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 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

In the network of resistors shown in the adjoining figure, the equivalent resistance between A and B is

(1) 54 ohm

(2) 18 ohm

(3) 36 ohm

(4) 9 ohm

A wire is broken in four equal parts. A packet is formed by keeping the four wires together. The resistance of the packet in comparison to the resistance of the wire will be 

(1) Equal

(2) One fourth

(3) One eight

(4) $\frac{1}{16}th$

Four resistances are connected in a circuit in the given figure. The electric current flowing through 4 ohm and 6 ohm resistance is respectively 

(1) 2 amp and 4 amp

(2) 1 amp and 2 amp

(3) 1 amp and 1 amp

(4) 2 amp and 2 amp

In the figure shown, the total resistance between A and B is

(1) 12 Ω

(2) 4 Ω

(3) 6 Ω

(4) 8 Ω

In the given figure, when key K is opened, the reading of the ammeter A will be

(1) 50 A

(2) 2 A

(3) 0.5 A

(4) $\frac{10}{9}A$

In the given circuit, the potential of the point E is 

(1) Zero

(2) –8 V

(3) –4/3 V

(4) 4/3 V

If a resistance R₂ is connected in parallel with the resistance R in the circuit shown, then possible value of current through R and the possible value of R₂ will be

(1) $\frac{I}{3},R$

(2) $I,2R$

(3) $\frac{I}{3},R$

(4) $\frac{I}{2},R$

A battery of emf \(10\) V is connected to resistance as shown in the figure below. The potential difference \(V_{A} - V_{B}\)
between the points \(A\) and \(B\) is:

1. \(-2\) V

2. \(2\) V

3. \(5\) V

4. \(\frac{20}{11}~\text{V}\)