the last two calculation of this lab have me stumped. after inserting the wattme

Question

the last two calculation of this lab have me stumped. after inserting the wattmeter what calculations would you use to prove the values of the inductor and resistor. Basically I need help with step 5 to the end of the lab

U1 AC te-009Ohm AC 1e-0090hm R1 LS 3 RS 1k0D 0.5H 2000 T1 6 LP RP R2 50? 60Hz 0s U4 U3 AC 10MOhm AC 10MOhm Figure 4. Multisim diagram incorporating the simplified non-ideal transformer model. Run another simulation and compare the voltages, currents, and load power with the values obtained using the ideal transformer model. Tabulate this comparison and include the table in your report. What general conclusions can you draw regarding the behavior of the non-ideal versus the ideal model? 4. measurements on the transformer. Consider the following approach that is often used in practice: Install a wattmeter in the primary circuit on the load side of the voltmeter and ammeter. Note that this instrumentation will allow you to determine the voltage and current as well as the real, reactive, and complex powers delivered to the transformer provided that you assume (correctly) that the load is inductive with a lagging power factor 5. 6 Remove all components on the secondary of the transformer 7. Run a simulation under these "open-circuit" conditions. Observe from the circuit diagram that the effets of effeur LS and RS will be negligible compared with those of LP and RP while the ideal transformer has no whatsoever because it draws no current.

Explanation / Answer

You are basically testing the transformer in this lab.You are conducting open circuit test and Short circuit test.

In the case of open circuit test.We are neglecting the effect of Ls and Rs as compared to Lp and Rp.(For more on that read any standard textbook)

I will help you in looking at the calculations.

First of all the resistance of ammeter is negligible.So there will be no voltage drop across it.Secondly the Voltmeter has very high resistance,so it will draw very negligible current.

Now considering all the above points.

The Voltmeter is effectively in parallel with Rp and Lp.(Rs and Ls neglected).

So the voltage across them will be 6.43 (RMS).

The wattmeter is reading 3.982 mW.Now the active power consumed will be equal to (Vrms measured)^2/Rp .

Therefore 6.43^2/Rp=3.982mW which gives us Rp =10.382kOhms.

Now for getting Lo,we need to know what is Xp.

The current drawn by Rp will be Vrms/Rp.=Irp

The Current Dewan by Xp will be Vrms/Xp=Ixp

There phasor sum(magnitude =square root of (Irp^2+Ixp^2)).

This magnitude will be equal to Current measured by Ammeter(As voltmeter draws no current).

So Ixp^2=Iammeter^2-Irp^2

Irp=Vrms/Rp=6.43/10.382kOhms

Which gives us Irp=0.619mA.

Therefore square root of (0.824^2-0.619^2)=0.543mA

Hence Ixp=0.543mA

Therefore Xp=Vrms/0.543mA=6.43/0.543mA

Which gives us Xp=11.822kohms

Therefore 2*?*f*L=11.822kohms

Which gives us L=31.35H

So we get our valurs of Rp and Xp but the are slightly more than the values given in ckt.But the inaccuracy is justified by the ease in obtaining the values with minimal error.

Similarly for Short circuit test you have got a Voltage of 1.609 and you know current (5.819mA)through the Rs and Xs.You know what is the power consumed(6.773mW). Use I^2R to get value of R.(200.0250)

You can also get the Value of Z by V/I i.e (1.609/5.819mA)=276.5079Ohms.

Use Z^2-R^2 to get X^2.Then find Ls using 2 *?*fLs=X

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.