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

What is the current (i) in the circuit as shown in figure 

(1) 2 A

(2) 1.2 A

(3) 1 A

(4) 0.5 A

n equal resistors are first connected in series and then connected in parallel. What is the ratio of the maximum to the minimum resistance 

(1) n

(2) $\frac{1}{n^2}$

(3) n²

(4) $\frac{1}{n}$

In the figure, current through the 3 Ω resistor is 0.8 ampere, then potential drop through 4 Ω resistor is 

(1) 9.6 V

(2) 2.6 V

(3) 4.8 V

(4) 1.2 V

What is the equivalent resistance between A and B in the figure below if R = 3 Ω 

(1) 9 Ω

(2) 12 Ω

(3) 15 Ω

(4) None of these

What is the equivalent resistance between A and B 

(1) $\frac{2}{3}R$

(2) $\frac{3}{2}R$

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

(4) 2 R

What is the equivalent resistance of the circuit​​​​​?

1. \(6~\Omega\)
2. \(7~\Omega\)
3. \(8~\Omega\)
4. \(9~\Omega\)