Unit 21 Capacitance in AC Circuits

1. Unit 21Capacitance in AC Circuits

2. Unit 21 Capacitance in AC Circuits Objectives: • Explain why current appears to flow through a capacitor in an AC circuit. • Discuss capacitive reactance. • Discuss the relationship of voltage and current in a pure capacitive circuit.

3. Unit 21 Capacitance in AC Circuits Capacitive Reactance Countervoltage limits the flow of current.

4. Unit 21 Capacitance in AC Circuits • Capacitors appear to pass alternating current (AC). • This is due to the capacitor charging for half of the cycle and discharging for half of the cycle. • This repetitive charging and discharging allows current to flow in the circuit.

5. Unit 21 Capacitance in AC Circuits Capacitive Reactance • Capacitive reactance opposes the flow of electricity in a capacitor circuit. • Capacitive reactance (XC) is measured in ohms (Ω). • XC = 1 /(2fC) • f = frequency • C = capacitance

6. Unit 21 Capacitance in AC Circuits Phase Relationships • Capacitive current leads the applied voltage by 90°. • Inductive current lags the applied voltage by 90°. • Resistive current is in phase with the applied voltage by 90.

7. Unit 21 Capacitance in AC Circuits Capacitive current leads the applied voltage by 90.

8. Unit 21 Capacitance in AC Circuits Power Relationships • Capacitive power is measured in VARsC. • Capacitive VARsC and inductive VARsL are 180° out of phase with each other. • Capacitive VARsC and inductive VARsL cancel each other.

9. Unit 21 Capacitance in AC Circuits Capacitive power phase relationships.

10. Unit 21 Capacitance in AC Circuits Inductive VARs and Capacitive VARs.

11. Unit 21 Capacitance in AC Circuits Frequency Relationships • Capacitive reactance (XC) is inversely proportional to frequency (f). • As frequency increases ↑ then capacitive reactance decreases ↓. • As frequency decreases ↓ then capacitive reactance increases ↑.

12. Unit 21 Capacitance in AC Circuits ET = 480 V F = 60 Hz C1 = 10 µF C2 = 30 µF C3 = 15 µF Solving a sample series capacitor circuit: Three capacitors (10µF, 30 µF, and 15 µF) are series connected to a 480-V, 60-Hz power source.

13. Unit 21 Capacitance in AC Circuits C1 = 10 µF XC1 = 265 Ω ET = 480 V F = 60 Hz C2 = 30 µF C3 = 15 µF First, calculate each capacitance reactance. Remember (XC) = 1/(2пFC) and 2пF = 377 at 60 Hz.

14. Unit 21 Capacitance in AC Circuits C1 = 10 µF XC1 = 265 Ω C2 = 30 µF XC2 = 88 Ω ET = 480 V F = 60 Hz C3 = 15 µF XC3 = 177 Ω XC1 = 1/ (377 x 0.000010) = 265.25 Ω XC2 = 1/ (377 x 0.000030) = 88.417 Ω XC3 = 1/ (377 x 0.000015) = 176.83 Ω

15. Unit 21 Capacitance in AC Circuits ET = 480 V F = 60 Hz XCT = 530.5 Ω C1 = 10 µF XC1 = 265 Ω C2 = 30 µF XC2 = 88 Ω C3 = 15 µF XC3 = 177 Ω Next, find the total capacitive reactance. XCT = XC1 + XC2 + XC3 XCT = 265.25 + 88.417 + 176.83 = 530.497 Ω

16. Unit 21 Capacitance in AC Circuits C2 = 30 µF XC2 = 88 Ω EC2 = ? C3 = 15 µF XC3 = 177 Ω EC3 = ? C1 = 10 µF XC1 = 265 Ω EC1 = ? ET = 480 V F = 60 Hz XCT = 530.5 Ω IT = 0.905 A Next, calculate the total current. IT = ECT / XCT IT = 480 V / 530.497 Ω = 0.905 A

17. Unit 21 Capacitance in AC Circuits C1 = 10 µF XC1 = 265 Ω I1 = .905 A EC1 = ? ET = 480 V F = 60 Hz XCT = 530.5 Ω IT = 0.905 A C2 = 30 µF XC2 = 88 Ω I2 = .905 A EC2 = ? C3 = 15 µF XC3 = 177 Ω I3 = .905 A EC3 = ? Next, calculate the component voltage drops. The total current flows through each component. EC = IT x XC

18. Unit 21 Capacitance in AC Circuits C1 = 10 µF XC1 = 265 Ω I1 = .905 A EC1 = 240 V ET = 480 V F = 60 Hz XCT = 530.5 Ω IT = 0.905 A C2 = 30 µF XC2 = 88 Ω I2 = .905 A EC2 = 80 V C3 = 15 µF XC3 = 177 Ω I3 = .905 A EC3 = 160 V EC1 = .905 x 265.25 = 240.051 V EC2 = .905 x 88.417 = 80.017 V EC3 = .905 x 176.83 = 160.031 V

19. Unit 21 Capacitance in AC Circuits ET = 480 V F = 60 Hz XCT = 530.5 Ω IT = 0.905 A 434 VARsCT C1 = 10 µF XC1 = 265 Ω I1 = .905 A EC1 = 240 V ? VARsC1 C2 = 30 µF XC2 = 88 Ω I2 = .905 A EC2 = 80 V ? VARsC2 C3 = 15 µF XC3 = 177 Ω I3 = .905 A EC3 = 160 V ? VARsC3 Now the reactive power can easily be computed! Use the Ohm’s law formula. VARsC = EC x IC = 480 x 0.905 = 434.4

20. Unit 21 Capacitance in AC Circuits ET = 480 V F = 60 Hz XCT = 530.5 Ω IT = 0.905 A 434 VARs C2 = 30 µF XC2 = 88 Ω I2 = .905 A EC2 = 80 V 72 VARsC2 C3 = 15 µF XC3 = 177 Ω I3 = .905 A EC3 = 160 V 144 VARsC3 C1 = 10 µF XC1 = 265 Ω I1 = .905 A EC1 = 240 V 217 VARsC1 VARsC1 = 240.051 x .905 = 217.246 VARsC2 = 80.017 x .905 = 72.415 VARsC3 = 160.031 x .905 = 144.828

21. Unit 21 Capacitance in AC Circuits C2 = 75 µF XC2 = ? Ω I2 = ? A EC2 = ? V ? VARsC2 C1 = 50 µF XC1 = ? Ω I1 = ? A EC1 = ? V ? VARsC1 ET = ? V F = 60 Hz XCT = ? Ω IT = ? A 787 VARsC C3 = 20 µF XC3 = ? Ω I3 = ? A EC3 = ? V ? VARsC3 Three capacitors (50 µF, 75 µF, and 20 µF) are connected to a 60-Hz line. The total reactive power is 787.08 VARsC.

22. Unit 21 Capacitance in AC Circuits ET = ? V F = 60 Hz XCT = ? Ω IT = ? A 787 VARsC C1 = 50 µF XC1 = 53 Ω I1 = ? A EC1 = ? V ? VARsC1 C2 = 75 µF XC2 = ? Ω I2 = ? A EC2 = ? V ? VARsC2 C3 = 20 µF XC3 = ? Ω I3 = ? A EC3 = ? V ? VARsC3 First, calculate the capacitive reactance (XC). XC1 = 1/ 2πFC = 1/ 377 x .000050 = 53.05 Ω XC1 = 53.05 Ω

23. Unit 21 Capacitance in AC Circuits ET = ? V F = 60 Hz XCT = ? Ω IT = ? A 787 VARsC C1 = 50 µF XC1 = 132 Ω I1 = ? A EC1 = ? V ? VARsC1 C2 = 75 µF XC2 = 35 Ω I2 = ? A EC2 = ? V ? VARsC2 C3 = 20 µF XC3 = 132 Ω I3 = ? A EC3 = ? V ? VARsC3 XC2 = 1/ 2πFC = 1/ 377 x .000075 = 35.367 Ω XC3 = 1/ 2πFC = 1/ 377 x .000020 = 132.63 Ω

24. Unit 21 Capacitance in AC Circuits C2 = 75 µF XC2 = 35 Ω I2 = ? A EC2 = ? V ? VARsC2 C3 = 20 µF XC3 = 132 Ω I3 = ? A EC3 = ? V ? VARsC3 C1 = 50 µF XC1 = 132 Ω I1 = ? A EC1 = ? V ? VARsC1 ET = ? V F = 60 Hz XCT = 18 Ω IT = ? A 787 VARsC 1/ XCT = 1/ XC1 + 1/ XC2 + 1/ XC3 1/ XCT = 1/ 53.05 + 1/ 35.367 + 1/ 132.63 XCT = 18.295 Ω

25. Unit 21 Capacitance in AC Circuits C3 = 20 µF XC3 = 132 Ω I3 = ? A EC3 = ? V ? VARsC3 C2 = 75 µF XC2 = 35 Ω I2 = ? A EC2 = ? V ? VARsC2 C1 = 50 µF XC1 = 132 Ω I1 = ? A EC1 = ? V ? VARsC1 ET = 120 V F = 60 Hz XCT = 18 Ω IT = ? A 787 VARsC Now use the formula: ET = √(VARsCT x XCT). ET = √(787.08 x 18.295) ET = 120 V

26. Unit 21 Capacitance in AC Circuits C1 = 50 µF XC1 = 132 Ω I1 = ? A EC1 = 120 V ? VARsC1 C2 = 75 µF XC2 = 35 Ω I2 = ? A EC2 = 120 V ? VARsC2 ET = 120 V F = 60 Hz XCT = 18 Ω IT = ? A 787 VARsC C3 = 20 µF XC3 = 132 Ω I3 = ? A EC3 = 120 V ? VARsC3 All the voltage drops equal the source voltage. ET = EC1 = EC2 = EC3 = 120 V

27. Unit 21 Capacitance in AC Circuits C2 = 75 µF XC2 = 35 Ω I2 = ? A EC2 = 120 V ? VARsC2 C3 = 20 µF XC3 = 132 Ω I3 = ? A EC3 = 120 V ? VARsC3 C1 = 50 µF XC1 = 132 Ω I1 = ? A EC1 = 120 V ? VARsC1 ET = 120 V F = 60 Hz XCT = 18 Ω IT = 6.559 A 787 VARsC Next, find the current: IT = ECT / XCT. IT = 120 / 18.295 IT = 6.559 A

28. Unit 21 Capacitance in AC Circuits C1 = 50 µF XC1 = 132 Ω I1 = 2.262 A EC1 = 120 V ? VARsC1 ET = 120 V F = 60 Hz XCT = 18 Ω IT = 6.559 A 787 VARsC C2 = 75 µF XC2 = 35 Ω I2 = ? A EC2 = 120 V ? VARsC2 C3 = 20 µF XC3 = 132 Ω I3 = ? A EC3 = 120 V ? VARsC3 Similarly: I1 = EC1 / XC1 I1 = 120 / 53.05 I1 = 2.262 A

29. Unit 21 Capacitance in AC Circuits ET = 120 V F = 60 Hz XCT = 18 Ω IT = 6.559 A 787 VARsC C1 = 50 µF XC1 = 132 Ω I1 = 2.262 A EC1 = 120 V ? VARsC1 C2 = 75 µF XC2 = 35 Ω I2 = 3.393 A EC2 = 120 V ? VARsC2 C3 = 20 µF XC3 = 132 Ω I3 = ? A EC3 = 120 V ? VARsC3 Similarly: I2 = EC2 / XC2 I2 = 120 / 35.367 I2 = 3.393 A

30. Unit 21 Capacitance in AC Circuits C1 = 50 µF XC1 = 132 Ω I1 = 2.262 A EC1 = 120 V ? VARsC1 C2 = 75 µF XC2 = 35 Ω I2 = 3.393 A EC2 = 120 V ? VARsC2 C3 = 20 µF XC3 = 132 Ω I3 = 0.905 A EC3 = 120 V ? VARsC3 ET = 120 V F = 60 Hz XCT = 18 Ω IT = 6.559 A 787 VARsC Similarly: I3 = EC3 / XC3 I3 = 120 / 132.63 I3 = 0.905 A

31. Unit 21 Capacitance in AC Circuits C1 = 50 µF XC1 = 132 Ω I1 = 2.262 A EC1 = 120 V 271 VARsC1 C2 = 75 µF XC2 = 35 Ω I2 = 3.393 A EC2 = 120 V ? VARsC2 C3 = 20 µF XC3 = 132 Ω I3 = 0.905 A EC3 = 120 V ? VARsC3 ET = 120 V F = 60 Hz XCT = 18 Ω IT = 6.559 A 787 VARsC Reactive power for each component is computed. VARsC1 = EC1 x IC1 VARsC1 = 120 x 2.262 = 271.442

32. Unit 21 Capacitance in AC Circuits C2 = 75 µF XC2 = 35 Ω I2 = 3.393 A EC2 = 120 V 407 VARsC2 C1 = 50 µF XC1 = 132 Ω I1 = 2.262 A EC1 = 120 V 271 VARsC1 C3 = 20 µF XC3 = 132 Ω I3 = 0.905 A EC3 = 120 V ? VARsC3 ET = 120 V F = 60 Hz XCT = 18 Ω IT = 6.559 A 787 VARsC Reactive power for each component is computed. VARsC2 = EC2 x IC2 VARsC2 = 120 x 3.393 = 407.159

33. Unit 21 Capacitance in AC Circuits C1 = 50 µF XC1 = 132 Ω I1 = 2.262 A EC1 = 120 V 271 VARsC1 ET = 120 V F = 60 Hz XCT = 18 Ω IT = 6.559 A 787 VARsC C2 = 75 µF XC2 = 35 Ω I2 = 3.393 A EC2 = 120 V 407 VARsC2 C3 = 20 µF XC3 = 132 Ω I3 = 0.905 A EC3 = 120 V 109 VARsC3 Reactive power for each component is computed. VARsC3 = EC3 x IC3 VARsC3 = 120 x 0.905 = 108.573

34. Unit 21 Capacitance in AC Circuits C3 = 20 µF XC3 = 132 Ω I3 = 0.905 A EC3 = 120 V 109 VARsC3 C1 = 50 µF XC1 = 132 Ω I1 = 2.262 A EC1 = 120 V 271 VARsC1 C2 = 75 µF XC2 = 35 Ω I2 = 3.393 A EC2 = 120 V 407 VARsC2 ET = 120 V F = 60 Hz XCT = 18 Ω IT = 6.559 A 787 VARsC A quick check is done by adding the individual VARs and comparing the value to the original VARs. VARsC = 271.442 + 407.129 + 108.573 = 787.174

35. Unit 21 Capacitance in AC Circuits Review: • Current appears to flow through a capacitor in an AC circuit. • A capacitor appears to allow current flow because of the periodic rise and fall of voltage and current. • Current flow in a pure capacitive circuit is only limited by capacitive reactance.

36. Unit 21 Capacitance in AC Circuits Review: • Capacitive reactance is proportional to the capacitance and frequency. • Capacitive reactance is measured in ohms. • Current flow in a pure capacitive circuit leads the voltage by 90.

37. Unit 21 Capacitance in AC Circuits Review: • In a pure capacitive circuit, there is no true power (watts). • Capacitive power is reactive and is measured in VARs, as is inductance. • Capacitive and inductive VARs are 180 out of phase with each other.