To send $10 \%$ of the main current through a moving coil galvanometer of resistance $99 \Omega$,the shunt required is . . . . . . . (in $\Omega$)

In the circuit shown below,the ammeter and the voltmeter readings are $3 \text{ A}$ and $6 \text{ V}$ respectively. Then the value of the resistance $R$ is

$A$ 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

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 $\Omega$)

The resistance of a galvanometer is $25\,\Omega$ and it requires $50\,\mu A$ for full-scale deflection. The value of the shunt resistance required to convert it into an ammeter of $5\,A$ range is:

Consider a moving coil galvanometer $(MCG)$ :
$A :$ The torsional constant in a moving coil galvanometer has dimensions $[ML^2 T^{-2}]$.
$B :$ Increasing the current sensitivity may not necessarily increase the voltage sensitivity.
$C :$ If we increase the number of turns $(N)$ to its double $(2N)$,then the voltage sensitivity doubles.
$D :$ $MCG$ can be converted into an ammeter by introducing a shunt resistance of a large value in parallel with the galvanometer.
$E :$ Current sensitivity of $MCG$ depends inversely on the number of turns of the coil.
Choose the correct answer from the options given below: