To reduce the current flowing through a coil of resistance $90\, \Omega$ by $90\%$,a resistor of what value in $\Omega$ must be connected in parallel?

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

The deflection in a moving coil galvanometer is reduced to half when it is shunted with a $40 \Omega$ coil. The resistance of the galvanometer is (in $\Omega$)

In the circuit shown below,the ammeter and the voltmeter readings are $3 \text{ A}$ and $6 \text{ V}$ respectively. Then the value of the resistance $R$ is

By connecting a resistance of $980 \Omega$ in series with a galvanometer, it is converted into a voltmeter of a certain range. When the resistance of $470 \Omega$ is connected in series, the range is halved. The resistance of the galvanometer is (in $\Omega$)

Current sensitivities of two galvanometers $G_1$ and $G_2$ of resistances $100 \ \Omega$ and $50 \ \Omega$ are $10^8 \ \text{div/A}$ and $0.5 \times 10^5 \ \text{div/A}$ respectively. The galvanometer in which the voltage sensitivity is more is