Question

When a resistance of 100$\Omega$ is connected in series with a galvanometer of resistance R, its range is V. To double its range, a resistance of 1000 $\Omega$ is connected in series. Find R.

• A. $700\Omega$
• B. $800\Omega$
• C. $900\Omega$
• D. $100\Omega$

Answer 1

The correct option is C $900\Omega$

When a resistance of $100\Omega$ is connected in series current,$i=\frac{V}{100+R}$
When a resistance of $1000\Omega$ is connected in series, the its range double current $i=\frac{2V}{1100+R}$

Thus $\frac{V}{100+R}=\frac{2V}{1100+R}$

$⟹R=900\Omega$