+91-8076753736[email protected]
logo
logo
-->

Electrostatic Potential And Capacitance

Question
CBSEENPH12039778
Wired Faculty App

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

  • 20%

  • 75%

  • 80%

  • 0%

Solution

C.

80%

Consider the figure given above.
When switch S is connected to point 1, then initial energy stored in the capacitor is given as,
straight E space equals space 1 half open parentheses 2 μF close parentheses xV squared
When the switch S is connected to point 2, energy dissipated on connection across 8μF will be,
straight E apostrophe space equals space 1 half open parentheses fraction numerator straight C subscript 1 straight C subscript 2 over denominator straight C subscript 1 plus straight C subscript 2 end fraction close parentheses. space straight V squared space space space space space equals space 1 half straight x fraction numerator 2 μFx 8 μF over denominator 10 μF end fraction space straight x space straight V squared space space space space space equals 1 half straight x space left parenthesis 1.6 space μF right parenthesis space straight x space straight V squared
Therefore, per centage loss of energy = fraction numerator 1.6 over denominator 2 end fraction x 100 space equals space 80 percent sign

Some More Questions From Electrostatic Potential and Capacitance Chapter

Mock Test Series

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation