When a resistance of $100 \Omega$ is connected in series with a galvanometer of resistance $G$, its range is…
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
$900 \Omega$
$300 \Omega$
$200 \Omega$
$100 \Omega$
Solution
When a resistance of $100 \Omega$ is connected in series
current, $i=\frac{2 V}{100+R}$ $\qquad$ when a resistance of $1000 \Omega$ is connected in series the its range double
current, $\quad i=\frac{2 V}{1100+R}$
From above equations
$\begin{aligned}
& \frac{V}{100+R}=\frac{2 V}{1100+R} \\
& R=900 \Omega
\end{aligned}$
^