The sensitivity of a galvanometer is $60 \text{ division/A}$. When a shunt is used,its sensitivity becomes $10 \text{ division/A}$. If the galvanometer is of resistance $20 \ \Omega$,the value of shunt used is (in $Omega$)

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

$A$ moving coil galvanometer has a resistance of $50\,\Omega$ and gives full-scale deflection for $10\,mA$. How could it be converted into an ammeter with a full-scale deflection for $1\,A$?

What is a shunt? Explain its function in a circuit,obtain the formula for the shunt,and explain its uses.

$A$ moving coil galvanometer of resistance $100 \ \Omega$ shows full-scale deflection when a current of $100 \ \mu A$ passes through it. If it is intended to show full-scale deflection when a current of $1 \ mA$ passes through it,the value of shunt resistance in ohms to be connected to the galvanometer is:

An ammeter of resistance $20 \Omega$ gives full scale deflection when $1 \text{ mA}$ current flows through it. What is the maximum current that can be measured by connecting $4$ resistors each of $16 \Omega$ in parallel with the meter (in $\text{ mA}$)?