$A$ galvanometer of $10 \,\Omega$ resistance gives full scale deflection with $0.01 \, A$ of current. It is to be converted into an ammeter for measuring $10 \, A$ current. The value of shunt resistance required will be

An ammeter has a range of $1 \ A$ without a shunt. The range of the ammeter can be changed by using different shunt resistances. What is the nature of the graph between the shunt resistance and the range?

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

$A$ milliammeter of range $10\, mA$ and resistance $9\, \Omega$ is joined in a circuit as shown. The meter gives full-scale deflection for current $I$ when $A$ and $B$ are used as its terminals,i.e.,current enters at $A$ and leaves at $B$ ($C$ is left isolated). The value of $I$ is

$A$ galvanometer of resistance $20 \Omega$ gives a deflection of $5$ divisions when $1 \text{ mA}$ current flows through it. The galvanometer scale has $50$ divisions. To convert the galvanometer into a voltmeter of range $25 \text{ V}$,we should connect a resistance of

In the circuit diagrams ($A, B, C$ and $D$) shown below,$R$ is a high resistance and $S$ is a resistance of the order of galvanometer resistance $G$. The correct circuit,corresponding to the half deflection method for finding the resistance and figure of merit of the galvanometer,is the circuit labelled as