The resistance of a galvanometer is $50\,\Omega$ and the current required to give full-scale deflection is $100\,\mu A$. In order to convert it into an ammeter reading up to $10\,A$,it is necessary to put a resistance of

Write the equation for the series resistance required to increase the voltage range of a galvanometer by a factor of $n$.

$100\,mA$ current gives a full scale deflection in a galvanometer of $2\,\Omega$ resistance. The resistance connected with the galvanometer to convert it into a voltmeter to measure $5\,V$ is ............... $\Omega$.

The maximum current that can be measured by a galvanometer of resistance $40 \,\Omega$ is $10 \,mA$. It is converted into a voltmeter that can read up to $50 \,V$. The resistance to be connected in series with the galvanometer is ... (in $\Omega$)

$A$ 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,

$A$ galvanometer,having a resistance of $50 \Omega$,gives a full scale deflection for a current of $0.05 \text{ A}$. The length in metre of a resistance wire of area of cross-section $2.97 \times 10^{-2} \text{ cm}^2$ that can be used to convert the galvanometer into an ammeter which can read a maximum of $5 \text{ A}$ current is: (Specific resistance of the wire $= 5 \times 10^{-7} \Omega\text{-m}$)