3-wire Edison Distribution Transformer Example

3-wire Edison Distribution Transformer Example - + P 4Ω +...

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EE369 Industrial Automation 3-Wire Edison Distribution Transformer Example Refer to the following diagram and calculate the currents I1, I2, I3 and In. Also determine the total secondary side (load) power. Solution: Vs1 = (208/2) / 4 = 26V Vs2 = (208/2) / 4 = 26V I3 = (Vs1+Vs2) / 10 = 5.2A Using mesh loop with Ia, Ib and Ic assumed in a clockwise direction around each of the three respective loops: Ic = I3 = 5.2A Loop a: 26 = 6 (Ia - Ic) Ia = (26/6) + 5.2 = 9.53A Loop b: 26 = 4 (Ib - Ic) Ib = (26/4) + 5.2 = 11.7A I1 = Ia = 9.53A I2 = Ib = 11.7A I3 = Ic = 5.2A In = Ib – Ia = 2.17A
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Power calculations: We find the individual power of each resistor and then we will sum the powers up to find the total power. P = I 2 R = (I1 – I3) 2 R = (9.53 – 5.2) 2 (6) = 112.5 W P = I 2 R = (I2 – I3) 2 R = (11.7 – 5.2) 2 (4) = 169 W P 10Ω = I 2 R = (I3) 2 R = (5.2) 2 (10) = 270.4W P total = P
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Unformatted text preview: + P 4Ω + P 10Ω = 551.9W NOTE: In this case, all the secondary loads were resistive. If there were reactive loads as well, we would have to find the corresponding reactive powers. This could be done similar to the above process, where we find the individual reactive powers and then sum them up. E.g.: If there was a capacitor in series with the 4Ω resistor, we would use Qc = I 2 Xc, where ‘I’ is the current going through the capacitor, (I2 – I3). Also, if there was an inductor in series with the 10Ω resistor, we would use Q L = I 2 X L , where ‘I’ would be the current going through the inductor, I3. The total reactive power is then simply Qc+Q L . The total power would then be the phasor sum of real and reactive powers, called the apparent power measured in VA: S Total = 2 2 total total Q P +...
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This note was uploaded on 03/21/2012 for the course EE 369 taught by Professor Khan during the Fall '10 term at Mohawk College.

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3-wire Edison Distribution Transformer Example - + P 4Ω +...

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