$A$ shunt of resistance $1\ \Omega$ is connected across a galvanometer of $120\ \Omega$ resistance. $A$ current of $5.5\ A$ flows through the circuit,and the galvanometer shows full-scale deflection. What is the current that would cause full-scale deflection in the galvanometer in the absence of the shunt (in $A$)?

In an experiment to determine the figure of merit of a galvanometer by the half-deflection method,a student constructed the following circuit. He unplugged a resistance of $5200 \ \Omega$ in $R$. When $K_1$ is closed and $K_2$ is open,the deflection observed in the galvanometer is $26 \ \text{div}$. When $K_2$ is also closed and a resistance of $90 \ \Omega$ is removed in $S$,the deflection becomes $13 \ \text{div}$. The resistance of the galvanometer is nearly: (in $\Omega$)

$A$ galvanometer of resistance $G$ is shunted with a resistance of $10 \%$ of $G$. The part of the total current that flows through the galvanometer is

$A$ small cylindrical soft iron piece is kept in a galvanometer so that

The scale of a galvanometer is divided into $160$ equal divisions. The galvanometer shows full scale deflection of $16$ mA and maximum voltage is $80$ mV. Now the range is changed so that the galvanometer reads $160$ $V$. The required resistance to be connected is

Derive the expression for the shunt resistance required to increase the current-measuring capacity of a galvanometer by a factor of $n$.