The sensitivity of a galvanometer that measures current is decreased by $\frac{1}{40}$ times by using a shunt resistance of $10 \Omega$. Then,the value of the resistance of the galvanometer is (in $\Omega$)

In an ammeter,$4 \%$ of the main current is passing through the galvanometer. If the shunt resistance is $5 \Omega$,then the resistance of the galvanometer will be: (in $\Omega$)

$A$ galvanometer having a coil resistance of $30\,\Omega$ shows full-scale deflection when a current of $2\,A$ passes through it. It can be converted into an ammeter to read currents up to $10\,A$ by:

To determine the resistance $G$ of a galvanometer by the half-deflection method,a battery of $emf$ $V$ and a series resistance $R$ are used to produce a deflection $\theta$ in the galvanometer. If a shunt resistance $S$ is connected in parallel to the galvanometer to reduce the deflection to $\theta/2$,then $G, R$,and $S$ are related by the equation:

$A$ galvanometer of resistance $50 \Omega$ is converted to an ammeter. After shunting,the effective resistance of the ammeter is $2.5 \Omega$. The value of the shunt is

The voltmeter has a range of $10 \ V$ and its internal resistance is $50 \ \Omega$. To increase the range of the voltmeter to $15 \ V$,the resistance which is to be connected is: