A wire has a resistance of 6 Ω. It is cut into two parts and both half values are connected in parallel. The new resistance is :

1. 12 Ω

2. 1.5 Ω

3. 3 Ω

4. 6 Ω

Six equal resistances are connected between points P, Q and R as shown in the figure. Then the net resistance will be maximum between

1. P and Q

2. Q and R

3. P and R

4. Any two points

The total current supplied to the circuit by the battery is:

1. \(1~\text{A}\)
2. \(2~\text{A}\)
3. \(4~\text{A}\)
4. \(6~\text{A}\)

An electric current is passed through a circuit containing two wires of the same material, connected in parallel. If the lengths and radii of the wires are in the ratio of 4/3 and 2/3, then the ratio of the currents passing through the wire will be 

1. 3

2. 1/3

3. 8/9

4. 2

In circuit shown below, the resistances are given in ohms and the battery is assumed ideal with emf equal to \(3\) volt. The voltage across the resistance \(R_4\) is:

1. \(0.4\) V
2. \(0.6\) V
3. \(1.2\) V
4. \(1.5\) V

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. $\frac{R}{2}$