Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
+1 vote
48.3k views
in Physics by (60.8k points)

A capacitor of capacitance 5.00μF is charged to 24.0V and another capacitor of capacitance 6.0μF is charged to 12.0V. (a) Find the energy stored in each capacitor, (b) The positive plate of the first capacitor is now connected to the negative plate of the second and vice versa. Find the. new charges on the capacitors, (c) Find the loss of electrostatic energy during the process, (d) Where does this energy go?

Please log in or register to answer this question.

1 Answer

+1 vote
by (60.8k points)
edited by

C1 = 5μf V1 = 24V

q1 = C1V1 = 5 × 24 = 120μc and C2 = 6μf  

V2 = R

q2 = C2V2 = 6 × 12 = 72

∴ Energy stored on first capacitor

(c) U1 = (1/2) C1V2

U2 = (1/2) C2V2

Uf = (1/2)V2 (C1 + C2) = (1/2) (4.36)2 (5 + 6) = 104.5 × 10–6J = 0.1045mJ

But Ui = 1.44 + 0.433 = 1.873

∴ The loss in KE = 1.873 – 0.1045 = 1.7687 = 1.77mJ

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...