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4.2.5 RC Example 2

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T
Talia Carbis
Technology
1 of 9
Find: (a) the current in the circuit (b) the total capacitive reactance (c) the impedance of the circuit (d) the applied voltage (e) the phase angle between the current and the applied voltage
I = 12 = (120 x 10¯³) = 120 mA 100 Answer I = 120 mA
C Z = √l00² + 46.43² = √12l55.75 Answer Z = 110.25 Ω
VC = I XC VC = (120 x 10¯³) x 46.43 VC = 5.57 V So,
V = √(12² + 5.57²) = √(144 + 31.02) = √175.02
∅ = cos¯¹ 12 = 24.8⁰ 13.22 Answer ∅ = 24.8°

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4.2.5 RC Example 2

  • 1. Find: (a) the current in the circuit (b) the total capacitive reactance (c) the impedance of the circuit (d) the applied voltage (e) the phase angle between the current and the applied voltage
  • 2.
  • 3. I = 12 = (120 x 10¯³) = 120 mA 100 Answer I = 120 mA
  • 4.
  • 5.
  • 6. C Z = √l00² + 46.43² = √12l55.75 Answer Z = 110.25 Ω
  • 7. VC = I XC VC = (120 x 10¯³) x 46.43 VC = 5.57 V So,
  • 8. V = √(12² + 5.57²) = √(144 + 31.02) = √175.02
  • 9. ∅ = cos¯¹ 12 = 24.8⁰ 13.22 Answer ∅ = 24.8°