The resistance of a galvanometer is $90\, \Omega$. If only $10\%$ of the main current is to flow through the galvanometer,how should a resistor be connected and what should be its value?

$A$ galvanometer has $36\,\Omega$ resistance. If a $4\,\Omega$ shunt is added to this,the fraction of current that passes through the shunt is

$A$ small cylindrical soft iron piece is kept in a galvanometer so that

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

If only $5 \%$ of the total current is to be passed through a galvanometer of resistance $G$,then the resistance of the shunt will be

When a resistance of $200 \Omega$ is connected in series with a galvanometer of resistance $G$,its range is $V$. To triple its range,a resistance of $2000 \Omega$ is connected in series. The value of $G$ is (in $\Omega$)