### Lab 1 Q1 AND 2 Solution

- Budget: $20 Normal
- Subjects: Computer Science | Painting |
- Due on 20 Aug, 2018 08:36:35
- Asked on 20 Aug, 2018 08:36:35
- Past Due (date has already expired)

**Lab 1 Q1 AND 2 Solution**

Measurements taken on a DC Machine during a lab: Rated at 125V, 2.4A

1. Shunt Field = .57A, 125.3V, 219.82 Ohms.

2. Series Field = 2.4 A, 19.008V, 7.92 Ohms.

3. Armature = 2.97 A, 21.48 V, 8.95 Ohms.

Explain the difference in the resistances found for the shunt field, series field, and armature circuits. In particular, why is the resistance of the shunt field so different from the resistances of the armature and series field? What would I expect the relative wire size to be (heavy or light gauge) and why, and the number of turns (many or few) on each of the three windings?

2. If the values obtained were taken at a temperature of 20 C, what would the resistance values above be at 55 C? could you please show the calculation, and how to do it. ( I researched and found that copper wire increases in resistance by .0039% for each degree rise in temperature, in Celsius.

3. If the shunt field windings are composed of #34 AWG copper wire, and each turn is approximately 12 inches long, how many turns of wire are there on the shunt field of the DC machine? (I looked this up and found the resistance coefficient of #34 copper wire to be 260.9 Ohms.

4. If the series field windings are #26 AWG and the turns are 13 inches long, how many turns of wire do I need. (for #26 copper wire, the resistance coefficient is 40.81 ohms).

5. Calculate the ampere-turns of the magnetomotive force generated by the shunt and series fields windings.

Is there anything glaring that I should notice during this experiment?

Please show work so I can follow and learn.