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6.5.3 Example I3
- 3. 1
XC =
2×∏×50×(25×10 −6)
Xc = 127 Ω
- 4. V
I3 =
Xc
240
I3 = = 1.88 A
127
I3 = 1.88 A
- 5. I3 leads V by 90⁰ (capacitance)
Therefore Φ3 = 90⁰ (leading)
- 8. V
Z=
Itotal
240
Z= = 34.8 Ω
6.9
Z = 34.8 Ω
- 9. The phase angle between the
total current flowing and the
applied voltage is 35⁰ lagging