JEE MAIN - Physics (2020 - 5th September Morning Slot - No. 5)
Two capacitors of capacitances C and 2C are
charged to potential differences V and 2V,
respectively. These are then connected in
parallel in such a manner that the positive
terminal of one is connected to the negative
terminal of the other. The final energy of this
configuration is :
Zero
$${3 \over 2}C{V^2}$$
$${9 \over 2}C{V^2}$$
$${{25} \over 6}C{V^2}$$
Explanation
_5th_September_Morning_Slot_en_5_1.png)
By conservation of charge,
$${q_1} + {q_2} = {q_1}' + {q_2}'$$
$$ \Rightarrow $$ $$ - CV + (2C)(2V) = (C + 2C)V'$$
$$V' = {{3CV} \over {3C}} = V$$
$${U_f} = {1 \over 2}C{v^2} + {1 \over 2}(2C){V^2}$$
$${U_f} = {3 \over 2}C{V^2}$$
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