In an ammeter,$5 \%$ of the main current passes through the galvanometer. If the resistance of the galvanometer is $G$,the resistance of the ammeter will be:

$A$ galvanometer having internal resistance $10 \Omega$ requires $0.01 \ A$ for a full scale deflection. To convert this galvanometer to a voltmeter of full-scale deflection at $120 \ V$, we need to connect a resistance of

$A$ galvanometer gives a full-scale reading of $50 \ mA$,when a potential difference across its terminals is $0.15 \ V$. It can be used as an ammeter of range $0 - 100 \ A$ by connecting a shunt resistance of: (in $\Omega$)

In the experiment to determine the galvanometer resistance by the half-deflection method,the plot of $\frac{1}{\theta}$ vs the resistance $(R)$ of the resistance box is shown in the figure. The figure of merit of the galvanometer is .............. $\times 10^{-1} \text{ A/division}$. [The source has emf $E = 2 \text{ V}$]

What is the function of the spiral spring in the construction of a galvanometer?

The deflection in a moving coil galvanometer is