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?

To find the resistance of a galvanometer by the half-deflection method,the experimental data obtained are given in the table below:

$S. No.$	Resistance $R \ (\Omega)$	Deflection $(\theta)$	Shunt $S \ (\Omega)$	Half deflection $(\theta / 2)$	Galvanometer resistance $(G)$
$1$	$3300$	$30$	$80$	$15$	$G_1$
$2$	$5000$	$20$	$80$	$10$	$G_2$

From the above data,the galvanometer resistance $G$ will be near to: (in $\Omega$)

Explain the equation of a shunt.

$20\%$ of the main current passes through the galvanometer. If the resistance of the galvanometer is $G$,then the resistance of the shunt will be

Two galvanometers $G_{1}$ and $G_{2}$ require $2 \ mA$ and $3 \ mA$ respectively to produce the same deflection. Then:

The galvanometer has a resistance of $1.8 \Omega$. Calculate the value of shunt to increase the range of galvanometer by $10$ times. (in $\Omega$)