A series combination of `n_(1)` capacitors, each of value `C_(1)`, is charged by a source of potential difference `4 V`. When another parallel combination of `n_(2)` capacitors, each of value `C_(2)`, is charged by a source of potential difference `V`, it has same (total) energy stored in it, as the first combination has. the value of `C_(2)`, in terms of `C_(1)`, is then

Options

A. `(2C_(1))/(n_(1)n_(2))`
B. `16(n_(2))/(n_(1))C_(1)`
C. `2(n_(2))/(n_(1))C`
D. `(16 C_(1))/(n_(1)n_(2))`

Answer

D

Explanation

A series combination of `n_(1)` capacitors each of capacitance `C_(1)` are connected to 4V source as shown in the figure.

Total capacitance of the series combination of the capacitors is
`(1)/(C_(s)) = (1)/(C_(1)) + (1)/(C_(1)) + (1)/(C_(1)) +`.... upto `n_(1)` terms `= (n_(1))/(C_(1))`
or `C_(s) = (C_(1))/(n_(1))` ...(i)
Total energy stored in a series combination of the capacitors is
`U_(s) = (1)/(2)C_(s)(4V)^(2) = (1)/(2)((C_(1))/(n_(1))) (4V)^(2)` (Using (i))
A parallel combination of `n_(2)` capacitors each of capacitance `C_(2)` are connected to V source as shown in figure.

Total capacitance of the parallel combination of capacitors is
`C_(p) = C_(2) + C_(2) +....+` upto `n_(2)` terms `= n_(2)C_(2)`
or `C_(p) = n_(2)C_(2)` ...(iii)
Total energy stored in a parallel combination of capacitors is
`U_(p) = (1)/(2)C_(p) V^(2) = (1)/(2)(n_(2)C_(2))(V)^(2)` (Using (iii)) ...(iv)
According to the given problem, `U_(s) = U_(p)`
Substituting the values of `U_(s)` and `U_(p)` from equations (ii) and (iv), we get
`(1)/(2)(C_(1))/(n_(1)) (4V)^(2) = (1)/(2)(n_(2)C_(2))(V)^(2)`
or `(C_(1)16)/(n_(1)) = n_(2)C_(2)` or `C_(2) = (16 C_(1))/(n_(1)n_(2))`