Class Management Help hapter 23ABegin Date: 3/19/2017 00:00 AM Due Date: 3/26/20

Question

Class Management Help hapter 23ABegin Date: 3/19/2017 00:00 AM Due Date: 3/26/201 11:59:00 PM End Date: 5/22/2017 12:00:00 AM 4 33% Part (a) What is their total self-inductance, assuming they act like a single solenoid, in H? Grade Summary Potential 100% sino coso tano 7 8 9 Attempts remaining: 8 4 5 6 cota no asino acoso (05 per attempt) 1 2 3 ata no acotan0 sinh0 ODegrees Radians I give up! Hints: 0% deduction per hint. Hints remaining: 2. Feedback 0% deduction per feedback. l A 33% Part (e) what average emf in mv opposes shutting them offif this is done in 5.00 ms (one-fourth of a cycle for 50 Hz AC)? All content 2017 Expert TA LLC

Explanation / Answer

Solenoid inductance is given by

L=uN*N*A/l

We knoe N=400

l=1.25

diameter=0.85 cm=0.0085 m

We got inductance=0.009127 mH or 9.127*10(-6) H

Energy=0.5*L*i*i=2.08*10(-4) J or 0.208 mJ

Emf=Ldi/dt=3.8*10(-4) V

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