An ideal ammeter and an ideal voltmeter have resistances of . . . . . . $\Omega$ and . . . . . . $\Omega$ respectively.

$A$ galvanometer coil has a resistance of $12\; \Omega$ and the meter shows full-scale deflection for a current of $3\; mA$. How will you convert the meter into a voltmeter of range $0$ to $18\; V$?

When a resistance of $100 \Omega$ is connected in series with a galvanometer of resistance $R$,its range is $V$. To double its range,a resistance of $1000 \Omega$ is connected in series. Find $R$. (in $\Omega$)

An ammeter with a resistance of $1\, \Omega$ can measure up to $10\, mA$. To convert it into a voltmeter that can measure up to $10\, V$,what resistance (in $\Omega$) must be connected in series?

$A$ galvanometer has $36\,\Omega$ resistance. If a $4\,\Omega$ shunt is added to this,the fraction of current that passes through the shunt is

$A$ moving coil galvanometer, when shunted with a $2\text{ }\Omega$ resistance, gives a full-scale deflection for a current of $500\text{ mA}$. When a resistance of $470\text{ }\Omega$ is connected in series, it gives a full-scale deflection for a potential of $10\text{ V}$ applied across it. The value of the resistance of the galvanometer coil is . . . . . . $\Omega$.