If only $1 \%$ of the total current is passed through a galvanometer of resistance $G$,then the resistance of the shunt is:

Explain the difficulties encountered when using a galvanometer directly as an ammeter.

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 $\Omega$)

The deflection in a moving coil galvanometer of resistance $45 \Omega$ falls from $30$ divisions to $3$ divisions. The length of the shunt wire required to convert the galvanometer into an ammeter is [specific resistance of the material of the shunt wire $= 5 \times 10^{-7} \Omega m$ and area of cross-section of the wire $= 4 \times 10^{-7} m^2$]. (in $m$)

$A$ 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

$A$ student uses the resistance of a known resistor $(1 \,\Omega)$ to calibrate a voltmeter and an ammeter using the circuits shown below. The student measures the ratio of the voltage to current to be $1 \times 10^3 \,\Omega$ in circuit $(a)$ and $0.999 \,\Omega$ in circuit $(b)$. From these measurements,the resistance (in $\Omega$) of the voltmeter and ammeter are found to be close to