A capacitor of is charged as shown in the diagram. When the switch S is turned to position 2, the percentage of its stored energy dissipated 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

75%

20%

80%

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Initial energy stored $= \frac{1}{2} (2 μF) \times V^{2}$
Energy dissipated on connection across 8µF
$= \frac{1}{2} \frac{c_{1} c_{2}}{c_{1} + c_{2}} V^{2} = \frac{1}{2} \times \frac{2 μF \times 8 μF}{10 μF} V^{2}$
$∴ % loss of energy = \frac{1 .6}{2} \times 100 = 80 %$

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

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

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE