$A$ 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 $A$)

Two identical fuses are rated at $10\,A$. If they are joined,

When subjected to a voltage of $10 \, V$, the current through a resistor at a temperature of $40^{\circ} C$ is $0.1 \, A$. The temperature coefficient of resistance of the material of the resistor is $2 \times 10^{-4} {}^{\circ} C^{-1}$. The temperature of the resistor in ${}^{\circ} C$ when the current drops to $0.098 \, A$ is

In the circuit shown in the figure,the potential difference between points $A$ and $B$ is $16\,V$. The current passing through the $2\,\Omega$ resistance will be $...........\,A$.

The resistance of an iron wire is $10\,\Omega$ and the temperature coefficient of resistivity is $5 \times 10^{-3}\,^{\circ}C^{-1}$. At $20\,^{\circ}C$,it carries $30\,mA$ of current. Keeping the potential difference between its ends constant,the temperature of the wire is raised to $120\,^{\circ}C$. The current in milliamperes that flows in the wire is:

The reading of the ammeter in the following figure will be: (in $\text{A}$)