A capacitor of capacitance 900µF is charged by a 100 V battery. The capacitor is disconnected from the battery and connected to another uncharged identical capacitor such that one plate of uncharged capacitor connected to positive plate and another plate of uncharged capacitor connected to negative plate of the charged capacitor. The loss of energy in this process is measured as x × × 10 -2 J. The value of x is
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(225)
Q = CV = 900 x 10 -6 x 100 = 9 x 10 -2 = 90 MC
Now
Common potential will be developed across both capacitors by kVL.
Total charge on left plates of capacitors should be conserved.
90 mc + 0 = 2cv 0
cv 0 = 45 mc
Heat dissipated = U i -U f
[Change in energy stored in the capacitors]
=2.25Joule
OR
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