Sponsor Area
C.
NeutronsSponsor Area
Three capacitors C1 = 6μF,C2 = 12μF and C3 = 20μF are connected to a 100 V battery, as shown in Figure 2 below :
Calculate:
(i) Charge on each plate of capacitor Cx
ii) Electrostatic potential energy stored in capacitor C3.
i)
Equivalent of C1 and C2 is
ii)
Energy on capacitor C3 
Capacitors each of capacitance C.
Series capacitance = Cs.
Parallel capacitance = Cp.
Cp= C + C + C = 3C
An isolated 16 μF parallel plate air capacitor has a potential differences of 1000 V (Figure 5 a). A dielectric slab having relative permittivity (i.e. dielectric constant) = 5 is introduced to fill the space between the two plates com pletely. (Figure 5 b). Calculate:
(i) The new capacitance of the capacitor.
(ii) The new potential differences between the two plates of the capacitor
C1 = 16 μ F where C1 → Capacitance of figure 5(a)
V1= 1000 V C2 → Capacitance of figure 5 (b)
(i) C2 / C1 = K
C2 = K C1
= 5 X 16
= 80 μF, is the new capacitance of the capacitor.
(ii) Similarly V2 / V1 = 1/K
V2 = V1 / K
= 1000 / 5
= 200 V.
Sponsor Area
Mock Test Series
