For the circuit shown in the figure, the current \(I\) will be:

              

1. \(0.75~\text{A}\)
2. \(1~\text{A}\) 
3. \(1.5~\text{A}\)
4. \(0.5~\text{A}\)

Two solid conductors are made up of the same material and have the same length and the same resistance. One of them has a circular cross-section of area $A_1$ and the other one has a square cross-section of area $A_2$. The ratio $A_1/A_2$ is:

For the circuit given below, Kirchhoff's loop rule for the loop \(BCDEB\) is given by the equation:

The equivalent resistance between \(A\) and \(B\) for the mesh shown in the figure is:

A cell having an emf \(\varepsilon\) and internal resistance \(r\) is connected across a variable external resistance \(R\). As the resistance \(R\) is increased, the plot of potential difference \(V\) across \(R\) is given by:

1.		2.
3.		4.

In the circuit shown in the figure below, if the potential at point \(\mathrm{A}\) is taken to be zero, the potential at point \(\mathrm{B}\) will be:

1. \(+1\) V
2. \(-1\) V
3. \(+2\) V
4. \(-2\) V

The net resistance of the circuit between \(A\) and \(B\) is:

Two batteries, one of emf \(18\) volts and internal resistance \(2~\Omega\) and the other of emf \(12\) V and internal resistance \(1~\Omega,\) are connected as shown. The voltmeter \(\mathrm{V}\) will record a reading of: 

1. \(18\) V
2. \(30\) V
3. \(14\) V
4. \(15\) V

A \(5-\)ampere fuse wire can withstand a maximum power of \(1\) watt in a circuit. The resistance of the fuse wire is:

For the network shown in the figure below, the value of the current i is:

1. $\frac{18\mathrm{V}}{5}$

2. $\frac{5\mathrm{V}}{9}$

3. $\frac{9\mathrm{V}}{35}$

4. $\frac{5\mathrm{V}}{18}$