Questions
A 2 F capacitor is charged as shown in the figure. The percentage of its stored energy dissipated after the switch S is turned to position 2 is :
detailed solution
Correct option is D
Ui=12CV2=12×2×V2=V2 qi=2VWhen switch S is turned to position 22V−q2=q8⇒q=8V5 ∴Energy dissipated =V2−64V22×25×8+4V22×25×2=4V25Talk to our academic expert!
Similar Questions
Two identical capacitors, have the same capacitance C. One of them is charged to potential V1 and the other to V2. The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is
799 666 8865
support@infinitylearn.com
6th Floor, NCC Building, Durgamma Cheruvu Road, Vittal Rao Nagar, HITEC City, Hyderabad, Telangana 500081.
free study material
JEE Mock Tests
JEE study guide
JEE Revision Notes
JEE Important Questions
JEE Sample Papers
JEE Previous Year's Papers
NEET previous year’s papers
NEET important questions
NEET sample papers
NEET revision notes
NEET study guide
NEET mock tests
