MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade: 11
        A capacitor of capacitance 10 μF is connected to an oscillator giving an output voltage ℰ = (10 V) sin ωt. Find the peak currents in the circuit for ω = 10 s^-1, 100 s^-1, 500 s^-1, 1000 s^-1.
5 years ago

Answers : (1)

Navjyot Kalra
askIITians Faculty
654 Points
							Sol. C = 10 μF = 10 * 10^-6 F = 10^-5 F
E = (10 V) Sin ωt
a) I = E base 0/Xc = E base 0/(1/ωC) = 10/(1/10*10^-5) = 1 * 10^-3 A
b) ω = 100 s^-1
I = E base 0/(1/ωC) = 10/(1/100 * 10^-5) = 1 10^-2 A = 0.01
c) ω = 500 s^-1
I = E base 0/(1/ωC) = 10/(1/500 * 10^-5) = 5 * 10^-2 A = 0.05 A
d) ω = 1000 s^-1
I = E base 0/(1/ωC) = 10/(1/1000 * 10^-5) = 1 * 10^-1 A = 0.1 A

						
5 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

  • 101 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 110 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details