Click to Chat

1800-2000-838

+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
Amit Saxena Grade: upto college level
`         The time constant of an LR circuit is 40 ms. The circuit is connected at t = 0 and the steady-state current is found to be 2.0 A. Find the current at (a) t = 10 ms (b) t = 20 ms, (c) t = 100 ms and (d) t = 1 s.`
3 years ago

## Answers : (1)

Jitender Pal
365 Points
```										Sol. τ = 40 ms
i0 = 2 A
a) t = 10 ms
i = i0 (1 – e^–t/τ) = 2(1 – e^–10/40) = 2(1 – e^–1/4)
= 2(1 – 0.7788) = 2(0.2211)A = 0.4422 A = 0.44 A
b) t = 20 ms
i = i0 (1 – e^–t/τ) = 2(1 – e^–20/40) = 2(1 – e^–1/2)
= 2(1 – 0.606) = 0.7869 A = 0.79 A
c) t = 100 ms
i = i0 (1 – e^–t/τ) = 2(1 – e^–100/40) = 2(1 – e^–10/4)
= 2(1 – 0.082) = 1.835 A =1.8 A
d) t = 1 s
i = i0 (1 – e^–t/τ) = 2(1 – e^-1/40x10^-3) = 2(1 – e^-10/40)
= 2(1 – e^–25) = 2 × 1 = 2 A

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

## Other Related Questions on Electromagnetic Induction

View all Questions »
• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details
• Electricity and Magnetism
• OFFERED PRICE: Rs. 1,696
• View Details