Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

Figure shows two identical parallel plate capacitors connected to a battery through a switch S. Initially, the switch is closed so that the capacitors are completely charged. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant 3. Find the ratio of the initial total energy stored in the capacitors to the final total energy stored

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya

a

9:16

b

5:9

c

2:3

d

3:5

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

Initial total energy =12CV2+12CV2=CV2When the switch is open, dielectric is introduced. Then capacitance, C = KC = 3C. Energy stored in C is =123CV2=32CV2Since the switch is open, charge will be the same in B so energy is B=12C3V2So total final energy =32CV2+16CV2=9CV2+1CV26=106CV2So required ratio =CV2106CV2=35=3:5
Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon
personalised 1:1 online tutoring