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
19.1k views
in JEE by (35 points)
A 5uf capacitor having a charge of 20u C is discharged through a wire of resistance 5 ohm. Find the heat dissipated in the wire between 25-50 u secafter the connections are made.
by (194 points)
Your question seems incomplete as the answer is in terms of e

Please log in or register to answer this question.

1 Answer

0 votes
by (1.9k points)

The potential difference across the capacitor

V0 = Q/C = 20/5 = 4V

The initial current,

i0 = V/R = 4/5 = 0.8A

τ = CR = 20*5 = 100 µs

The current in the resistor at any time t

i = i0e-t/τ

The heat dissipated,

After substituting and simplifying, we get

H = 4.7 µJ

by (194 points)
You answer could be probably wrong as time constant is not equal to 100 µs, it is 25 µs

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

...