In the circuit shown, the value of each of the resistances is \(r\). The equivalent resistance of the circuit between terminals \(A\) and \(B\) will be:

1.	\(\dfrac{4r}{3}\)	2.	\(\dfrac{3r}{2}\)
3.	\(\dfrac{r}{3}\)	4.	\(\dfrac{8r}{7}\)

Drift velocity \(v_d\) varies with the intensity of the electric field as per the relation:
1. \(v_{d} \propto E\)
2. \(v_{d} \propto \frac{1}{E}\)
3. \(v_{d}= \text{constant}\)
4. \(v_{d} \propto E^2\)

A wire of resistance \(R\) is divided into \(10\) equal parts. These parts are connected in parallel, the equivalent resistance of such connection will be:
1. \(0.01R\)
2. \(0.1R\)
3. \(10R\)
4. \(100R\)

In the figure, the value of resistors to be connected between \(C\) and \(D\) so that the resistance of the entire circuit between \(A\) and \(B\) does not change with the number of elementary sets used is:

A battery of emf \(10\) V is connected to resistance as shown in the figure below. The potential difference \(V_{A} - V_{B}\)
between the points \(A\) and \(B\) is:

1. \(-2\) V

2. \(2\) V

3. \(5\) V

4. \(\frac{20}{11}~\text{V}\)

What is the equivalent resistance of the circuit​​​​​?

1. \(6~\Omega\)
2. \(7~\Omega\)
3. \(8~\Omega\)
4. \(9~\Omega\)

If each resistance in the figure is \(9~\Omega\), then the reading of the ammeter is:

1. \(5~\text{A}\)
2. \(8~\text{A}\)
3. \(2~\text{A}\)
4. \(9~\text{A}\)

Equivalent resistance across terminals \(A\) and \(B\) will be:

The total current supplied to the circuit by the battery is:

1. \(1~\text{A}\)
2. \(2~\text{A}\)
3. \(4~\text{A}\)
4. \(6~\text{A}\)