JUST CHOOSE THE RIGHT CHOICE Since it is known that the magnetic permeability of the steel rod and the core of the electromagnet and the cross-sections are the same, the steps to be passed in order to find the current value that will lift the steel rod with a weight of G = 1 ton are correctly expressed in which option. μrpol=1;μrnuve=200;μr steel=200 1-) S=75.10-4; m2 Rnuve+Rsteel=(450+2*75+500)10-3/(200μoS) ;Rpoly=10-3/(200μoS); 2-) S=75.10-4; m2 Rnuve+Rsteel=(450+2*75+500)10-3/(200μoS) ;Rpoly=2*10-3/μoS); 3-) S=75.10-4; m2 Rnuve+Rsteel=(2*500+2*100)10-3/(200μoS) ;Rpoly=2*10-3/(200μoS); 4-)Rtotal=Rnuve+Rcelik+Rpoly; 5-) Φ is found from G=Φ2/(μoS). 6-) Φ is found from G=0.5o2/(μoS). 7-) There is I current from Φ=NI/Rtotal. 8-) Φ=NIRtotal I current is found. A. 2,4,5,7 B. 2,4,6,8 C. 3,4,6,8 D. 1,4,5,7
JUST CHOOSE THE RIGHT CHOICE
Since it is known that the magnetic permeability of the steel rod and the core of the electromagnet and the cross-sections are the same, the steps to be passed in order to find the current value that will lift the steel rod with a weight of G = 1 ton are correctly expressed in which option. μrpol=1;μrnuve=200;μr steel=200
1-) S=75.10-4; m2 Rnuve+Rsteel=(450+2*75+500)10-3/(200μoS) ;Rpoly=10-3/(200μoS);
2-) S=75.10-4; m2 Rnuve+Rsteel=(450+2*75+500)10-3/(200μoS) ;Rpoly=2*10-3/μoS);
3-) S=75.10-4; m2 Rnuve+Rsteel=(2*500+2*100)10-3/(200μoS) ;Rpoly=2*10-3/(200μoS);
4-)Rtotal=Rnuve+Rcelik+Rpoly;
5-) Φ is found from G=Φ2/(μoS).
6-) Φ is found from G=0.5o2/(μoS).
7-) There is I current from Φ=NI/Rtotal.
8-) Φ=NIRtotal I current is found.
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