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

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