1.

A series combination of `N_(1)` capacitors (each of capacity `C_(1)`) is charged to potential difference 3V. Another parallel combination of `N_(2)` capacitors (each of capacity `C_(2)`) is charged to potential difference V. The total energy stored in both the combinations is same, The value of `C_(1)` in terms of `C_(2)` isA. `(C_(2)N_(1)N_(2))/(9)`B. `(C_(2)N_(1)^(2)N_(2)^(2))/(9)`C. `(C_(2)N_(1))/(9N_(2))`D. `(C_(2)N_(2))/(9N_(1))`

Answer» Correct Answer - A
In the first condition,
`C_(eq)=(C_(1))/(N_(1))`
Potential difference `(V)=3V`
`:.` Energy stored `(E_(1))=(1)/(2)CV^(2)`
`=(1)/(2)((C_(1))/(N_(1)))(3V)^(2)`
`=(9C_(1))/(2N_(1))` . . . (i)
In the second condition,
`C_(eq)=N_(2)C_(2)`, potential difference =V
energy stored `(E_(2))=(1)/(2)CV^(2)`
`=(1)/(2)N_(2)C_(2)V^(2)` . . . (ii)
According to the question,
`E_(1)=E_(2)`
From Eqs. (i) and (ii), we get
`(9)/(2)(C_(1))/(N_(1))V^(2)(1)/(2)N_(2)C_(2)V^(2)`
`C_(1)=C_(2)(N_(2)N_(1))/(9)`


Discussion

No Comment Found

Related InterviewSolutions