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_(1)`
D. `(16C_(1))/(n_(1)n_(2))`

Answer

D

Explanation

A series combination of `n_(1) ` capacitors each of capacitance `C_(1)` are connecte to 4V source as shown in the figure .
Total capaciture of the series combination of the capacitors is given by .
`U_(s) = (1)/(2)C_(s)(4V)^(2) = (1)/(2) ((C_(1))/(n_(1)))(4V)^(2) " "("Using (i))......(ii)")`
A parallel combination of `n_(2)` capacitors each of capacitance `C_(2)` are connected to V source to V source as shown in the figure.
Total energy stored in a parallel combintion 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)`
`(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) = (16C_(1))/(n_(1)n_(2))`