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

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. \(\frac{16}{3}\times 10^{-6}~\text{J}\)