Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Suppose the initial charge on the capacitor in Exercise 7.7 is 6 mC. What is the total energy stored in the circuit initially? What is the total energy at later time?

Q7.8  Suppose the initial charge on the capacitor in Exercise 7.7 is  6mC . What is the total energy stored in the circuit initially? What is the total energy at later time?

Answers (1)

best_answer

Given

Capacitance  C=30\mu F=30*10^{-6}

Inductance L = 27mH = 27*10^{-3}H

Charge on the capacitor  Q=6mC=6*10^{-3}C

Now,

The total energy stored in Capacitor :

E=\frac{Q^2}{2C}=\frac{(6*10^{-3})^2}{2*30*10^{-6}}=\frac{6}{10}=0.6J

Total energy later will be same because energy is being shared with capacitor and inductor and none of them loses the energy, they just store it and transfer it.

Posted by

Pankaj Sanodiya

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Single Girl Child Scholarship 2024: Registration extended till January 10; eligibility criteria, amount
anam-khan

CBSE wants international boards reined in; letter to education ministry seeks directions for AIU
anam-khan

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14

Latest Question