The resistance between points $A$ and $B$ is

In the circuit shown in the adjoining figure,the reading of ammeter $A$ is ................ $A$.

$A$ uniform metallic wire having resistance $4 \ \Omega$ is bent to form a square loop $(ABCD)$ (see figure). $A$ resistance of $2 \ \Omega$ is connected between points $B$ and $D$ and a battery of $2 \ V$ is connected across points $A$ and $C$ as shown in the figure. Now the value of current $(I)$ is: (in $A$)

$A$ cell of negligible internal resistance and $e.m.f.$ $2 \ V$ is connected to a series combination of $2 \ \Omega$,$3 \ \Omega$,and $5 \ \Omega$ resistors. The potential difference in volts between the terminals of the $3 \ \Omega$ resistor is:

If two bulbs of wattage $25\, W$ and $100\, W$ respectively,each rated at $220\, V$,are connected in series with a supply of $440\, V$,then which bulb will fuse?

The reading of the ammeter $(A)$ in steady state in the following circuit (assuming negligible internal resistance of the ammeter) is . . . . . . $A$.