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Two condensers `C_(1)` and `C_(2)` in a circuit are jhioned as shown in . The potential fo point `A` is ` V_1` and that of `B` is ` V_2` . The potential of point `D` will be ` A. `(1)/(2) (V_(1) + V_(2))`B. `(C_(2)V_(1) + C_(1)V_(2))/(C_(1) + C_(2))`C. `(C_(1)V_(1) + C_(2)V_(2))/(C_(1) + C_(2))`D. `(C_(2)V_(1) - C_(1)V_(2))/(C_(1) + C_(2))` |
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Answer» Correct Answer - C Charge on `C_(1) =` charge on `C_(2)` `rArr C_(1) (V_(A - V_(D)) = C_(2) (V_(D) - V_(B))` `rArr C_(1) (V_(1) - V_(D)) = C_(2) (V_(D) - V_(2)) rArr V_(D) = (C_(1)V_(1) + C_(2)V_(2))/(C_(1) + C_(2))` |
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