$A$ galvanometer of $50 \, \Omega$ resistance has $25$ divisions. $A$ current of $4 \times 10^{-4} \, A$ gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of $25 \, V$, it should be connected with a resistance of

$A$ galvanometer having a resistance $G$ and current $I_g$ flowing in it,produces full-scale deflection. If $S_1$ is the value of the shunt which converts it into an ammeter of range $0-I$ and $S_2$ is the value of the shunt for range $0-8I$,then the ratio $\frac{S_1}{S_2}$ will be:

The scale of a galvanometer is divided into $100$ equal divisions. It has a current sensitivity of $10 \text{ div./mA}$ and a voltage sensitivity of $4 \text{ div./mV}$. The resistance of the galvanometer is: (in $\Omega$)

An ammeter whose resistance is $180\,\Omega$ gives full scale deflection when the current is $2\,mA$. The shunt required to convert it into an ammeter reading $20\,mA$ (in $\Omega$) is

$A$ moving coil galvanometer is converted into an ammeter reading up to $0.03\,A$ by connecting a shunt of resistance $4r$ across it,and into an ammeter reading up to $0.06\,A$ when a shunt of resistance $r$ is connected across it. What is the maximum current which can be sent through this galvanometer if no shunt is used?

$A$ moving coil galvanometer, when shunted with a $2\text{ }\Omega$ resistance, gives a full-scale deflection for a current of $500\text{ mA}$. When a resistance of $470\text{ }\Omega$ is connected in series, it gives a full-scale deflection for a potential of $10\text{ V}$ applied across it. The value of the resistance of the galvanometer coil is . . . . . . $\Omega$.