$A$ circuit contains an ammeter,a battery of $30\,V$ and a resistance of $40.8\,\Omega$ all connected in series. If the ammeter has a coil of resistance $480\,\Omega$ and a shunt of $20\,\Omega$,the reading in the ammeter will be .................. $A$.

$A$ galvanometer has resistance '$G$' and range '$V_g$'. How much resistance is required to read voltage up to '$V$' volt?

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 $A$?

An ammeter with a resistance of $1\, \Omega$ can measure up to $10\, mA$. To convert it into a voltmeter that can measure up to $10\, V$,what resistance (in $\Omega$) must be connected in series?

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 $Omega$)

$A$ 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