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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))` |
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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)` |
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