Two dielectric slabs of constant K₁ and K₂ have been filled in between the plates of a capacitor as shown below. What will be the capacitance of the capacitor 

(1) $\frac{2{\epsilon }_{0}A}{2}(K_1+K_2)$

(2) $\frac{2{\epsilon }_{0}A}{2}\left(\frac{{\displaystyle K_1+K_2}}{{\displaystyle K_1\times K_2}}\right)$

(3) \(\frac{2\epsilon _{0}A}{d}(\frac{k_{1}+k_{2}}{k_{1}-k_{2}})\)

(4) $\frac{2{\epsilon }_{0}A}{d}\left(\frac{{\displaystyle K_1\times K_2}}{{\displaystyle K_1+K_2}}\right)$

What is the equivalent capacitance between A and B in the given figure (all are in farad) 

(1) $\frac{13}{18}F$

(2) $\frac{48}{13}F$

(3) $\frac{1}{31}F$

(4) $\frac{240}{71}F$

Four capacitors are connected as shown in the figure. Their capacities are indicated in the figure. The effective capacitance between points x and y is (in μF) 

(1) $\frac{5}{6}$

(2) $\frac{7}{6}$

(3) $\frac{8}{3}$

(4) 2

A 10 μF capacitor and a 20 μF capacitor are connected in series across a 200 V supply line. The charged capacitors are then disconnected from the line and reconnected with their positive plates together and negative plates together and no external voltage is applied. What is the potential difference across each capacitor 

(1) $\frac{400}{9}V$

(2) $\frac{800}{9}V$

(3) 400 V

(4) 200 V

Two condensers C₁ and C₂ in a circuit are joined as shown in figure. The potential of point A is V₁ and that of B is V₂. The potential of point D will be 

(1) $\frac{1}{2}(V_1+V_2)$

(2) $\frac{C_2{V}_1+C_1{V}_2}{C_1+C_2}$

(3) $\frac{C_1{V}_1+C_2{V}_2}{C_1+C_2}$

(4) $\frac{C_2{V}_1-C_1{V}_2}{C_1+C_2}$

The combined capacity of the parallel combination of two capacitors is four times their combined capacity when connected in series. This means that 

(1) Their capacities are equal

(2) Their capacities are 1 μF and 2 μF

(3) Their capacities are 0.5 μF and 1 μF

(4) Their capacities are infinite

In the given network capacitance, C₁ = 10 μF, C₂ = 5 μF and C₃ = 4 μF. What is the resultant capacitance between A and B 

(1) 2.2 μF

(2) 3.2 μF

(3) 1.2 μF

(4) 4.7 μF

The equivalent capacitance between A and B is:

In the circuit shown in figure, each capacitor has a capacity of 3 μF. The equivalent capacity between A and B is 

(1) $\frac{3}{4}\mu F$

(2) 3 μF

(3) 6 μF

(4) 5 μF

Two capacitors A and B are connected in series with a battery as shown in the figure. When the switch S is closed and the two capacitors get charged fully, then 

1. The potential difference across the plates of A is 4V and across the plates of B is 6V

2. The potential difference across the plates of A is 6V and across the plates of B is 4V

3. The ratio of electrical energies stored in A and B is 2 : 3

4. The ratio of charges on A and B is 3 : 2