If you are provided three resistances 2 Ω, 3 Ω and 6 Ω. How will you connect them so as to obtain the equivalent resistance of 4 Ω 

1. 

2. 

3. 

4. None of these

The equivalent resistance and potential difference between A and B for the circuit is respectively 

1. 4 Ω, 8 V

2. 8 Ω, 4 V

3. 2 Ω, 2 V

4. 16 Ω, 8 V

Five equal resistances each of resistance R are connected as shown in the figure. A battery of V volts is connected between A and B. The current flowing in AFCEB will be 

1. $\frac{3V}{R}$

2. $\frac{V}{R}$

3. $\frac{V}{2R}$

4. $\frac{2V}{R}$

For the network shown in the figure the value of the current i is 

1. $\frac{9V}{35}$

2. $\frac{5V}{18}$

3. $\frac{5V}{9}$

4. $\frac{18V}{5}$

When a wire of uniform cross-section a, length l and resistance R is bent into a complete circle, the resistance between any two of diametrically opposite points will be :

1. $\frac{R}{4}$

2. $\frac{R}{8}$

3. 4R

4. $\frac{R}{2}$ 

In the circuit given E = 6.0 V, R₁ = 100 ohms, R₂ = R₃ = 50 ohms, R₄ = 75 ohms. The equivalent resistance of the circuit, in ohms, is 

1. 11.875

2. 26.31

3. 118.75

4. None of these

By using only two resistance coils-singly, in series, or in parallel one should be able to obtain resistances of 3, 4, 12, and 16 ohms. The separate resistances of the coil are :

1. 3 and 4

2. 4 and 12

3. 12 and 16

4. 16 and 3

In the adjoining circuit, the battery E₁ has an e.m.f. of 12 volts and zero internal resistance while the battery E has an e.m.f. of 2 volts. If the galvanometer G reads zero, then the value of the resistance X in ohm is 

1. 10

2. 100

3. 500

4. 200

The magnitude and direction of the current in the circuit shown will be 

1. $\frac{7}{3}$A from a to b through e

2. $\frac{7}{3}$A from b to a through e

3. 1A from b to a through e

4. 1A from a to b through e

The e.m.f. of a cell is E volts and internal resistance is r ohm. The resistance in external circuit is also r ohm. The p.d. across the cell will be 

1. E/2

2. 2E

3. 4E

4. E/4