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Question Number 12683 by tawa last updated on 29/Apr/17

A coil of inductance 0.12Hz and resistance 4 Ω is connected across a 240 v,   50 Hz is supplied . Calculate the current on the load (π = 3.142).

$$\mathrm{A}\:\mathrm{coil}\:\mathrm{of}\:\mathrm{inductance}\:\mathrm{0}.\mathrm{12Hz}\:\mathrm{and}\:\mathrm{resistance}\:\mathrm{4}\:\Omega\:\mathrm{is}\:\mathrm{connected}\:\mathrm{across}\:\mathrm{a}\:\mathrm{240}\:\mathrm{v},\: \\ $$$$\mathrm{50}\:\mathrm{Hz}\:\mathrm{is}\:\mathrm{supplied}\:.\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{current}\:\mathrm{on}\:\mathrm{the}\:\mathrm{load}\:\left(\pi\:=\:\mathrm{3}.\mathrm{142}\right). \\ $$

Answered by ajfour last updated on 29/Apr/17

I_(rms) =(E_(rms) /Z)  Z=(√(R^2 +4π^2 ν^2 L^2 ))      =(√(16+4(9.86)(2500)(0.12)^2 ))      =37.9 Ω  I_(rms) =((240V)/(37.9Ω)) =6.33 A .

$${I}_{{rms}} =\frac{{E}_{{rms}} }{{Z}} \\ $$$${Z}=\sqrt{{R}^{\mathrm{2}} +\mathrm{4}\pi^{\mathrm{2}} \nu^{\mathrm{2}} {L}^{\mathrm{2}} }\: \\ $$$$\:\:\:=\sqrt{\mathrm{16}+\mathrm{4}\left(\mathrm{9}.\mathrm{86}\right)\left(\mathrm{2500}\right)\left(\mathrm{0}.\mathrm{12}\right)^{\mathrm{2}} }\: \\ $$$$\:\:\:=\mathrm{37}.\mathrm{9}\:\Omega \\ $$$${I}_{{rms}} =\frac{\mathrm{240}{V}}{\mathrm{37}.\mathrm{9}\Omega}\:=\mathrm{6}.\mathrm{33}\:{A}\:. \\ $$

Commented by tawa last updated on 29/Apr/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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