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Exam (elaborations)

Mod EM1 Practice Exam

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Mod EM1 Practice Exam

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Mod EM1 Practice Exam


Why cdoes cSwinburne's ctest cgive ca chigher cefficiency cthan
cHopkinson's ctest?


Answer:

Swinburne’s ctest cis ca cno-load ctest cperformed con cDC cmachines cto cdetermine
cconstant closses c(iron cand cmechanical), cassuming cthem cto cbe cthe csame cat
cfull cload. cHowever, cit cdoesn't cconsider cload-dependent closses, clike:


● Increase cin carmature ccopper closses cdue cto carmature creaction cand
cdistortion.


● Increase cin cstray cload closses.


Hopkinson’s ctest, con cthe cother chand, cis ca cfull-load ctest cwith cboth cmachines
cunder coperating cconditions, cso:


● It cconsiders call creal closses cduring cactual coperation.

● Hence, cit cshows ca clower cefficiency cthan cSwinburne’s ctest, cwhich
cslightly coverestimates cefficiency.




2. Given:

● Supply cVoltage c= c220V

● Armature cresistance c(Ra) c= c0.2Ω

● Field cresistance c(Rsh) c= c110Ω


(i) Starting ccurrent cwithout ca cstarter:

, At cstarting, cEb=0E_b c= c0Eb=0

Istart=VRa=2200.2=1100 cAI_{\text{start}} c= c\frac{V}{R_a} c= c\frac{220}
{0.2} c= c1100 c\text{
A}Istart=RaV=0.2220=1100 c A

(ii) Running carmature ccurrent cat cfull cload:

Assuming cback cEMF cEbE_bEb cis cgiven c(not creadable cin cthe cscan, cassuming
cEb=210VE_b c= c210VEb=210V):


Ia=V−EbRa=220−2100.2=50 cAI_a c= c\frac{V c- cE_b}{R_a} c= c\frac{220 c-
c210}{0.2} c= c50 c\text{ cA}Ia=RaV−Eb=0.2220−210=50 cA

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