The resistance of a $1\, A$ ammeter is $0.018\,\Omega$. To convert it into a $10\, A$ ammeter,the shunt resistance required will be:

An ammeter gives full-scale deflection when a current of $2 \, A$ flows through it. The resistance of the ammeter is $12 \, \Omega$. If the same ammeter is to be used for measuring a maximum current of $5 \, A$,then the ammeter must be connected with a resistance of:

$A$ galvanometer has resistance $G \ \Omega$ and $I_g$ is the current flowing through it which produces full-scale deflection. $S_1$ is the value of the shunt which converts it into an ammeter of range $0$ to $3I$,and $S_2$ is the shunt value which converts it into an ammeter of range $0$ to $4I$. The ratio $S_2:S_1$ is:

$A$ galvanometer of resistance $10 \Omega$ changes its range from $1 \text{ mA}$ to $101 \text{ mA}$ when a resistive wire is connected in parallel. If the resistivity of the material of the wire and its area of cross-section are $1 \times 10^{-6} \Omega \text{ m}$ and $1 \text{ mm}^2$ respectively,then the length of the wire is (in $cm$)

In the adjoining circuit diagram,the readings of the ammeter and voltmeter are $2 \, A$ and $120 \, V$,respectively. If the value of $R$ is $75 \, \Omega$,then the voltmeter resistance will be (in $\Omega$):

In an ammeter,$4 \%$ of the main current is passing through the galvanometer. If the shunt resistance is $5 \Omega$,then the resistance of the galvanometer will be: (in $Omega$)