A 5μF capacitor is charged fully by a 220V supply. It is then disconnected from the supply and is connected in series to another uncharged 2.5μF capacitor. If the energy change during the charge redistribution is X100J then value of X to the nearest integer is_______.

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

answer is 0004.00.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Energy lost during charge redistribution =12C1C2C1+C2V1−V22=125×2.55+2.5×220−02×10−6∵For uncharged capacitance V=0=56×2.22×10−2=X100⇒x=5×2.226=4.03

Watch 3-min video & get full concept clarity