webassign.net Inbox 56@gmail.com-Gmail HW#1 4/15 Inductor circuit Course: EC 202
ID: 1791680 • Letter: W
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webassign.net Inbox 56@gmail.com-Gmail HW#1 4/15 Inductor circuit Course: EC 202 (001) Fall 2017 Principles of Ma ics 1. -1 points GianPSE4 29.P038. My Notes A simple generator has a 430 loop square coil 21.0 cm on a side. How fast must it turn in a 0.630 T field to produce a 120 V peak output? rev/s 2. -13 points GianPSEWA4 29.P040.Tutorial My Notes (a) What is the voltage output of a transformer used for rechargeable flashlight batteries if its primary has 481 turns, its secondary has 8 turns, and the input voltage is 120 V? (b) What input current is required to produce a 4.00 A output? mA (c) What is the power input? W Tutorial 3. -12 points GianPSE4 30.P.001 My Notes A 2.28 m long coil containing 300 loops is wound on an iron core (average -2000 0) along with a second coil of 100 loops. The loops of each coil have a radius of 3.00 cm. Assume the current in the first coil drops uniformly from 12.0 A to zero in 98.0 ms. (a) Determine the mutual inductance M (b) Determine the emf induced in the second coilExplanation / Answer
magnetic fleux = N*B*A*coswt
emf = rate of change in flux = N*B*A*sinwt*w
peak emf = N*B*A*w
120 = 430*0.63*0.21^2*w
w = 10 rad/s
w = 1.6 rev/s <<<<----ANSWER
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(a)
Vin/Vout = Np/Ns
120/Vout = 481/8
vout = 1.99 volts
(b)
Iin/Iout = Ns/Np
Iin/4 = 8/481
Iinput = 66.5 mA
(c)
power = Vin*Iin = 120*66.5*10^-3 = 7.98 W
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3)
mutual inductance M = u*N1*N2*pi*r2^2/L
M = 2000*4*pi*10^-7*300*100*pi*0.03^2/2.28
M = 0.0935 H
M = 93.5 mH
(b)
emf = M*dI/dt
emf = 0.0935*(12-0)/0.098
emf = 11.45 V
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