$A$ galvanometer of resistance $50 \Omega$ is converted to an ammeter. After shunting,the effective resistance of the ammeter is $2.5 \Omega$. The value of the shunt is

$A$ galvanometer of resistance $50 \Omega$ gives a full-scale deflection for a current of $5 \times 10^{-4} \text{ A}$. The resistance that should be connected in series with the galvanometer to read $3 \text{ V}$ is: (in $\Omega$)

$A$ galvanometer of resistance $50 \ \Omega$ is connected to a battery of $8 \ V$ along with a resistance of $3950 \ \Omega$ in series. $A$ full-scale deflection of $30$ divisions is obtained in the galvanometer. In order to reduce this deflection to $15$ divisions,the resistance in series should be . . . . . . $\Omega$.

$A$ galvanometer of resistance $G \ \Omega$ is shunted by a resistance $S \ \Omega$. To keep the main current in the circuit unchanged,the resistance to be connected in series with the galvanometer is

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

$A$ moving coil galvanometer,having a resistance $G$,produces full scale deflection when a current $I_g$ flows through it. This galvanometer can be converted into $(i)$ an ammeter of range $0$ to $I_0$ $(I_0 > I_g)$ by connecting a shunt resistance $R_A$ to it and $(ii)$ into a voltmeter of range $0$ to $V$ $(V = GI_0)$ by connecting a series resistance $R_V$ to it. Then,