ToughSTEM
Sign Up
Log In
ToughSTEM
A question answer community on a mission
to share Solutions for all STEM major Problems.
Cant find a problem on ToughSTEM?
0
The maximum allowable potential difference across a 200mH inductor is 350V . You need to raise the current through the inductor from 1.5A to 3.5A .

What is the minimum time you should allow for changing the current?

Express your answer to two significant figures and include the appropriate units.

t_min= ____________________
Edit
Added Thu, 06 Aug '15
Community
1
Comment
Add a Comment
Solutions
0
apply induced emf e = L di/dt

so di/dt = 350/0.2

di/dt = 1750

dt = (3.5-1.5)/1750

dt = 1.14 msecs
Edit
Added Thu, 06 Aug '15
Community
1
Comment
Add a Comment
Add Your Solution!
Close

Click here to
Choose An Image
or
Get image from URL
GO
Close
Back
Add Image
Close
What URL would you like to link?
GO
α
β
γ
δ
ϵ
ε
η
ϑ
λ
μ
π
ρ
σ
τ
φ
ψ
ω
Γ
Δ
Θ
Λ
Π
Σ
Φ
Ω
Copied to Clipboard

Add Your Solution
Sign Up
to interact with the community. (That's part of how we ensure only good content gets on ToughSTEM)
OR
OR
ToughSTEM is completely free, and its staying that way. Students pay way too much already.
Almost done!
Please check your email to finish creating your account!
Welcome to the Club!
Choose a new Display Name
Only letters, numbers, spaces, dashes, and underscores, are allowed. Can not be blank.
Great! You're all set, .
A question answer community on a mission
to share Solutions for all STEM major Problems.
Why
The Purpose
How
The Community
Give Feedback
Tell us suggestions, ideas, and any bugs you find. Help make ToughSTEM even better.