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. $\frac{4C}{3}$

In the given figure the capacitors C₁, C₃, C₄, C₅ have a capacitance 4 μF each and if the capacitor C₂ has a capacitance 10 μF, then effective capacitance between A and B will be 

1. 2 μF

2. 4 μF

3. 6 μF

4. 8 μF

Two identical capacitors, have the same capacitance C. One of them is charged to potential V₁ and the other to V₂. The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is 

1. $\frac{1}{4}C(V_1^2-V_2^2)$

2. $\frac{1}{4}C(V_1^2+V_2^2)$

3. $\frac{1}{4}C{(V_1-V_2)}^2$

4. $\frac{1}{4}C{(V_1+V_2)}^2$