1.

Two capacitors A and B with capacities C_1 and C_2 are charged to potential difference of V_1 and V_2 respectively. The plates of the capacitors are connected as shown in the figure with one wire free from each capacitor. The upper plate of A is positive and that of B is negative. An uncharged capacitor of capacitance C_3 and lead wires falls on the free ends to complete circuit, then

Answer»

the final charge on each capacitor are same to each other
the final sum of charge on PLATES a and d is `C_1V_1`.
the final sum of charge on plates B and g is `C_2V_2 - C_1V_1`
both (b) and(c ) are correct.

Solution :d. Charge on positive plate of `A` is `C_(1)V_(1)`.
Charge on negative plate of `B` is `-C_(2)V_(3)`. When `d` plate of capacitor `C` is connected with the plate of ``, then the total charge of `(d,a)` plate system will be `C_(1)V_(1)` (conservation of charge). SIMILARTY on `(c,s)` plates, total charge will be `-C_(2)V_(2)`. The total charge on `(h,g)` plate system will be `+C_(2)V_(2)-C_(1)V_(1)`.


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