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. \(100R\)

Three resistors each of 2 ohm are connected together in a triangular shape. The resistance between any two vertices will be 

1. 4/3 ohm

2. 3/4 ohm

3. 3 ohm

4. 6 ohm

There are n similar conductors each of resistance R. The resultant resistance comes out to be x when connected in parallel. If they are connected in series, the resistance comes out to be :

1. x/n²

2. n²x

3. x/n

4. nx

Equivalent resistance between A and B will be 

1. 2 ohm

2. 18 ohm

3. 6 ohm

4. 3.6 ohm

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

1. 2r

2. 4r

3. 10r

4. 5r/2

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

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

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

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

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