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6.3.3 Example 2

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6.3.3 Example 2

  1. 1. XL = 78.52Ω IR = 0.4A RL = 30Ω R = l00Ω V = 40V F = 40V
  2. 2. ZL = √ (30² + 78.5²) ZL = 84.04 Ω
  3. 3. V IL = ZL 40 IL = = 0.48 A 84.04 IL = 0.48 A
  4. 4. ΦL = cos ¯¹ 30 84.04 ΦL = 69.1º (Lagging)
  5. 5. To find I, you need to draw a phasor diagram. Draw a phasor diagram of the currents, and find I by measurement. At this stage, this is the only method you can use
  6. 6. IL = 0.48A (lagging by 69.1°) IR = 0.4A (in phase with the voltage as branch is resistive)
  7. 7. = (0.73A) I = O.73A
  8. 8. = (0.73A) Answer Φ = 38.5° (with current lagging the voltage as circuit is inductive).

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