Referring to the figure below, the effective resistance of the network is 

1. 2r

2. 4r

3. 10r

4. 5r/2

The equivalent resistance of the arrangement of resistances shown in adjoining figure between the points A and B is 

1. 6 ohm

2. 8 ohm

3. 16 ohm

4. 24 ohm

Two resistors are connected (a) in series (b) in parallel. The equivalent resistance in the two cases is 9 ohm and 2 ohms respectively. Then the resistances of the component resistors are :

1. 2 ohm and 7 ohm

2. 3 ohm and 6 ohm

3. 3 ohm and 9 ohm

4. 5 ohm and 4 ohm

Resistances of 6 ohm each are connected in the manner shown in adjoining figure. With the current 0.5 ampere as shown in figure, the potential difference $V_P-V_Q$ is 

1. 3.6 V

2. 6.0 V

3. 3.0 V

4. 7.2 V

The equivalent resistance of resistors connected in series is always :

1. Equal to the mean of component resistors

2. Less than the lowest of component resistors

3. In between the lowest and the highest of component resistors

4. Equal to the sum of component resistors

Four wires of equal length and of resistances 10 ohms each are connected in the form of a square. The equivalent resistance between two opposite corners of the square is :

1. 10 ohm

2. 40 ohm

3. 20 ohm

4. 10/4 ohm

A cell of negligible resistance and e.m.f. 2 volts is connected to a series combination of 2, 3, and 5 $\Omega$. The potential difference in volts between the terminals of 3 $\Omega$ resistance will be :

1. 0.6 V

2. 2/3 V

3. 3 V

4. 6 V

Two resistances are joined in parallel whose resultant is $\frac{6}{8}$ ohm. One of the resistance wire is broken and the effective resistance becomes 2Ω. Then the resistance in ohm of the wire that got broken was 

1. 3/5

2. 2

3. 6/5

4. 3

Resistors of 1, 2, 3 ohm are connected in the form of a triangle. If a 1.5-volt cell of negligible internal resistance is connected across a 3-ohm resistor, the current flowing through this resistance will be :

1. 0.25 amp

2. 0.5 amp

3. 1.0 amp

4. 1.5 amp