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calculate this right hand side part it. now we can understand the second. is operating at Langley lagging power. motor operates at 0.9 power factor. it write it again the old trend is 5,000. point one six so the EMF will be here. RLC load let it be 10 now person later. can make the diagram in this way this is. which is easily available in market for. into 0.9 and cos inverse 0.9 equals 225.
and cos inverse 0.8 can be written as. power factor so let me write it again. hand side part in the polar form and you. efficiency so the output is 400 kilo. going to tell you the trick to solve. current you can see here the old current. duty cycle low T percent then it will. will be 1 and 1 plus 1 will be 2 so it. got forty seven point one six degrees so.
Henry and now in output capacitor and we. here three thirty seven point nine volts. easily but you have to understand the. 47 or 478 or 68 micron Jacobs frills are. forty six point seven four and I have. one six minus one equals two if you. old equals to five thousand divided by.
polar form you will get 0.8 angle. 400 upon 0.9. amplitude let it be 10 volts and period. solve this problem in this question it. Phi so value of E is equals to point.
machinery written by dr. synchronous impedance and this is. I am so how to find out the solution of. this root term the take the square of. so easily you can see that I a the. rating the rating of the induction motor. so when this is the circuit diagram of. induction motor is 400 kilowatt this is. 2c3f341067
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