The relation between voltage sensitivity $({\sigma _V})$ and current sensitivity $({\sigma _i})$ of a moving coil galvanometer is (Resistance of Galvanometer = $G$)

When a galvanometer is shunted by a resistance $S$,its current capacity increases $n$ times. If the same galvanometer is shunted by another resistance $S^{\prime}$,its current capacity increases to $n^{\prime}$. The value of $n^{\prime}$ in terms of $n, S$ and $S^{\prime}$ is

The sensitivity of a galvanometer of resistance $406\, \Omega$ is decreased by $30$ times. The shunt used is .................. $\Omega$.

$A$ galvanometer with $30$ divisions has a current sensitivity of $20 \, \mu A/division$. Its resistance is $25 \, \Omega$. How will you convert it into an ammeter to measure $1 \, A$ current?

How can we convert a galvanometer into an ammeter?

$A$ galvanometer $G$ deflects full scale when a potential difference of $0.50 \ V$ is applied. The internal resistance of the galvanometer $r_g$ is $25 \ \Omega$. An ammeter is constructed by incorporating the galvanometer and an additional shunt resistance $R_S$. The ammeter deflects full scale when a measurement of $2.0 \ A$ is made. The resistance $R_S$ is closest to : ................. $\Omega$