Unit 21 Capacitance in AC Circuits

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Transcription:

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.

Capacitive Reactance Countervoltage limits the flow of current.

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.

Capacitive Reactance Capacitive reactance opposes the flow of electricity in a capacitor circuit. Capacitive reactance (X C ) is measured in ohms (Ω). X C = 1 /(2 fc) f = frequency C = capacitance

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.

Capacitive current leads the applied voltage by 90.

Power Relationships Capacitive power is measured in VARs C. Capacitive VARs C and inductive VARs L are 180 out of phase with each other. Capacitive VARs C and inductive VARs L cancel each other.

Capacitive power phase relationships.

Inductive VARs and Capacitive VARs.

Frequency Relationships Capacitive reactance (X C ) is inversely proportional to frequency (f). As frequency increases then capacitive reactance decreases. As frequency decreases then capacitive reactance increases.

E T = 480 V C 1 = 10 µf C 2 = 30 µf C 3 = 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.

E T = 480 V C 1 = 10 µf X C1 = 265 Ω C 2 = 30 µf C 3 = 15 µf First, calculate each capacitance reactance. Remember (X C ) = 1/(2пFC) and 2пF = 377 at 60 Hz.

E T = 480 V C 1 = 10 µf X C1 = 265 Ω C 2 = 30 µf X C2 = 88 Ω C 3 = 15 µf X C3 = 177 Ω X C1 = 1/ (377 x 0.000010) = 265.25 Ω X C2 = 1/ (377 x 0.000030) = 88.417 Ω X C3 = 1/ (377 x 0.000015) = 176.83 Ω

E T = 480 V X CT = 530.5 Ω C 1 = 10 µf X C1 = 265 Ω C 2 = 30 µf X C2 = 88 Ω C 3 = 15 µf X C3 = 177 Ω Next, find the total capacitive reactance. X CT = X C1 + X C2 + X C3 X CT = 265.25 + 88.417 + 176.83 = 530.497 Ω

E T = 480 V X CT = 530.5 Ω I T = 0.905 A C 1 = 10 µf X C1 = 265 Ω E C1 =? C 2 = 30 µf X C2 = 88 Ω E C2 =? C 3 = 15 µf X C3 = 177 Ω E C3 =? Next, calculate the total current. I T = E CT / X CT I T = 480 V / 530.497 Ω = 0.905 A

E T = 480 V X CT = 530.5 Ω I T = 0.905 A C 1 = 10 µf X C1 = 265 Ω I 1 =.905 A E C1 =? C 2 = 30 µf X C2 = 88 Ω I 2 =.905 A E C2 =? C 3 = 15 µf X C3 = 177 Ω I 3 =.905 A E C3 =? Next, calculate the component voltage drops. The total current flows through each component. E C = I T x X C

E T = 480 V X CT = 530.5 Ω I T = 0.905 A C 1 = 10 µf X C1 = 265 Ω I 1 =.905 A E C1 = 240 V C 2 = 30 µf X C2 = 88 Ω I 2 =.905 A E C2 = 80 V C 3 = 15 µf X C3 = 177 Ω I 3 =.905 A E C3 = 160 V E C1 =.905 x 265.25 = 240.051 V E C2 =.905 x 88.417 = 80.017 V E C3 =.905 x 176.83 = 160.031 V

E T = 480 V X CT = 530.5 Ω I T = 0.905 A 434 VARs CT C 1 = 10 µf X C1 = 265 Ω I 1 =.905 A E C1 = 240 V? VARs C1 C 2 = 30 µf X C2 = 88 Ω I 2 =.905 A E C2 = 80 V? VARs C2 C 3 = 15 µf X C3 = 177 Ω I 3 =.905 A E C3 = 160 V? VARs C3 Now the reactive power can easily be computed! Use the Ohm s law formula. VARs C = E C x I C = 480 x 0.905 = 434.4

E T = 480 V X CT = 530.5 Ω I T = 0.905 A 434 VARs C 1 = 10 µf X C1 = 265 Ω I 1 =.905 A E C1 = 240 V 217 VARs C1 C 2 = 30 µf X C2 = 88 Ω I 2 =.905 A E C2 = 80 V 72 VARs C2 C 3 = 15 µf X C3 = 177 Ω I 3 =.905 A E C3 = 160 V 144 VARs C3 VARs C1 = 240.051 x.905 = 217.246 VARs C2 = 80.017 x.905 = 72.415 VARs C3 = 160.031 x.905 = 144.828

E T =? V X CT =? Ω I T =? A 787 VARs C C 1 = 50 µf X C1 =? Ω I 1 =? A E C1 =? V? VARs C1 C 2 = 75 µf X C2 =? Ω I 2 =? A E C2 =? V? VARs C2 C 3 = 20 µf X C3 =? Ω I 3 =? A E C3 =? V? VARs C3 Three capacitors (50 µf, 75 µf, and 20 µf) are connected to a 60-Hz line. The total reactive power is 787.08 VARs C.

E T =? V X CT =? Ω I T =? A 787 VARs C C 1 = 50 µf X C1 = 53 Ω I 1 =? A E C1 =? V? VARs C1 C 2 = 75 µf X C2 =? Ω I 2 =? A E C2 =? V? VARs C2 C 3 = 20 µf X C3 =? Ω I 3 =? A E C3 =? V? VARs C3 First, calculate the capacitive reactance (X C ). X C1 = 1/ 2πFC = 1/ 377 x.000050 = 53.05 Ω X C1 = 53.05 Ω

E T =? V X CT =? Ω I T =? A 787 VARs C C 1 = 50 µf X C1 = 132 Ω I 1 =? A E C1 =? V? VARs C1 C 2 = 75 µf X C2 = 35 Ω I 2 =? A E C2 =? V? VARs C2 C 3 = 20 µf X C3 = 132 Ω I 3 =? A E C3 =? V? VARs C3 X C2 = 1/ 2πFC = 1/ 377 x.000075 = 35.367 Ω X C3 = 1/ 2πFC = 1/ 377 x.000020 = 132.63 Ω

E T =? V X CT = 18 Ω I T =? A 787 VARs C C 1 = 50 µf X C1 = 132 Ω I 1 =? A E C1 =? V? VARs C1 C 2 = 75 µf X C2 = 35 Ω I 2 =? A E C2 =? V? VARs C2 C 3 = 20 µf X C3 = 132 Ω I 3 =? A E C3 =? V? VARs C3 1/ X CT = 1/ X C1 + 1/ X C2 + 1/ X C3 1/ X CT = 1/ 53.05 + 1/ 35.367 + 1/ 132.63 X CT = 18.295 Ω

E T = 120 V X CT = 18 Ω I T =? A 787 VARs C C 1 = 50 µf X C1 = 132 Ω I 1 =? A E C1 =? V? VARs C1 C 2 = 75 µf X C2 = 35 Ω I 2 =? A E C2 =? V? VARs C2 Now use the formula: E T = (VARs CT x X CT ). E T = (787.08 x 18.295) E T = 120 V C 3 = 20 µf X C3 = 132 Ω I 3 =? A E C3 =? V? VARs C3

E T = 120 V X CT = 18 Ω I T =? A 787 VARs C C 1 = 50 µf X C1 = 132 Ω I 1 =? A E C1 = 120 V? VARs C1 C 2 = 75 µf X C2 = 35 Ω I 2 =? A E C2 = 120 V? VARs C2 All the voltage drops equal the source voltage. E T = E C1 = E C2 = E C3 = 120 V C 3 = 20 µf X C3 = 132 Ω I 3 =? A E C3 = 120 V? VARs C3

E T = 120 V X CT = 18 Ω I T = 6.559 A 787 VARs C C 1 = 50 µf X C1 = 132 Ω I 1 =? A E C1 = 120 V? VARs C1 C 2 = 75 µf X C2 = 35 Ω I 2 =? A E C2 = 120 V? VARs C2 Next, find the current: I T = E CT / X CT. I T = 120 / 18.295 I T = 6.559 A C 3 = 20 µf X C3 = 132 Ω I 3 =? A E C3 = 120 V? VARs C3

E T = 120 V X CT = 18 Ω I T = 6.559 A 787 VARs C C 1 = 50 µf X C1 = 132 Ω I 1 = 2.262 A E C1 = 120 V? VARs C1 C 2 = 75 µf X C2 = 35 Ω I 2 =? A E C2 = 120 V? VARs C2 C 3 = 20 µf X C3 = 132 Ω I 3 =? A E C3 = 120 V? VARs C3 Similarly: I 1 = E C1 / X C1 I 1 = 120 / 53.05 I 1 = 2.262 A

E T = 120 V X CT = 18 Ω I T = 6.559 A 787 VARs C C 1 = 50 µf X C1 = 132 Ω I 1 = 2.262 A E C1 = 120 V? VARs C1 C 2 = 75 µf X C2 = 35 Ω I 2 = 3.393 A E C2 = 120 V? VARs C2 C 3 = 20 µf X C3 = 132 Ω I 3 =? A E C3 = 120 V? VARs C3 Similarly: I 2 = E C2 / X C2 I 2 = 120 / 35.367 I 2 = 3.393 A

E T = 120 V X CT = 18 Ω I T = 6.559 A 787 VARs C C 1 = 50 µf X C1 = 132 Ω I 1 = 2.262 A E C1 = 120 V? VARs C1 C 2 = 75 µf X C2 = 35 Ω I 2 = 3.393 A E C2 = 120 V? VARs C2 C 3 = 20 µf X C3 = 132 Ω I 3 = 0.905 A E C3 = 120 V? VARs C3 Similarly: I 3 = E C3 / X C3 I 3 = 120 / 132.63 I 3 = 0.905 A

E T = 120 V X CT = 18 Ω I T = 6.559 A 787 VARs C C 1 = 50 µf X C1 = 132 Ω I 1 = 2.262 A E C1 = 120 V 271 VARs C1 C 2 = 75 µf X C2 = 35 Ω I 2 = 3.393 A E C2 = 120 V? VARs C2 C 3 = 20 µf X C3 = 132 Ω I 3 = 0.905 A E C3 = 120 V? VARs C3 Reactive power for each component is computed. VARs C1 = E C1 x I C1 VARs C1 = 120 x 2.262 = 271.442

E T = 120 V X CT = 18 Ω I T = 6.559 A 787 VARs C C 1 = 50 µf X C1 = 132 Ω I 1 = 2.262 A E C1 = 120 V 271 VARs C1 C 2 = 75 µf X C2 = 35 Ω I 2 = 3.393 A E C2 = 120 V 407 VARs C2 C 3 = 20 µf X C3 = 132 Ω I 3 = 0.905 A E C3 = 120 V? VARs C3 Reactive power for each component is computed. VARs C2 = E C2 x I C2 VARs C2 = 120 x 3.393 = 407.159

E T = 120 V X CT = 18 Ω I T = 6.559 A 787 VARs C C 1 = 50 µf X C1 = 132 Ω I 1 = 2.262 A E C1 = 120 V 271 VARs C1 C 2 = 75 µf X C2 = 35 Ω I 2 = 3.393 A E C2 = 120 V 407 VARs C2 C 3 = 20 µf X C3 = 132 Ω I 3 = 0.905 A E C3 = 120 V 109 VARs C3 Reactive power for each component is computed. VARs C3 = E C3 x I C3 VARs C3 = 120 x 0.905 = 108.573

E T = 120 V X CT = 18 Ω I T = 6.559 A 787 VARs C C 1 = 50 µf X C1 = 132 Ω I 1 = 2.262 A E C1 = 120 V 271 VARs C1 C 2 = 75 µf X C2 = 35 Ω I 2 = 3.393 A E C2 = 120 V 407 VARs C2 C 3 = 20 µf X C3 = 132 Ω I 3 = 0.905 A E C3 = 120 V 109 VARs C3 A quick check is done by adding the individual VARs and comparing the value to the original VARs. VARs C = 271.442 + 407.129 + 108.573 = 787.174

Review: Unit 21 Capacitance in AC Circuits 1. Current appears to flow through a capacitor in an AC circuit. 2. A capacitor appears to allow current flow because of the periodic rise and fall of voltage and current. 3. Current flow in a pure capacitive circuit is only limited by capacitive reactance.

Review: 4. Capacitive reactance is proportional to the capacitance and frequency. 5. Capacitive reactance is measured in ohms. 6. Current flow in a pure capacitive circuit leads the voltage by 90.

Review: 7. In a pure capacitive circuit, there is no true power (watts). 8. Capacitive power is reactive and is measured in VARs, as is inductance. 9. Capacitive and inductive VARs are 180 out of phase with each other.

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