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 $Omega$)

When a shunt resistance of $4r$ is connected to a galvanometer,it becomes an ammeter that can measure $0.03 \, A$. When a shunt resistance of $r$ is connected to the same galvanometer,it becomes an ammeter that can measure $0.06 \, A$. What is the current capacity $(i_g)$ of the galvanometer in $A$?

$A$ galvanometer of resistance $100\,\Omega$ has $50$ divisions on its scale and a sensitivity of $20\,\mu A/\text{division}$. It is to be converted into a voltmeter with three ranges: $0-2\,V$,$0-10\,V$,and $0-20\,V$. The appropriate circuit to do so is:

$A$ multirange current meter can be constructed by using a galvanometer circuit as shown in the figure. We want a current meter that can measure $10 \text{ mA}$,$100 \text{ mA}$,and $1 \text{ A}$ using a galvanometer of resistance $10 \text{ } \Omega$ that produces maximum deflection for a current of $1 \text{ mA}$. Find the values of $S_1, S_2$,and $S_3$ that have to be used.

$A$ galvanometer coil has a resistance of $15\; \Omega$ and the meter shows full-scale deflection for a current of $4\; mA$. How will you convert the meter into an ammeter of range $0$ to $6\; A$?

$A$ 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 $= 5 \times 10^{-7} \Omega\text{-m}$)