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) $\frac{1}{4}C{(V_1+V_2)}^2$

Three capacitors each of capacity 4 μF are to be connected in such a way that the effective capacitance is 6 μF. This can be done by 

(1) Connecting them in parallel

(2) Connecting two in series and one in parallel

(3) Connecting two in parallel and one in series

(4) Connecting all of them in series

Three capacitors of capacitance 3 μF are connected in a circuit. Then their maximum and minimum capacitances will be

(1) 9 μF, 1 μF

(2) 8 μF, 2 μF

(3) 9 μF, 0 μF

(4) 3 μF, 2 μF

A capacitor of capacity C₁ is charged upto V volt and then connected to an uncharged capacitor of capacity C₂. Then final potential difference across each will be 

(1) $\frac{C_{2}V}{C_1+C_2}$

(2) $\left(1+\frac{{\displaystyle C_2}}{{\displaystyle C_1}}\right)V$

(3) $\frac{C_{1}V}{C_1+C_2}$

(4) $\left(1-\frac{{\displaystyle C_2}}{{\displaystyle C_1}}\right)V$

Four identical capacitors are connected as shown in diagram. When a battery of 6 V is connected between A and B, the charge stored is found to be 1.5 μC. The value of C₁ is 

(1) 2.5 μF

(2) 15 μF

(3) 1.5 μF

(4) 0.1 μF