- #1
Ajay.makhecha
- 10
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Homework Statement
A capacitor of capacitance 5µF is charged to 24 V and another capacitor of capacitance 6µF is charged to 12 V. The positive plate of first capacitor is now connected to negative plate of second and vice versa.Fid ew charges on capacitor.
Homework Equations
Q=CV
E= QV/2 = CV^2 / 2
The Attempt at a Solution
After the capacitors are connected , finally potential difference between both the capacitors would be same. Hence final charges would be in the ratio of their capacitances. Hence it must be 960/11 µC and 1152/11 µC.
But the answer is given as 21.8µC and 26.2µC