CapacitanceHard
Question
A capacitor of capacitance C0 is charged to a potential V0 and then isolated. A capacitor C is then charged from C0, discharged and charged again ; the process is repeated n times. Due to this, potential of the larger capacitor is decreased to V, then value of C is :
Options
A.C0 [V0/V]1/n
B.C0 [(V0/V)1/n − 1]
C.C0 [(V0/V) − 1]
D.C0 [(V0/V)n − 1]
Solution
Charge on C0, Q1 = C0V0,
Initial charge on C1, Q2 = 0
Common potential V1 =
⇒ Q1 = C0V1 = 
V0
Similarly V2 =
V0 ⇒ Q2 = C0V2 = 
V0
for n times Vn =
V0 = V ⇒ C =
C0 Ans
Initial charge on C1, Q2 = 0
Common potential V1 =
⇒ Q1 = C0V1 = V0Similarly V2 =
V0 ⇒ Q2 = C0V2 = V0for n times Vn =
V0 = V ⇒ C =C0 Ans Create a free account to view solution
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