 Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

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
`        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 Navjyot Kalra
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 ?

## Other Related Questions on Electric Current

View all Questions »  ### 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