Management i Help Homework Ch.23 Begin Date: 3/23/201 12:00:00 AM Due Date: 4ni2

Question

Management i Help Homework Ch.23 Begin Date: 3/23/201 12:00:00 AM Due Date: 4ni2017 00:00 AM End Date: 4/7/2017 1200 00AM 7%) Problem 10: Aprecision laboratory resistoris made of a coil of wire. The coilis 1.65 cm in diameter 275 cm long, and has 500 tums. 4 33% Part (a) What selfinductance in mH? t average emfin v is induced if the 12.0 Acurrent through it is turned on in 500 ms (one-fourth of cycle for 50 Hz AC)? A 33% Part (e) What is its inductance if it is shortened to half its length and counterwound (two layers of 250 tums in opposite directions? Grade Summary Potential 100% 7 8 9 cotano asino acos() Attempts remaining: 4 5 6 per attempt) atano acotano sinho cotanh0 Degrees ORadians give up! Hints: 1 deduction per hint. Hints remaining: Feedback deduction per feedback.

Explanation / Answer

Given that

Length L=1.65*10^-2 m

diameter D=2.75*10^-2 m

number of turns N=500 turns

now we find the self inductance

self inductance L=4*3.14*10^-7*500^2*3.14*(2.75*10^-2/2)^2*1.65*10^-2=30.8mH

now we find the induced emf

the induced emf e=30.8*10^-3*12/5*10^-3=73.92 v

now we find the self inductance

self inductance L=4*3.14*10^-7*250^2*3.14*(2.75*10^-2/2)^2*(1.65*10^-2/2)

=3.845 mH

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