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

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

answer is 0004.00.

(Unlock A.I Detailed Solution for FREE)

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