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$A$ galvanometer of resistance $G$ is shunted with a resistance of $10 \%$ of $G$. The part of the total current that flows through the galvanometer is

What is the value of resistance to double the value of number of turns of coil for a galvanometer? Why?

The scale of a galvanometer is divided into $160$ equal divisions. The galvanometer shows full scale deflection of $16$ mA and maximum voltage is $80$ mV. Now the range is changed so that the galvanometer reads $160$ $V$. The required resistance to be connected is

$A$ galvanometer of resistance $50 \Omega$ is connected to a battery of $3 \text{ V}$ along with a resistance of $2950 \Omega$ in series. $A$ full-scale deflection of $30$ divisions is obtained in the galvanometer. In order to reduce this deflection to $20$ divisions,the resistance in series should be: (in $Omega$)

If an ammeter is to be used in place of a voltmeter,then we must connect with the ammeter a