Question 486: Current Electricity

Question

When a resistance of $100 \Omega$ is connected in series with a galvanometer of resistance ' G ', its range is ' $V$ '. To double its range, a resistance of $1000 \Omega$ is connected in series. The value of ' $G$ ' is

Accepted Answer

Using, $\mathrm{R}_{\mathrm{s}}=\frac{\mathrm{V}}{\mathrm{I}_{\mathrm{g}}}-\mathrm{G}$ we get,
for $1^{	ext {st }}$ case, $100=\frac{\mathrm{V}}{\mathrm{I}_{\mathrm{g}}}-\mathrm{G}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\,\,and$
for $2^{	ext {nd }}$ case, $1000=\frac{2 \mathrm{~V}}{\mathrm{I}_{\mathrm{g}}}-\mathrm{G}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$
By subtracting equation (i) from (ii) we get, $\frac{\mathrm{V}}{\mathrm{I}_{\mathrm{g}}}=900\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$
$	herefore \quad \mathrm{G}=800 \Omega\,\,\,\,\,Ã¢ÂÂ¦[From (i) and (iii)]$

Answer

$400 \Omega$

Answer

$800 \Omega$

Answer

$1000 \Omega$

Answer

$1200 \Omega$

Current Electricity

When a resistance of 100 \Omega is connected in series with a galvanometer of resistance ' G ', its range is ' V '. To double its range, a resistance of 1000 \Omega is connected in series. The value of ' G ' is

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