IIT JEE , AIEEE Forum - Physics , Electricity

karthik2789

a capacitor of capacitance C is charged to potential V and then isolate.A small capacitor c is charged frm C ,discharged and charged again & process is repeated n times.due to this potential of lager capacitor is reduced to v . value of c is ????? plz give d deatailed solution!

iitkgp_bipin

Initially charge on C, q₀ = CV

When c is charged through C, charge remaining on C is :
q₁ = q₀C/(C+c) = (CV)C/(C+c)

Now c is discharged and again charged through C.
Charge remaining, q₂ = q₁C/(C+c) = q₀C²/(C+c)² = (CV)C²/(C+c)²

If this process is repeated n times then charge remaining on C is :
q_n = (CV)Cⁿ/(C+c)ⁿ

Potential of C is v now.
So, q_n = Cv
(CV)Cⁿ/(C+c)ⁿ = Cv
(1 + c/C)ⁿ = V/v

c = C{(V/v)^1/n - 1}

Ask & Discuss Questions with Community & Experts