A parallel plate capacitor is made by stacking n equally spaced plates connected alternately. If the capacitance between any two plates is C then the resultant capacitance is
1. C
2. nC
3. (n – 1)C
4. (n + 1)C
Four plates of equal area A are separated by equal distances d and are arranged as shown in the figure. The equivalent capacity is
1.
2.
3.
4.
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 and qf respectively, then
1.
2.
3.
4.
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 VP – VQ = 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