Lab 8

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Oklahoma State University *

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3114

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Electrical Engineering

Date

Jan 9, 2024

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docx

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Lab 8 Generator and AC Current Part I Generator Open this electric generator simulation • Keep all unchanged and select “the motion and both graphs” option (located in the bottom of the simulation). Now, click on play button, observe the coil moving in the magnetic field. Is there any change in flux inside the coil? Yes, there is a change in the flux • What kind of power is generated, AC or DC? AC • Now Reset and Pause the simulation just when both plots are done, so you can see both the flux and current plots. • What is the rotation period? 4.0s • What is the rotation frequency? 0.25 Hz • The current is proportional to change of flux (slop), what is the current when flux is at maximum? 0A • Use the maximum flux; calculate the maximum generated emf (revisit the section on this topic if necessary). please show your calculation in detail emf(max)=N(phi)w=N(phi)2(pi)f emf(max)=1*1*2(pi)*.25=1.57V The simulation doesn’t list exactly how many turns there are so I used 1 turn. Max flux was 1 because of the graph that indicated the max is 1. Part II Open EMF in inductors simulation • Turn the switch on and off a few times to observe the light bulb. Read through the tutorial to see if what you have observed makes sense. • Now concentrate on turn off the switch action, at the instance the switch is off, do you see the light bulb light up? What provided the emf for the light bulb to light up, why the light bulb went off eventually? When the switch is turned off, the lightbulb flashes. The back emf from the resistor as a result of

Lenz’s law which causes a current to flow in a direction that opposes the change in flux from the resistor. This means that the electricity will briefly pass through the lightbulb causing it to flash. Once the magnetic field in the inductor is steady, it puts an end to the back emf and the electricity goes back to taking the path of least resistance, meaning it will avoid the lightbulb. Part III Open the Transformer and Transmission Line Simulation You can read through the tutorial see if you can understand the need for stepping up the voltage on the transmission line. See if you understand why the light bulb is not as bright when you switch the position to low voltage as compare to switch the position to high voltage. After you get some general idea, do the following calculations ( please use all the values given here to do all the calculations, not the value in the simulation!!! ). • Set the switch position to low voltage (let the coil count at the step-up transformer end to be 1:1) • The input voltage (read from the simulation): I=P/V =60/110 =0.55 A • Assume the light bulb at the input end is 60 watts, calculate the input current or the current on the light bulb (left side, input side) • Output voltage (at the input end): Power output =60*0.70= 42 W V=Pl =42W * 0.55A = 23.1 V • Output current: Vp/Vs=Is/Ip 110V/23.1v=Is/0.55A 4.76=Is/0.55A Is=2.62A • Let the resistor on the transmission wire to be 40Ω, what is potential drop on the wire, what is the power wasted on the transmission wire: V=IR

=2.62*40 =104.8V P(lost)=I^2*R 2.62^2*40 = 274.57W • The potential difference on the step-down end (the output subtracts out the potential difference on the transmission line): 110V-13.1V=96.9V • The step-down end coil count is also (1:1), find the power output on the light bulb on the step-down end (right side). 60W-274.5=-214 • Does your result agree with what brightness you see on the bulb? Yes. • Set the switch position to the high voltage, and assume the coil count at step up transformer • Output voltage at the step-up transformer. 750V • The input and output coil ratio: (hint: calculate from the given voltage on the simulation) • Output current (hint: power can never step up, when you step up voltage, the current must step down): Vp/Vs=IS/IP =110/750=Is/0.55 • Let the resistor on the transmission wire to be 5Ω, what is potential drop on the wire? what is the power wasted on the transmission wire? V=IR =0.081*5 =0.405V P(LOST)=I^2*R =(0.081^2)*5 =0.033 W • The potential difference on the step-down end (the output subtracts out the part on the transmission line):

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750V-0.405V =749.6V • With the same ratio as you calculated in b), the voltage on the light bulb on the step- down end: • The current on the step-down end light bulb 749.6/Vs=10/1 0.81A=Is • Power output on the step-down end light bulb P=IV 0.81^2*74.95 =60.7W • Does your result agree with what brightness you see on the bulb? Yes