A capacitor of capacitance C 0 is charged to a voltage V 0 and then isolated. An uncharged capacitor C is then charged from C 0 , discharged and charged again; the process is repeated n times. Due to this, potential of the C 0 is decreased to V, then value of C is:
Text Solution
Verified by ExpertsThe correct answer is:
B
Charge on C 0 , Q 1 = C 0 V 0 ,
Initial charge on C 1 , Q 2 = 0
Common potential V 1 = 
⇒ Q 1 = C 0 V 1 = 
V 0
Similarly V 2 = 
V 0
⇒ Q 2 = C 0 V 2 = 
V 0
for n times n V n = 
V 0 = V
⇒ C = 
C 0
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