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

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. $\frac{R}{n^2}$