is to be used as an ammeter to measure current through the resistance, for this a resistance $r_s$ is connected to the galvanometer. Which of the following connections should be employed if the measured current is within $1\% $ of the current without the ammeter in the circuit ?
JEE MAIN 2016, Diffcult
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As we know, $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{5}{50}=0.1$
$I' = 0.099$
When Galvanometer is connected
$R_{e q}=50+\frac{100 S}{100+S}=\frac{V}{I}$
$\Rightarrow \frac{100 \mathrm{S}}{100+\mathrm{S}}=\frac{5}{0.099}-50$
$\Rightarrow \frac{100 \mathrm{S}}{100+\mathrm{S}}=50.50-50 \Rightarrow \frac{100 \mathrm{S}}{100+\mathrm{S}}=0.5$
$\Rightarrow 100 \mathrm{S}=50+0.55 \Rightarrow 99.5 \mathrm{S}=50$
$S=\frac{50}{99.05}=0.5 \,\Omega$
So, shunt of resistance $=0.5\, \Omega$ is onnected in parallel with the galvanometer.
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