The current sensitivity of a moving coil galvanometer increases by $20 \%$ when its resistance is doubled. Calculate,by what factor does the voltage sensitivity change?

$A$ galvanometer with $30$ divisions has a current sensitivity of $20 \, \mu A/division$. Its resistance is $25 \, \Omega$. How will you convert it into an ammeter to measure $1 \, A$ current?

$A$ galvanometer having a resistance $G$ and current $I_g$ flowing in it,produces full-scale deflection. If $S_1$ is the value of the shunt which converts it into an ammeter of range $0-I$ and $S_2$ is the value of the shunt for range $0-8I$,then the ratio $\frac{S_1}{S_2}$ will be:

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

Two tangent galvanometers $A$ and $B$ are identical except in their number of turns. They are connected in series. On passing a current through them,deflections of $60^{\circ}$ and $30^{\circ}$ are produced. The ratio of the number of turns in $A$ and $B$ is

Explain the equation of a shunt.