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

In the figure, three capacitors each of capacitance 6 pF are connected in series. The total capacitance of the combination will be 

1. 9 × 10^–12 F

2. 6 × 10^–12 F

3. 3 × 10^–12 F

4. 2 × 10^–12 F

Equivalent capacitance between A and B is 

1. 8 μF

2. 6 μF

3. 26 μF

4. 10/3 μF

In the figure a capacitor is filled with dielectrics. The resultant capacitance is 

1. $\frac{2{\epsilon }_{0}A}{d}\left[\frac{{\displaystyle 1}}{{\displaystyle k_1}}+\frac{{\displaystyle 1}}{{\displaystyle k_2}}+\frac{{\displaystyle 1}}{{\displaystyle k_3}}\right]$

2. $\frac{{\epsilon }_{0}A}{d}\left[\frac{{\displaystyle 1}}{{\displaystyle k_1}}+\frac{{\displaystyle 1}}{{\displaystyle k_2}}+\frac{{\displaystyle 1}}{{\displaystyle k_3}}\right]$

3. $\frac{2{\epsilon }_{0}A}{d}[k_1+k_2+k_3]$

4. None of these

Three capacitors of capacitance 3 μF, 10 μF and 15 μF are connected in series to a voltage source of 100V. The charge on 15 μF is 

1. 50 μC

2. 100 μC

3. 200 μC

4.) 280 μC

Two capacitors $C_1 = 2~\mu\text{F}$ and $C_2 = 6~\mu \text{F}$ in series, are connected in parallel to a third capacitor $C_3= 4~\mu\text{F}$. This arrangement is then connected to a battery of $\text{emf}= 2~\text{V}$, as shown in the figure. How much energy is lost by the battery in charging the capacitors? 

1. $22\times 10^{-6}~\text{J}$
2. $11\times 10^{-6}~\text{J}$
3. $\frac{32}{3}\times 10^{-6}~\text{J}$
4. $\frac{16}{3}\times 10^{-6}~\text{J}$

A parallel plate capacitor has capacitance $C$. If it is equally filled with parallel layers of materials of dielectric constants $K_1$ and $K_2$, its capacity becomes $C_1$. The ratio of $C_1$ to $C$ is:

The equivalent capacitance in the circuit between A and B will be 

1. 1 μF

2. 2 μF

3. 3 μF

4. $\frac{1}{3}\mu F$

The equivalent capacitance between A and B is 

1. $\frac{C}{4}$

2. $\frac{3C}{4}$

3. $\frac{C}{3}$

4. $\frac{4C}{3}$