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A DC self excited shunt generator has a field resistance of 160ohms generated emf is 420v and supplies a load of 55kw how would u go about calculating armature resistance when terminal voltage is 408v I cant seem to get my head around it. Would the following be correct

I=P/V = 55000/408 = 134.8a

E=V+IaRa  therefore

420=408+IaRa

Voltage drop between E and V = IaRa so

420-408 =IaRa

12=IaRa therefore

Ra=12/Ia

Ra=12/134.8

Ra=0.089 

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    \$\begingroup\$ This seems to be a study or homework problem that may be missing some details and shows no documentation of an effort to solve. The first step would be to draw and label the equivalent circuit. See what you can calculate with what you know. \$\endgroup\$ – Charles Cowie Oct 3 '19 at 21:39
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    \$\begingroup\$ I'm voting to close this question as off-topic because it appears to be a study or homework problem with no effort to solve demonstrated. \$\endgroup\$ – Charles Cowie Oct 3 '19 at 21:40
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    \$\begingroup\$ Welcome to StackExchange. We will help, but we do not do the heavy lifting. You learn nothing by one of us doing your homework. Follow Charles' advice and add a schematic putting what you know on it. There is a schematic editor built into the site. A schematic of a DC circuit is a great way to organize your thoughts. Then explain what you understand and where you are stuck. Otherwise I predict your question will be closed. Clock is ticking. \$\endgroup\$ – StainlessSteelRat Oct 3 '19 at 22:31
  • \$\begingroup\$ I do apologise I have not given a lot of info. I would not want anybody to do it for me because as you say I will not learn from it. I have been working on it and will put what I have so far on asap. Thank you \$\endgroup\$ – Oli Oct 4 '19 at 22:24

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