unit 19 - capacitors (1)

8/6/2019 Unit 19 - Capacitors (1)

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Unit 19 Capacitors

Objectives:� List three factors that determine

capacitance.

� Discuss the electrostatic charge.� Discuss polarized and nonpolarized

capacitors.

� Compute capacitive circuit values.� Compute an RC time constant.



Unit 19 Capacitors

Objectives:� Explain current flow in an AC capacitive

circuit.

� Discuss capacitive reactance.� Compute values of capacitance and

capacitive reactance.

� Discuss voltage and current phaserelationships.



Unit 19 Capacitors

� The simplest type of capacitor is made byseparating two metal plates with aninsulating material called the dielectric.

Plates Dielectric



Unit 19 Capacitors

� Three factors determine the capacitance of acapacitor:1. The surface area of the plates.2. The distance between the plates.3. The type of dielectric.

Distance

between

plates

Type of

Dielectric

Area of Plates

Area of

Plates



Unit 19 Capacitors

� A capacitor is charged by removing electronsfrom one plate and depositing electrons on theother plate.

� Capacitors oppose a change of voltage.

PlatesDielectric

Electrons



Unit 19 Capacitors

� Current flows until the voltage across thecapacitor is equal to the voltage of the battery.

� Current flows only during charging anddischarging!

PlatesDielectric

Electrons



Unit 19 Capacitors

� The capacitor remains charged until after thebattery is removed from the circuit.� Current flows only when there is a pathway from

plate to plate!

PlatesDielectric

Electrons



Unit 19 Capacitors

� Leakage current of a few electrons through thedielectric will eventually discharge a capacitor.

Positive Plate

Dielectric

Negative Plate



Unit 19 Capacitors

Electron orbits in an uncharged capacitor.

Positive Plate

Dielectric

Negative Plate



Unit 19 Capacitors

Electron orbits are stretched in a charged capacitor.

Positive Plate

Dielectric

Negative Plate



Unit 19 Capacitors

� Dielectric stress issimilar to drawing back

a bowstring with anarrow and holding it.This is similar to a fullycharged capacitor.



Unit 19 Capacitors

� Capacitors connected inparallel.

CT = C1 + C2 + C3

CT = 20 + 30 + 60 = 110 µF

C3 = 60 µFC2 = 30 µFC1 = 20 µFCT = 110 µF



Unit 19 Capacitors

� Capacitors connected in series.1/CT = 1/C1 + 1/C2 + 1/C3

1/CT = 1/20 + 1/30 + 1/60CT = 10 µF

C3 = 60 µFC2 = 30 µFC1 = 20 µF

CT = 10 µF



Unit 19 Capacitors

� 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.



Unit 19 Capacitors

� Capacitors produce a counter-voltage that limitsthe flow of electricity.

� This is called capacitive reactance (XC).� Capacitive reactance (XC) is measured in ohms.



Unit 19 Capacitors

� The formula to calculatecapacitive reactance is:

(XC) = 1/(2fC)� F is frequency in hertz.� C is capacitance in farads.



Unit 19 Capacitors

� 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.

ET = 480 V

F = 60 Hz

C1 = 10 µF C2 = 30 µF C3 = 15 µF



Unit 19 Capacitors

� First, calculate each capacitance reactance.Remember (XC) = 1/(2FC) and 2F = 377 at 60 Hz.

ET = 480 V

F = 60 Hz

C1 = 10 µF

XC1 = 265 �

C2 = 30 µF C3 = 15 µF



Unit 19 Capacitors

XC1 = 1/ (377 x 0.000010) = 265.25 �XC2 = 1/ (377 x 0.000030) = 88.417 �XC3 = 1/ (377 x 0.000015) = 176.83 �

ET = 480 V

F = 60 Hz

C1 = 10 µF

XC1 = 265 �

C2 = 30 µF

XC2 = 88 �

C3 = 15 µF

XC3 = 177 �



Unit 19 Capacitors

� Next, find the total capacitive reactance.XCT = XC1 + XC2 + XC3

XCT = 265.25 + 88.417 + 176.83 = 530.497 �

ET = 480 V

F = 60 Hz

XCT = 530.5 �

C1 = 10 µF

XC1 = 265 �

C2 = 30 µF

XC2 = 88 �

C3 = 15 µF

XC3 = 177 �



Unit 19 Capacitors

� Next, calculate the total current.IT = ECT / XCT

IT = 480 V / 530.497 � = 0.905 A

ET = 480 V

F = 60 Hz

XCT = 530.5 �IT = 0.905 A

C1 = 10 µF

XC1 = 265 �

EC1 = ?

C2 = 30 µF

XC2 = 88 �

EC2 = ?

C3 = 15 µF

XC3 = 177 �

EC3 = ?



Unit 19 Capacitors

� Next, calculate the component voltage drops.The total current flows through each component.

EC = IT x XC

ET = 480 V

F = 60 Hz

XCT = 530.5 �IT = 0.905 A

C1 = 10 µF

XC1 = 265 �

I1 = .905 AEC1 = ?

C2 = 30 µF

XC2 = 88 �

I2 = .905 AEC2 = ?

C3 = 15 µF

XC3 = 177 �

I3 = .905 AEC3 = ?



Unit 19 Capacitors

� Now the reactive power can easily be computed!Use the Ohm¶s law formula.

VARsC = EC x IC = 480 x 0.905 = 434.4

ET = 480 V

F = 60 Hz

XCT = 530.5 �IT = 0.905 A

434 VARsCT

C1 = 10 µF

XC1 = 265 �

I1 = .905 AEC1 = 240 V

? VARsC1

C2 = 30 µF

XC2 = 88 �

I2 = .905 AEC2 = 80 V

? VARsC2

C3 = 15 µF

XC3 = 177 �

I3 = .905 AEC3 = 160 V

? VARsC3



Unit 19 Capacitors

VARsC1 = 240.051 x .905 = 217.246VARsC2 = 80.017 x .905 = 72.415VARsC3 = 160.031 x .905 = 144.828

ET = 480 V

F = 60 Hz

XCT = 530.5 �IT = 0.905 A

434 VARs

C1 = 10 µF

XC1 = 265 �

I1 = .905 AEC1 = 240 V

217 VARsC1

C2 = 30 µF

XC2 = 88 �

I2 = .905 AEC2 = 80 V

72 VARsC2

C3 = 15 µF

XC3 = 177 �

I3 = .905 AEC3 = 160 V

144 VARsC3



Unit 19 Capacitors

Three capacitors (50 µF, 75 µF, and 20 µF) areconnected to a 60-Hz line. The total reactive power

is 787.08 VARsC.

ET = ? VF = 60 Hz

XCT = ? �

IT = ? A

787 VARsC

C1 = 50 µFXC1 = ? �

I1 = ? A

EC1 = ? V

? VARsC1

C2 = 75 µFXC2 = ? �

I2 = ? A

EC2 = ? V

? VARsC2

C3 = 20 µFXC3 = ? �

I3 = ? A

EC3 = ? V

? VARsC3



Unit 19 Capacitors

� First, calculate the capacitive reactance (XC).XC1 = 1/ 2FC = 1/ 377 x .000050 = 53.05 �

XC1 = 53.05 �

ET = ? VF = 60 Hz

XCT = ? �

IT = ? A

787 VARsC

C1 = 50 µFXC1 = 53 �

I1 = ? A

EC1 = ? V

? VARsC1

C2 = 75 µFXC2 = ? �

I2 = ? A

EC2 = ? V

? VARsC2

C3 = 20 µFXC3 = ? �

I3 = ? A

EC3 = ? V

? VARsC3



Unit 19 Capacitors

XC2 = 1/ 2FC = 1/ 377 x .000075 = 35.367 �XC3 = 1/ 2FC = 1/ 377 x .000020 = 132.63 �

ET = ? VF = 60 Hz

XCT = ? �

IT = ? A

787 VARsC

C1 = 50 µFXC1 = 132 �

I1 = ? A

EC1 = ? V

? VARsC1

C2 = 75 µFXC2 = 35 �

I2 = ? A

EC2 = ? V

? VARsC2

C3 = 20 µFXC3 = 132 �

I3 = ? A

EC3 = ? V

? VARsC3



Unit 19 Capacitors

1/ XCT = 1/ XC1 + 1/ XC2 + 1/ XC3

1/ XCT = 1/ 53.05 + 1/ 35.367 + 1/ 132.63XCT = 18.295 �

ET = ? VF = 60 Hz

XCT = 18 �

IT = ? A

787 VARsC

C1

= 50 µF

XC1 = 132 �

I1 = ? A

EC1 = ? V

? VARsC1

C2

= 75 µF

XC2 = 35 �

I2 = ? A

EC2 = ? V

? VARsC2

C3

= 20 µF

XC3 = 132 �

I3 = ? A

EC3 = ? V

? VARsC3



Unit 19 Capacitors

� Now use the formula: ET = ¥(VARsCT x XCT).ET = ¥(787.08 x 18.295)

ET = 120 V

ET

= 120 V

F = 60 Hz

XCT = 18 �

IT = ? A

787 VARsC

C1

= 50 µF

XC1 = 132 �

I1 = ? A

EC1 = ? V

? VARsC1

C2

= 75 µF

XC2 = 35 �

I2 = ? A

EC2 = ? V

? VARsC2

C3

= 20 µF

XC3 = 132 �

I3 = ? A

EC3 = ? V

? VARsC3



Unit 19 Capacitors

� All the voltage drops equal the source voltage.ET = EC1 = EC2 = EC3 = 120 V

ET

= 120 V

F = 60 Hz

XCT = 18 �

IT = ? A

787 VARsC

C1

= 50 µF

XC1 = 132 �

I1 = ? A

EC1 = 120 V

? VARsC1

C2

= 75 µF

XC2 = 35 �

I2 = ? A

EC2 = 120 V

? VARsC2

C3

= 20 µF

XC3 = 132 �

I3 = ? A

EC3 = 120 V

? VARsC3



Unit 19 Capacitors

� Next, find the current: IT = ECT / XCT.IT = 120 / 18.295

IT = 6.559 A

ET

= 120 V

F = 60 Hz

XCT = 18 �

IT = 6.559 A

787 VARsC

C1

= 50 µF

XC1 = 132 �

I1 = ? A

EC1 = 120 V

? VARsC1

C2

= 75 µF

XC2 = 35 �

I2 = ? A

EC2 = 120 V

? VARsC2

C3

= 20 µF

XC3 = 132 �

I3 = ? A

EC3 = 120 V

? VARsC3



Unit 19 Capacitors

� Similarly: I1 = EC1 / XC1I1 = 120 / 53.05

I1 = 2.262 A

ET

= 120 V

F = 60 Hz

XCT = 18 �

IT = 6.559 A

787 VARsC

C1

= 50 µF

XC1 = 132 �

I1 = 2.262 A

EC1 = 120 V

? VARsC1

C2

= 75 µF

XC2 = 35 �

I2 = ? A

EC2 = 120 V

? VARsC2

C3

= 20 µF

XC3 = 132 �

I3 = ? A

EC3 = 120 V

? VARsC3



Unit 19 Capacitors

� Similarly: I3 = EC3 / XC3I3 = 120 / 132.63

I3 = 0.905 A

ET

= 120 V

F = 60 Hz

XCT = 18 �

IT = 6.559 A

787 VARsC

C1

= 50 µF

XC1 = 132 �

I1 = 2.262 A

EC1 = 120 V

? VARsC1

C2

= 75 µF

XC2 = 35 �

I2 = 3.393 A

EC2 = 120 V

? VARsC2

C3

= 20 µF

XC3 = 132 �

I3 = 0.905 A

EC3 = 120 V

? VARsC3



Unit 19 Capacitors

� Reactive power for each component is computed.VARsC1 = EC1 x IC1

VARsC1 = 120 x 2.262 = 271.442

ET

= 120 V

F = 60 Hz

XCT = 18 �

IT = 6.559 A

787 VARsC

C1

= 50 µF

XC1 = 132 �

I1 = 2.262 A

EC1 = 120 V

271 VARsC1

C2

= 75 µF

XC2 = 35 �

I2 = 3.393 A

EC2 = 120 V

? VARsC2

C3

= 20 µF

XC3 = 132 �

I3 = 0.905 A

EC3 = 120 V

? VARsC3



Unit 19 Capacitors


VARsC2 = 120 x 3.393 = 407.159

ET

= 120 V

F = 60 Hz

XCT = 18 �

IT = 6.559 A

787 VARsC

C1

= 50 µF

XC1 = 132 �

I1 = 2.262 A

EC1 = 120 V

271 VARsC1

C2

= 75 µF

XC2 = 35 �

I2 = 3.393 A

EC2 = 120 V

407 VARsC2

C3

= 20 µF

XC3 = 132 �

I3 = 0.905 A

EC3 = 120 V

? VARsC3



Unit 19 Capacitors


VARsC3 = 120 x 0.905 = 108.573

ET

= 120 V

F = 60 Hz

XCT = 18 �

IT = 6.559 A

787 VARsC

C1

= 50 µF

XC1 = 132 �

I1 = 2.262 A

EC1 = 120 V

271 VARsC1

C2

= 75 µF

XC2 = 35 �

I2 = 3.393 A

EC2 = 120 V

407 VARsC2

C3

= 20 µF

XC3 = 132 �

I3 = 0.905 A

EC3 = 120 V

109 VARsC3



Unit 19 Capacitors

� A quick check is done by adding the individual VARsand comparing the value to the original VARs.

VARsC = 271.442 + 407.129 + 108.573 = 787.174

ET

= 120 V

F = 60 Hz

XCT = 18 �

IT = 6.559 A

787 VARsC

C1

= 50 µF

XC1 = 132 �

I1 = 2.262 A

EC1 = 120 V

271 VARsC1

C2

= 75 µF

XC2 = 35 �

I2 = 3.393 A

EC2 = 120 V

407 VARsC2

C3

= 20 µF

XC3 = 132 �

I3 = 0.905 A

EC3 = 120 V

109 VARsC3



Unit 19 Capacitors

RC Time Constants� Capacitors charge and discharge atexponential rates.

� A curve of the charge or discharge rate isdivided into five time constants.

� Each time constant is equal to 63.2% of the remaining voltage.



Unit 19 Capacitors

RC Time Constants� The formula for determining charge timeis:

� = R x C, where

� (tau) = the time for one time constant inseconds

� R = resistance in ohms

� C = capacitance in farads



Unit 19 Capacitors

Capacitors charge at an exponential rate.



Unit 19 Capacitors

Capacitor discharge curve.



Unit 19 Capacitors

Oil-filled paper capacitor.



Unit 19 Capacitors

Marks indicate plate nearest capacitor case.



Unit 19 CapacitorsP

olarized Capacitors� Polarized capacitors are generallyreferred to as electrolytic capacitors.

� These capacitors have one terminalidentified as positive or negative.

� Polarized capacitors are only to be usedin DC circuits.

� Electrolytic capacitors have very highcapacitance in a small case.



Unit 19 Capacitors

Wet-type electrolytic capacitor.



Unit 19 Capacitors

Identification of mica and tubular capacitors.



Unit 19 Capacitors

Color codes for ceramic capacitors.



Unit 19 Capacitors

Film-type capacitors.



Unit 19 Capacitors

Testing a capacitor with an ohmmeter.



Unit 19 Capacitors

Determining the capacitance value.



Unit 19 Capacitors

Review:1. Capacitors are devices that oppose achange of voltage.

2. Three factors that determine thecapacitance of a capacitor are:

a. the surface area of the plates.

b. the distance between the plates.c. the type of dielectric.



Unit 19 Capacitors

Review:3. A capacitor stores energy in anelectrostatic field.

4. Current can flow only during the time acapacitor is charging or discharging.

5. Capacitors charge and discharge at anexponential rate.

6. The basic unit of capacitance is thefarad.



Unit 19 Capacitors

Review:7. Capacitors are generally rated inmicrofarads, nanofarads, or picofarads.

8.W

hen capacitors are connected inparallel, their capacitance values add.

9. When capacitors are connected inseries, the reciprocal of the totalcapacitance is equal to the sum of thereciprocals of all the capacitors.



Unit 19 Capacitors

Review:10.The charge and discharge times of acapacitor are proportional to the amountof capacitance and resistance in thecircuit.

11.Five time constants are required tocharge or discharge a capacitor.

12.Nonpolarized capacitors are often called AC capacitors.



Unit 19 Capacitors

Review:13.Nonpolarized capacitors can beconnected to direct or alternating currentcircuits.

14.Polarized capacitors are often referred toas electrolytic capacitors.

15.Polarized capacitors can only beconnected to direct current circuits.



Unit 19 CapacitorsReview:

19.Capacitors are often marked with color codes or with numbers and letters.

20.To test a capacitor for leakage, amicroammeter should be connected inseries with the capacitor and the ratedvoltage applied to the circuit.

unit 19 - capacitors (1)

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