$20\%$ of the main current passes through the galvanometer. If the resistance of the galvanometer is $G$,then the resistance of the shunt will be

$A$ galvanometer coil has a resistance of $10 \ \Omega$ and the meter shows full-scale deflection for $3 \ \text{mA}$. The value of the shunt required to convert this meter into an ammeter of range $0$ to $10 \ \text{A}$ is . . . . . . $\Omega$.

$A$ milliammeter of range $10\, mA$ and resistance $9\, \Omega$ is joined in a circuit as shown. The meter gives full-scale deflection for current $I$ when $A$ and $B$ are used as its terminals,i.e.,current enters at $A$ and leaves at $B$ ($C$ is left isolated). The value of $I$ is

The deflection in a galvanometer falls to $\left(\frac{1}{4}\right)^{th}$ of its initial value when it is shunted by a resistor of $3 \Omega$. If an additional shunt of $2 \Omega$ is connected in parallel to the earlier shunt,the deflection in the galvanometer falls to:

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?

The maximum current that can be measured by a galvanometer of resistance $40 \,\Omega$ is $10 \,mA$. It is converted into a voltmeter that can read up to $50 \,V$. The resistance to be connected in series with the galvanometer is ... (in $\Omega$)