In an ammeter,$0.25 \%$ of the main current passes through the galvanometer. If the resistance of the galvanometer is $G$,the resistance of the ammeter will be:

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

In the diagram shown,the reading of the voltmeter is $20\, V$ and that of the ammeter is $4\, A$. The value of $R$ should be (Consider the given ammeter and voltmeter are not ideal).

$A$ galvanometer of resistance $100 \Omega$ when connected in series with $400 \Omega$ measures a voltage of up to $10 \ V$. The value of resistance required to convert the galvanometer into an ammeter to read up to $10 \ A$ is $x \times 10^{-2} \Omega$. The value of $x$ is:

Consider the following two statements $A$ and $B$ and identify the correct choice given in the answers.
$A$. Duddell's thermo-galvanometer is suitable to measure direct current only.
$B$. Thermopile can measure temperature differences of the order of $10^{-3} \, ^\circ C$.

$A$ galvanometer has a resistance of $100 \Omega$ and a current of $10 \text{ mA}$ produces full-scale deflection in it. The resistance to be connected to it in series,to convert it into a voltmeter of range $50 \text{ V}$,is: (in $\Omega$)