In an adjoining figure are shown three capacitors C₁, C₂ and C₃ joined to a battery. The correct condition will be (Symbols have their usual meanings) :

(1) Q₁ = Q₂ = Q₃ and V₁ = V₂ = V₃ = V

(2) Q₁ = Q₂ + Q₃ and V = V₁ + V₂ + V₃

(3) Q₁ = Q₂ + Q₃ and V = V₁ + V₂

(4) Q₂ = Q₃ and V₂ = V₃

In the circuit diagram shown in the adjoining figure, the resultant capacitance between P and Q is 

(1) 47 μF

(2) 3 μF

(3) 60 μF

(4) 10 μF

Two capacitances of capacity C₁ and C₂ are connected in series and potential difference V is applied across it. Then the potential difference across C₁ will be:

(1) $V\frac{C_2}{C_1}$

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

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

(4) $V\frac{C_1}{C_1+C_2}$

A capacitor of capacity C₁ is charged to the potential of V₀. On disconnecting with the battery, it is connected with a capacitor of capacity C₂ as shown in the adjoining figure. The ratio of energies before and after the connection of switch S will be

(1) (C₁ + C₂)/C₁

(2) C₁/(C₁ + C₂)

(3) C₁C₂

(4) C₁/C₂

Four capacitors each of capacity \(3~\mu\text{F}\) are connected as shown in the adjoining figure. The ratio of equivalent capacitance between \(A\) and \(B\) and between \(A\) and \(C\) will be:

1. \(4:3\)

2. \(3:4\)

3. \(2:3\)

4. \(3:2\)

A parallel plate condenser is filled with two dielectrics as shown. Area of each plate is \(A\) metre² and the separation is \(t\) metre. The dielectric constants are \(k_1\) and \(k_2\) respectively. Its capacitance in farad will be:

        
1. \(\frac{\varepsilon_{0} A}{t} \left( k_{1} + k_{2}\right)\)
2. \(\frac{\varepsilon_{0} A}{t} \frac{\left( k_{1} + k_{2}\right)}{2}\)
3. \(\frac{2\varepsilon_{0} A}{t} \left( k_{1} + k_{2}\right)\)
4. \(\frac{\varepsilon_{0} A}{t} \frac{\left( k_{1} - k_{2}\right)}{2}\)

Three condensers each of capacitance 2F are put in series. The resultant capacitance is 

(1) 6F

(2) $\frac{3}{2}F$

(3) $\frac{2}{3}F$

(4) 5F

Four condensers are joined as shown in the adjoining figure. The capacity of each is 8 μF. The equivalent capacity between the points A and B will be

(1) 32 μF

(2) 2 μF

(3) 8 μF

(4) 16 μF

The capacities and connection of five capacitors are shown in the adjoining figure. The potential difference between the points A and B is 60 volts. Then the equivalent capacity between A and B and the charge on 5 μF capacitance will be respectively

(1) 44 μF; 300 μC

(2) 16 μF; 150 μC

(3) 15 μF; 200 μC

(4) 4 μF; 50 μC

Four plates of the same area of cross-section are joined as shown in the figure. The distance between each plate is d. The equivalent capacity across A and B will be

(1) $\frac{2{\epsilon }_{0}A}{d}$

(2) $\frac{3{\epsilon }_{0}A}{d}$

(3) $\frac{3{\epsilon }_{0}A}{2d}$

(4) $\frac{{\epsilon }_{0}A}{d}$