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

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## 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

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

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## 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

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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
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