Two wires of same metal have the same length but their cross-sections are in the ratio 3 : 1. They are joined in series. The resistance of the thicker wire is 10 Ω. The total resistance of the combination will be 

1. 40 Ω

2. $\frac{40}{3}\Omega$

3. $\frac{5}{2}\Omega$  

4. 100 Ω

The equivalent resistance of the following infinite network of resistances is 

1. Less than 4 Ω

2. 4 Ω

3. More than 4 Ω but less than 12 Ω

4. 12 Ω

In the figure given below, the current passing through 6 Ω resistor is 

1. 0.40 ampere

2. 0.48 ampere

3. 0.72 ampere

4. 0.80 ampere

The equivalent resistance between points A and B of an infinite network of resistances each of 1 Ω connected as shown, is 

1. Infinite

2. 2 Ω

3. $\frac{1+\sqrt{5}}{2}\Omega$

4. Zero

In the circuit shown, the point ‘B’ is earthed. The potential at the point ‘A’ is :

1. 14 V

2. 24 V

3. 26 V

4. 50 V

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. $\frac{1}{n}$