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

A student has 10 resistors of resistance ‘r’. The minimum resistance made by him from given resistors is :

(1) 10 r

(2) $\frac{r}{10}$

(3) $\frac{r}{100}$

(4) $\frac{r}{5}$

Two wires of same metal have the same length but their cross-sections are in the ratio 3 : 1. They are joined in series. The resistance of the thicker wire is 10 Ω. The total resistance of the combination will be 

(1) 40 Ω

(2) $\frac{40}{3}\Omega$

(3) $\frac{5}{2}\Omega$  

(4) 100 Ω

The equivalent resistance of the following infinite network of resistances is 

(1) Less than 4 Ω

(2) 4 Ω

(3) More than 4 Ω but less than 12 Ω

(4) 12 Ω

In the figure given below, the current passing through 6 Ω resistor is 

(1) 0.40 ampere

(2) 0.48 ampere

(3) 0.72 ampere

(4) 0.80 ampere

The equivalent resistance between points A and B of an infinite network of resistances each of 1 Ω connected as shown, is 

(1) Infinite

(2) 2 Ω

(3) $\frac{1+\sqrt{5}}{2}\Omega$

(4) Zero

In the circuit shown, the point ‘B’ is earthed. The potential at the point ‘A’ is :

(1) 14 V

(2) 24 V

(3) 26 V

(4) 50 V

In the circuit shown below, the cell has an e.m.f. of 10 V and internal resistance of 1 ohm. The other resistances are shown in the figure. The potential difference $V_A-V_B$ is :

(1) 6 V

(2) 4 V

(3) 2 V

(4) –2 V

A wire of resistance R is cut into ‘n’ equal parts. These parts are then connected in parallel. The equivalent resistance of the combination will be :

(1) nR

(2) $\frac{R}{n}$

(3) $\frac{n}{R}$

(4) $\frac{R}{n^2}$ 

The current in the given circuit is 

(1) 8.31 A

(2) 6.82 A

(3) 4.92 A

(4) 2 A