A parallel plate capacitor with air as medium between the plates has a capacitance of 10 μF. The area of capacitor is divided into two equal halves and filled with two media as shown in the figure having dielectric constant k₁ = 2 and k₂ = 4. The capacitance of the system will now be 

(1) 10 μF

(2) 20 μF

(3) 30 μF

(4) 40 μF

Three capacitors are connected to D.C. source of 100 volts shown in the adjoining figure. If the charge accumulated on plates of C₁, C₂ and C₃ are $q_a,q_b,q_c,q_d.q_e$ and q_f respectively, then 

(1) $q_b+q_d+q_f=\frac{100}{9}C$

(2) $q_b+q_d+q_f=0$

(3) $q_a+q_c+q_e=50C$

(4) $q_b=q_d=q_f$

n identical condensers are joined in parallel and are charged to potential V. Now they are separated and joined in series. Then the total energy and potential difference of the combination will be 

(1) Energy and potential difference remain the same

(2) Energy remains the same and the potential difference is nV

(3) Energy increases n times and potential difference is nV

(4) Energy increases n times and potential difference remains the same

Five capacitors of 10 μF capacity each are connected to a d.c. potential of 100 volts as shown in the adjoining figure. The equivalent capacitance between the points A and B will be equal to 

(1) 40 μF

(2) 20 μF

(3) 30 μF

(4) 10 μF

Three capacitors of capacitances \(3~\mu\text{F}\), \(9~\mu\text{F}\) and \(18~\mu\text{F}\) are connected once in series and another time in parallel. The ratio of equivalent capacitance in the two cases \(\frac{C_s}{C_p}\) will be:
1. \(1:15\)
2. \(15:1\)
3. \(1:1\)
4. \(1:3\)

Four condensers each of capacity 4 μF are connected as shown in figure and V_P – V_Q = 15 volts. The energy stored in the system is 

(1) 2400 ergs

(2) 1800 ergs

(3) 3600 ergs

(4) 5400 ergs

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