$A$ galvanometer of resistance $G$ is shunted by a resistance $S$. To keep the main current in the circuit unchanged,what value of resistance should be connected in series with the galvanometer?

What is the value of shunt resistance required to convert a galvanometer of resistance $ 100 \Omega $ into an ammeter of range $ 1 \text{ A} $? Given: Full scale deflection of the galvanometer is $ 5 \text{ mA} $.

$A$ galvanometer of resistance $20 \Omega$ gives a deflection of $5$ divisions when $1 \text{ mA}$ current flows through it. The galvanometer scale has $50$ divisions. To convert the galvanometer into a voltmeter of range $25 \text{ V}$,we should connect a resistance of

$A$ galvanometer having a coil of resistance $30 \ \Omega$ needs $20 \ \text{mA}$ of current for full-scale deflection. If a maximum current of $3 \ \text{A}$ is to be measured using this galvanometer,the resistance of the shunt to be added to the galvanometer should be $\frac{30}{X} \ \Omega$,where $X$ is

$A$ multirange current meter can be constructed by using a galvanometer circuit as shown in the figure. We want a current meter that can measure $10 \text{ mA}$,$100 \text{ mA}$,and $1 \text{ A}$ using a galvanometer of resistance $10 \text{ } \Omega$ that produces maximum deflection for a current of $1 \text{ mA}$. Find the values of $S_1, S_2$,and $S_3$ that have to be used.

When a resistance of $200 \Omega$ is connected in series with a galvanometer of resistance $G$,its range is $V$. To triple its range,a resistance of $2000 \Omega$ is connected in series. The value of $G$ is (in $Omega$)