PDA

View Full Version : A 5 H inductor is subjected to an electric current that changes at a rate of 4.5 amps



cnowak
September 13, 2021, 01:37 PM
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

Kalbi_Rob
September 14, 2021, 06:46 AM
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

briantilley123
September 16, 2021, 05:18 AM
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 :)

Izeldeen
November 7, 2021, 08:16 PM
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

👍