×

# A 5 H inductor is subjected to an electric current that changes at a rate of 4.5 amps

###### Jump to latest post
1.  Junior Member
Join Date
Jun 2021
Posts
17
Reputation  ## A 5 H inductor is subjected to an electric current that changes at a rate of 4.5 amps

A 5 H inductor is subjected to an electric current that changes at a rate of 4.5 amps per second. How much voltage will be dropped by the inductor?

11.25
22.50
37.50
45.00  Reply With Quote

2. Originally Posted by cnowak A 5 H inductor is subjected to an electric current that changes at a rate of 4.5 amps per second. How much voltage will be dropped by the inductor?

11.25
22.50
37.50
45.00
V = L (di/dt) or L*((I1-I0)/(t1-t0))

V = 5H ((4.5A-0)/(1s-0s)) = 5H*(4.5A/1s) = 5H*4.5(A/s) = 22.5 V  Reply With Quote

3.  Junior Member Pro Subscriber
Join Date
Jun 2021
Posts
2
Reputation ## Sep 21' NETA Level II Test Topics

Topics not covered in/not covered enough the practice tests include:

Common IEEE Device number names

Recloser functions and definitions

MEDIUM VOLTAGE BREAKERS!!!!!

Operations and construction of amp/volt/ohm/mego meters

Oil sample analysis elements/ symptoms

LOTS OF ARC FLASH ratings and boundaries.

Confined spaces definitions/ rules and regulations/ person in charge

schematic diagrams/ diagnosing interlocks

Happy studying   Reply With Quote

4.  Member
Join Date
Jul 2020
Location
Washington, United State
Posts
44
Reputation   Originally Posted by Kalbi_Rob V = L (di/dt) or L*((I1-I0)/(t1-t0))

V = 5H ((4.5A-0)/(1s-0s)) = 5H*(4.5A/1s) = 5H*4.5(A/s) = 22.5 V
👍  Reply With Quote

electrical theory, neta exam questions, neta level 2  