In the circuit shown below, the cell has an e.m.f. of 10 V and internal resistance of 1 ohm. The other resistances are shown in the figure. The potential difference $V_A-V_B$ is :

1. 6 V

2. 4 V

3. 2 V

4. –2 V

A wire of resistance R is cut into ‘n’ equal parts. These parts are then connected in parallel. The equivalent resistance of the combination will be :

1. nR

2. $\frac{R}{n}$

3. $\frac{n}{R}$

4. $\frac{R}{n^2}$ 

The current in the given circuit is 

1. 8.31 A

2. 6.82 A

3. 4.92 A

4. 2 A

What is the current (i) in the circuit as shown in figure 

1. 2 A

2. 1.2 A

3. 1 A

4. 0.5 A

n equal resistors are first connected in series and then connected in parallel. What is the ratio of the maximum to the minimum resistance 

1. n

2. $\frac{1}{n^2}$

3. n²

4. $\frac{1}{n}$

In the figure, current through the 3 Ω resistor is 0.8 ampere, then potential drop through 4 Ω resistor is 

1. 9.6 V

2. 2.6 V

3. 4.8 V

4. 1.2 V

What is the equivalent resistance between A and B in the figure below if R = 3 Ω 

1. 9 Ω

2. 12 Ω

3. 15 Ω

4. None of these

What is the equivalent resistance between A and B 

1. $\frac{2}{3}R$

2. $\frac{3}{2}R$

3. $\frac{R}{2}$

4. 2 R

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

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