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Unit 21 Capacitance in AC Circuits

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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.

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

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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.

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Unit 21 Capacitance in AC Circuits 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

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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 .

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

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Unit 21 Capacitance in AC Circuits 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.

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Unit 21 Capacitance in AC Circuits Capacitive power phase relationships.

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

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Unit 21 Capacitance in AC Circuits 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 ↑.

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Unit 21 Capacitance in AC Circuits 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 F = 60 Hz C 1 = 10 µF C 2 = 30 µF C 3 = 15 µF

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Unit 21 Capacitance in AC Circuits First, calculate each capacitance reactance. Remember (X C ) = 1/(2пFC) and 2пF = 377 at 60 Hz. E T = 480 V F = 60 Hz C 1 = 10 µF X C1 = 265 Ω C 2 = 30 µF C 3 = 15 µF

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Unit 21 Capacitance in AC Circuits X C1 = 1/ (377 x ) = Ω X C2 = 1/ (377 x ) = Ω X C3 = 1/ (377 x ) = Ω E T = 480 V F = 60 Hz C 1 = 10 µF X C1 = 265 Ω C 2 = 30 µF X C2 = 88 Ω C 3 = 15 µF X C3 = 177 Ω

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Unit 21 Capacitance in AC Circuits Next, find the total capacitive reactance. X CT = X C1 + X C2 + X C3 X CT = = Ω E T = 480 V F = 60 Hz X CT = 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

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Unit 21 Capacitance in AC Circuits Next, calculate the total current. I T = E CT / X CT I T = 480 V / Ω = A E T = 480 V F = 60 Hz X CT = Ω I T = 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 = ?

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Unit 21 Capacitance in AC Circuits Next, calculate the component voltage drops. The total current flows through each component. E C = I T x X C E T = 480 V F = 60 Hz X CT = Ω I T = 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 = ?

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Unit 21 Capacitance in AC Circuits E C1 =.905 x = V E C2 =.905 x = V E C3 =.905 x = V E T = 480 V F = 60 Hz X CT = Ω I T = 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

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Unit 21 Capacitance in AC Circuits Now the reactive power can easily be computed! Use the Ohm’s law formula. VARs C = E C x I C = 480 x = E T = 480 V F = 60 Hz X CT = Ω I T = 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

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Unit 21 Capacitance in AC Circuits VARs C1 = x.905 = VARs C2 = x.905 = VARs C3 = x.905 = E T = 480 V F = 60 Hz X CT = Ω I T = 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

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Unit 21 Capacitance in AC Circuits Three capacitors (50 µF, 75 µF, and 20 µF) are connected to a 60-Hz line. The total reactive power is VARs C. E T = ? V F = 60 Hz 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

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Unit 21 Capacitance in AC Circuits First, calculate the capacitive reactance (X C ). X C1 = 1/ 2πFC = 1/ 377 x = Ω X C1 = Ω E T = ? V F = 60 Hz 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

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Unit 21 Capacitance in AC Circuits X C2 = 1/ 2πFC = 1/ 377 x = Ω X C3 = 1/ 2πFC = 1/ 377 x = Ω E T = ? V F = 60 Hz 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

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Unit 21 Capacitance in AC Circuits 1/ X CT = 1/ X C1 + 1/ X C2 + 1/ X C3 1/ X CT = 1/ / / X CT = Ω E T = ? V F = 60 Hz 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

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Unit 21 Capacitance in AC Circuits Now use the formula: E T = √(VARs CT x X CT ). E T = √( x ) E T = 120 V F = 60 Hz 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

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Unit 21 Capacitance in AC Circuits All the voltage drops equal the source voltage. E T = E C1 = E C2 = E C3 = 120 V E T = 120 V F = 60 Hz 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 C 3 = 20 µF X C3 = 132 Ω I 3 = ? A E C3 = 120 V ? VARs C3

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Unit 21 Capacitance in AC Circuits Next, find the current: I T = E CT / X CT. I T = 120 / I T = A E T = 120 V F = 60 Hz 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 C 3 = 20 µF X C3 = 132 Ω I 3 = ? A E C3 = 120 V ? VARs C3

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Unit 21 Capacitance in AC Circuits Similarly: I 1 = E C1 / X C1 I 1 = 120 / I 1 = A E T = 120 V F = 60 Hz 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 C 3 = 20 µF X C3 = 132 Ω I 3 = ? A E C3 = 120 V ? VARs C3

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Unit 21 Capacitance in AC Circuits Similarly: I 2 = E C2 / X C2 I 2 = 120 / I 2 = A E T = 120 V F = 60 Hz 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 C 3 = 20 µF X C3 = 132 Ω I 3 = ? A E C3 = 120 V ? VARs C3

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Unit 21 Capacitance in AC Circuits Similarly: I 3 = E C3 / X C3 I 3 = 120 / I 3 = A E T = 120 V F = 60 Hz 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 C 3 = 20 µF X C3 = 132 Ω I 3 = A E C3 = 120 V ? VARs C3

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Unit 21 Capacitance in AC Circuits Reactive power for each component is computed. VARs C1 = E C1 x I C1 VARs C1 = 120 x = E T = 120 V F = 60 Hz 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 271 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

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Unit 21 Capacitance in AC Circuits Reactive power for each component is computed. VARs C2 = E C2 x I C2 VARs C2 = 120 x = E T = 120 V F = 60 Hz 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 271 VARs C1 C 2 = 75 µF X C2 = 35 Ω I 2 = A E C2 = 120 V 407 VARs C2 C 3 = 20 µF X C3 = 132 Ω I 3 = A E C3 = 120 V ? VARs C3

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Unit 21 Capacitance in AC Circuits Reactive power for each component is computed. VARs C3 = E C3 x I C3 VARs C3 = 120 x = E T = 120 V F = 60 Hz 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 271 VARs C1 C 2 = 75 µF X C2 = 35 Ω I 2 = A E C2 = 120 V 407 VARs C2 C 3 = 20 µF X C3 = 132 Ω I 3 = A E C3 = 120 V 109 VARs C3

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Unit 21 Capacitance in AC Circuits A quick check is done by adding the individual VARs and comparing the value to the original VARs. VARs C = = E T = 120 V F = 60 Hz 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 271 VARs C1 C 2 = 75 µF X C2 = 35 Ω I 2 = A E C2 = 120 V 407 VARs C2 C 3 = 20 µF X C3 = 132 Ω I 3 = A E C3 = 120 V 109 VARs C3

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Unit 21 Capacitance in AC Circuits Review: 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.

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Unit 21 Capacitance in AC Circuits 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 .

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Unit 21 Capacitance in AC Circuits 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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