Recent content by Kev1n

Is that what you meant

The Rp and Rs should be equal?

1. A large power transformer has a turns ratio N1 / N2 of 1:5. When the primary voltage is 400v and it is connected to a resistive load its secondary current is 100A. Calculate the required value of source resistance of the supply to give a maximum power trnasfer 2. Vp = 400v, Ratio 1:5...

1. Does this formula make sense 2B=Ф / A 3. need to check please

1. A loop of wire is placed between the poles of a magnet. The magnetic field can be assumed uniform between the poles. The formed by the loop within the field of the magnet is 20cm2. (i). Determine the magnitude of the flux density required to induce a maximum voltage of 1V when the loop is...

1. A loop of wire between the poles of a magnet, the magnetic field is uniform between the poles. The are of the loop of wire within the field is 20cm sqaured. Determine the magnitude of flux density required to induce a mx voltage of 1v when the loop is rotated at 600rpm 2. B= Flux x A...

Roger, of course - thanks for pointing that out, appreciated

Homework Statement A non ideal capacitor having a 'leaky' dielectric where RL is the leakage resistance measured across the dielectric. A 100pF parallel plate capacitor has a mica dielectric of relative permittivelty Er = 12 and resistivity of p = 1014 ohm m. Calculate the leakage resistance...

Dan, Did you ever get the formula as I am stuck on asimilar question

Should I have used d = Eo x Er x A / C 12 x 8.85x10-12 x 50 x 10-3 / 100 x 10-9 = 5.31x10-5 = 0.0513mm

1. A capacitor is formed by 2 parallel plates separated by a dielectric of relative permattivity er = 12, the capacitor plates are 5mm x 10mm. Calculate the thickness of the dielectric required to create a capacitor of 100pF 2. C = eA / d therefore d=eA/C, we have C 100 x 10-9, area of...

Hi, thanks for the reply, think I am having a christmas moment. The area should be 10mm x 20mm = 200mm in m 200 x 10[SIZE="2"]-3 therefore 0.2m Thickness = 5,000 x 0.2/ 1,000 = 1mm How does this seem

Thanks: Thickness = 5,000 x 0.0002/ 1,000 = 1 x 10[SIZE="2"]-3m therefore 1mm or do I have to sq the .ooo2 giving: =5,000 x 4x10[SIZE="2"]-8 / 1,000 = 2 x 10[SIZE="2"]-7

1. A conductive sheet of material has a CSA of 10mm x 20mm. Reisistivity of 103 (10 to the 3, 10,000) ohms M. Calculate thickness of the material required to give resistance of 5k ohms 2. Ro = RA/l , where l would be thickness, l = RA/Ro 3. Thickness = 5,000 x 0.2 / 10,000 = o.1m...

Thanks for advise