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,rotate
After connecting them distribute charges such that the potential drop of 1 capacitor=potential increase accross 2nd capacitor. Further, keep in mind the law of conservation of charge.
-3CV and +3CV total charges on left and right (say) part of the system. So, distribute it as -2CV and -CV on the negative parts.
Net energy is (final) $\frac{CV^{2}+2CV^{2}}{2}=\frac{3CV^{2}}{2}$
SirLancelotDuLac
@Gyro Gearloose
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.I didnt understand what to do after this step
By charge conservation, total charge on left hand side must be equal to -3CV and smlrly, on right hand side it should be +3CV right?
Then V=q/c must be constant for both the capacitors. Since capacitances are in ratio 1:2, charge must also be in the raio 1:2. So charges on both capacitors are CV and 2CV.
Oh okay i got it
If this is cleared you can close the thread.
+solved @SirLancelotDuLac
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