outline e. electricity and magnetism...

E. ELECTRICITY AND MAGNETISM

18. Alternating currents

1

Outline 18.1 Alternating current through a resistor 18.2 Alternating currents through an inductor 18.3 Alternating currents through a capacitor 18.4 R-C and R-L circuits in series

2

Objectives (a) explain the concept of the r.m.s. value of an alternating

current, and calculate its value for the sinusoidal case only (b) derive an expression for the current from V = V0 sin t (c) explain the phase difference between the current and voltage

for a pure resistor (d) derive and use the formula for the power in an alternating

current circuit which consists only of a pure resistor (e) derive an expression for the current from V = V0 sin t (f) explain the phase difference between the current and voltage

for a pure inductor (g) define the reactance of a pure inductor (h) use the formula XL = L (i) derive and use the formula for the power in an alternating

current circuit which consists only of a pure inductor 3

Objectives j) derive an expression for the current from V = V0 sin t k) explain the phase difference between the current and voltage for

a pure capacitor l) define the reactance of a pure capacitor m) use the formula XC = 1/ C n) derive and use the formula for the power in an alternating

current circuit which consists only of a pure capacitor o) define impedance p) use the formula q) sketch the phasor diagrams of R-C and R-L circuits

4

Introduction Alternating currents and voltages vary with time and periodically change their direction

5

Sine waves by far the most important form of alternating quantity

important properties are shown below

6

Instantaneous value shape of the sine wave is defined by the sine function

y = A sin in a voltage waveform

v = Vp sin

7

Angular frequency frequency f (in hertz) is a measure of the number of cycles per second each cycle consists of 2 radians therefore there will be 2 f radians per second this is the angular frequency (units are rad/s)

= 2 f

8

the angular frequency can be thought of as the rate at which the angle of the sine wave changes at any time

= t therefore

v = Vp sin t or v = Vp sin 2 ft

similarly

i = Ip sin t or i = Ip sin 2 ft

Equation of a sine wave

9

Example Determine the equation of the following voltage

signal. From diagram:

Period is 50 ms = 0.05 s Thus f = 1/T =1/0.05 = 20 Hz Peak voltage is 10 V Therefore

tt

ftpVv

126sin10202sin10

2sin

10

18.1 Alternating currents through a resistor

11

If alternating voltage is applied across a resistor, the voltage and current across the resistor obeys

V = IR

12

Phase angles the expressions given above assume the angle of the sine wave is zero at t = 0 if this is not the case the expression is modified by adding the angle at t = 0

13

Phase difference two waveforms of the same frequency may have a constant phase difference we say that one is phase-shifted with respect to the other

14

Average value of a sine wave average value over one (or more) cycles is clearly zero however, it is often useful to know the average magnitude of the waveform independent of its polarity

we can think of this as the average value over

of the rectified signal p

p

p

pav

VV

V

VV

637.02

cos

dsin1

0

0

15

Average value of a sine wave

16

r.m.s. value of a sine wave

the instantaneous power (p) in a resistor is given by

therefore the average power is given by

where is the mean-square voltage

R

vp

2

2v

R

v

R

vavP

2] of mean) (or average[ 2

17

r.m.s. value of a sine wave While the mean-square voltage is useful, more often we use the square root of this quantity, namely the root-mean-square voltage Vrms

where Vrms =

we can also define Irms =

it is relatively easy to show that

2v

2i

pprms VVV 707.02

1pprms III 707.0

21

18

r.m.s. value of a sine wave r.m.s. values are useful because their relationship to average power is similar to the corresponding DC values

rmsrmsavIVP

RIPrmsav

2

RV

P rmsav

2

19

Form factor for any waveform the form factor is defined as

for a sine wave this gives

value averagevalue r.m.s.factor Form

11.1 0.637 0.707

factor FormpVpV

20

Peak factor for any waveform the peak factor is defined as

for a sine wave this gives

value r.m.s.value peakfactor Peak

414.1 0.707factor PeakpV

pV

21

Square Waves Frequency, period, peak value and peak-to-peak value have the same meaning for all repetitive waveforms

22

Phase angle we can divide the period into 360 or 2 radians useful in defining phase relationship between signals in the waveforms shown here, B lags A by 90 we could alternatively give the time delay of one with respect to the other

23

Average and r.m.s. values the average value of a symmetrical waveform is its average value over the positive half-cycle

thus the average value of a symmetrical square wave is equal to its peak value similarly, since the instantaneous value of a square wave is either its peak positive or peak negative value, the square of this is the peak value squared, and

pVavV

pVrmsV24

Form factor and peak factor from the earlier definitions, for a square wave

0.1value averagevalue r.m.s.factor Form

pVpV

0.1value r.m.s.value peakfactor Peak

pVpV

25

Power The instantaneous power (for some frequency, w) delivered at time t is given by:

The most useful quantity to consider here is not the instantaneous power but rather the average power delivered in a cycle.

To evaluate the average on the right, we first expand the sin( t- ) term.

Power in Alternating Current Circuits

26

Power (Product of even and odd function = 0)

sin tcos t

t 0

0

+1

-1

sin2 t

t 0

0

+1

-1 tttItP pp cossinsinsincos)( 2

cos21)( ppItP

0 1/2

Expanding,

Taking the averages,

)sin(sin tt

0cossin ttGenerally: 2

0

22

21sin

21sin xdxx

Putting it all back together again,


)sincoscos(sinsin ttt

27

Power

prms2

1prms II

21

cos)( rmsrmsItP

Phase depends on the values of L, C, R, and w Therefore...

This result is often rewritten in terms of rms values:

cos)( rmsrms ItP


28

Power

cosR

I mp

22 2

rms( ) cosrmsP t I RR

We can write this in the following manner (which we

2222

22

)1()(

xQxx

RtP rms

Power, as well as current, peaks at w = w0. The sharpness of the resonance depends on the values of the components. Recall:


29

Average and r.m.s. values Let us first consider the general case of a current the waveform of which cannot be represented by a simple mathematical expression. For instance, the waveform shown in Fig. below.

i1 i2 i3 i4

i7 i8 i9 in

t

Fig.4.4 30

i1 i2 i3 i4

i7 i8 i9 in

t

Fig.4.4

Average value of current

= Iav = nni3i2i1i

For a pure sinusoidal wave, Iav = 0.637 Im Vav = 0.637 Vm

Average and r.m.s. values

31

RI smr2

...

2n

22

21 )(

nRiRiRi

Average and r.m.s. values In a.c. work, however, the average value is of comparatively little importance. This is due to the fact that it is the power produced by the current that usually matters. Thus, if the current represented in Fig.4.4 is passed through a resistor having resistance R ohms, the heating effect of i1 is (i1)2R, that of i2 is (i2)2R etc., as shown in Fig. 4.5. Therefore, the average heating effect in half-cycle is :

32

So, Ir.m.s. = root-mean-square of the current

niiii 2

n23

22

21

For a pure sinusoidal wave,

Ir.m.s. = 0.707 Im

Vr.m.s. = 0.707 Vm


33

t

Heating effect

(i3)2 (i8)2

r.m.s. value

(i7)2

(i9)2 (i2)2 (i4)2


34

Average v value of a sine wave average value over one (or more) cycles is clearly zero

however, it is often useful to know the average magnitude of the waveform independent of its polarity we can think of this as the average value over

of the rectified signal p

p

p

pav

VV

V

VV

637.02

cos

dsin1

0

0

35

Average v value of a sine wave

36

r.m.s. voltage value of a sine wave the instantaneous power (p) in a resistor is given by

therefore the average power is given by

where is the mean-square voltage

R

vp

2

2v

R

v

R

vavP

2] of mean) (or average[ 2

37

Average v value of a sine wave While the mean-square voltage is useful, more often we use the square root of this quantity, namely the root-mean-square voltage Vrms

where Vrms =

we can also define Irms =

it is relatively easy to show that (see text for

analysis)

2v2i

pprms VVV 707.02

1

pprms III 707.02

1

38

Average v value of a sine wave r.m.s. values are useful because their relationship to average power is similar to the corresponding DC values

rmsrmsavIVP

RIPrmsav

2

RV

P rmsav

2

39

Average v value of a sine wave Form factor

for any waveform the form factor is defined as for a sine wave this gives

value averagevalue r.m.s.factor Form

11.1 0.637 0.707

factor FormpVpV

40

Average v value of a sine wave Peak factor

for any waveform the peak factor is defined as for a sine wave this gives

value r.m.s.value peakfactor Peak

414.1 0.707factor PeakpV

pV

41

Square Waves Frequency, period, peak value and peak-to-peak value have the same meaning for all repetitive waveforms

42

Phase angle we can divide the period into 360 or 2 radians

useful in defining phase relationship between signals in the waveforms shown here, B lags A by 90 we could alternatively give the time delay of one with respect to the other

43

Average and r.m.s. values the average value of a symmetrical waveform is its average value over the positive half-cycle

thus the average value of a symmetrical square wave is equal to its peak value similarly, since the instantaneous value of a square wave is either its peak positive or peak negative value, the square of this is the peak value squared, and

pVavV

pVrmsV

44

Form factor and peak factor from the earlier definitions, for a square wave

0.1value averagevalue r.m.s.factor Form

pVpV

0.1value r.m.s.value peakfactor Peak

pVpV

45

Pure Resistor in the AC Circuit

46

I V

Phasor diagram

VR I

47

The current flows in the resistor is II 0 sin

The voltage across the resistor VR at any instant is

IRVR

RIV 0R sin 00 VRI

VVV 0R sin voltageSupply :V

The phase difference between V and I is

0 In pure resistor, the voltage V is in phase with the

current I and constant with time.(the current and the voltage reach their maximum values at the same time). 48

The resistance in a pure resistor is The instantaneous power, o

o

rms

rms

IV

IVR

tVIP

tVtIPR

VRIIVP

oo

oo

2

22

sin

sinsin

The average power,

o

oo

o

rmsave

P

IV

RI

RIP

212121 2

2

A resistor in ac circuit dissipates energy in the form of heat

t0

0P

TT21 T2T

23

)(PPower

18.2 Alternating currents through an inductor

49

Inductor Circuits An inductor is a device that produces a uniform magnetic field when a current passes through it. A solenoid is an inductor.

L| = L dI/dt

50

Pure Inductor in the AC Circuit

51

IV

Phasor diagram

II VL

rad2

Pure inductor means that no resistance and capacitance effect in the a.c. circuit.

52

When a sinusoidal voltage is applied across a inductor, the voltage reaches its maximum value one quarter of a cycle before the current reaches its maximum value,( ) rad

2 The current flows in the ac circuit is II 0 sin When the current flows in the inductor, the back emf

caused by the self induction is produced and given by

dtdILB

IdtdL 0B sin

LI 0B cos

53

At each instant the supply voltage V must be equal to the back e.m.f B (voltage across the inductor) but the back e.m.f always oppose the supply voltage V.

Hence, the magnitude of V and B , LIV 0B cos or

2sin

2sin0

VV

LIV

o

where oo LIV

B

B

VIRV0

54


2

2

In pure inductor, the voltage V leads the current I by /2 radians or the current I lags behind the voltage V by /2 radians.

The inductive reactance in a pure inductor is

2

rms o oL

rms o o

L

V V LIXI I I

X L fL

55

The inductive reactance is defined as 2LX L fL

The instantaneous power, The average power, 0aveP

tPP

tVIP

tVtIPR

VRIIVP

o

oo

oo

2sin21

2sin21

cossin

22

t0

2P0

2P0

TT21 T2T

23

)(PPower

For the first half of the cycle where the power is positive, the inductor is saving the power. For the second half cycle where the power is negative, the power is returned to the circuit.

Example 1 A coil having an inductance of 0.5 H is connected to a 120 V, 60 Hz power source. If the resistance of the coil is neglected, what is the effective current through the coil.

56

A 0.642 fLV

XVI rms

L

rmsrms

Example 2 A 240 V supply with a frequency of 50 Hz causes a current of 3.0 A to flow through an pure inductor. Calculate the inductance of the inductor.

57

H 0.25 2

80

LfLXIVX

L

L

18.3 Alternating currents through a capacitor

58

Pure capacitor means that no resistance and self-inductance effect in the a.c. circuit.

Pure Capacitor in the AC Circuit

59

I

VPhasor diagram

I I

rad2

VR

60

CVVV )2

sin(0

When an alternating voltage is applied across a capacitor, the voltage reaches its maximum value one quarter of a cycle after the current reaches its maximum value,( ) rad

2 The voltage across the capacitor VC at any instant

is equal to the supply voltage V and is given by

The charge accumulates on the plates of the capacitor is

CCVQ)

2sin(0CVQ

dtdQI The current flows in the ac circuit is

61

2sin0CV

dtdI

2sin0 dt

dCVI

)2

cos(0CVI 00 IC Vand

II sin0

or


2

2

62

In pure capacitor, the voltage V lags behind the current I by /2 radians or the current I leads the voltage V by /2 radians.

The capacitive reactance in a pure capacitor is

fCCX

CVV

IV

IVX

C

o

o

o

o

rms

rmsC

211

The capacitive reactance is defined as

fCCX C 2

11

63

The instantaneous power,

tPP

tVIP

tVtIPR

VRIIVP

o

oo

oo

2sin21

2sin21

cossin

22

t0

2P0

TT21 T

23

)(PPower

2P0

T2

The average power,

0aveP

For the first half of the cycle where the power is negative, the power is returned to the circuit. For the second half cycle where the power is positive, the capacitor is saving the power.

Example An 8.00 F capacitor is connected to the terminals of an AC generator with an rms voltage of 150 V and a frequency of 60.0 Hz. Find the capacitive reactance rms current and the peak current in the circuit.

64

Capacitive reactance, 332fCC

X c 211

Rms current,

A0.452C

rmsrms X

VI

Peak current

AIpI rms

320.0452.0707.02

18.4 R-C and R-L circuits in series

65

RC in series circuit

In the circuit diagram: VR and VC represent the instantaneous voltage across the resistor and the capacitor. In the phasor diagram: VR and VC represent the peak voltage across the resistor and the capacitor. 66

R

IRV

V

CV

C

CVI

RV

V

Phasor diagram

voltagesupply angle phase:

V

67

Note CV

IRV

V

Phasor diagram

2

222

22

22

22

22

222

222

1C

RIV

XRIV

XRIV

XRIV

XIRIV

XIRIV

VVV

C

Crmsrms

Coo

Looo

Looo

CoRoo

2by side both ...divide

68

R

IRV

V

CV

C

CVI

RV

V

Phasor diagram


V

The total p.d (supply voltage), V across R and C is equal to the vector sum of VR and VC as shown in the phasor diagram.

2C

2R

2 VVV2

C22 IXIRV

2C

222 XRIV X C1

222

C1RIV

where

IRVR

CC IXV

69

CVI

RV

V

Phasor diagram

CX

R

Z

The impedance in RC circuit,

222

222

1

1

CRZ

IC

RI

IVZ

rms

rmsR

C

VVtan

IRIX Ctan

RX Ctan

1tan

From the phasor diagrams,

or

I leads V by

Impedance diagram

70

fCX C 2

1

Graph of Z against f

Z

f 0

R

Example An alternating current of angular frequency of 1.0 x 104 rad s-1 flows through a 10 k resistor and a 0.10 F capacitor which are connected in series. Calculate the rms voltage across the capacitor if the rms voltage across the resistor is 20 V.

71

R

C

VVtan

RX CtanFrom and

0.1 tanCRR

X C 1

V2.00.120

0.1

C

R

C

VVV

RL in series circuit

72

R

IRV

V

L

LVLV

IRV

V

Phasor diagram

tagesupply vol angle phase:

V

The voltage across the resistor VR and the inductor VL are

IRVR

LL IXV

73

R

IRV

V

L

LV LV

IRV

V

Phasor diagram


V

The total p.d (supply voltage), V across R and L is equal to the vector sum of VR and VL as shown in the phasor diagram.

2L

2R

2 VVV2

L22 IXIRV

2L

222 XRIV X L

222 LRIVwhere

74

LV

IRV

V

Phasor diagram

LXZ

RImpedance diagram

The impedance in RC circuit,


222

222

LRZ

ILRI

IVZ

rms

rmsR

L

VVtan

IRIX Ltan

RX Ltan

Rtan

or

V leads I by

75

fLX L 2

Graph of Z against f

Z

f 0

R

RLC in series circuit

76

I

V

R

RV

L

LVCV

C

77

I

V

R

RV

L

LV CV

CLV

IRV

V

Phasor diagram CV

CL VV

78

I

V

R

RV

L

LV CV

C LV

IRV

V

Phasor diagram CV

CL VV

The voltage across the inductor VL , resistor VR and capacitor VC are

LL IXV IRVR CC IXV

79

I

V

R

RV

L

LV CV

CLV

IRV

V

Phasor diagram CV

CL VV

The total p.d (supply voltage), V across L, R and C is equal to the vector sum of VL ,VR and VC as shown in the phasor diagram.

2CL

2R

2 VVVV2

CL22 IXIXIRV

2222CL XXRIV

2CL

2 XXRIV

80

CL XX

LX

R

Z

CX

LV

IRV

V

Phasor diagram CV

CL VV

Impedance diagram

The impedance in RLC circuit,


22

22

CL

CL

rms

rms

XXRZ

IXXRI

IVZ

V leads I by

R

CL

VVVtan

IRXXI CLtan

RXX CLtan

R

1

tan

VS

R L C

tot L CX X X

RLC in series circuit RLC in series circuit When a circuit contains an inductor and capacitor in series, the reactance of each tend to cancel. The total reactance is given by The total impedance is given by The phase angle is given by 1tan totX

R

2 2tot totZ R X

81

Rea

ctan

ce

f

XC XL

Series resonance

XC=XL

XC>XL XL>XC

Variation of XL and XC In a series RLC circuit, the circuit can be capacitive or inductive, depending on the frequency. At the frequency where XC=XL, the circuit is at series resonance. Below the resonant frequency, the circuit is predominantly capacitive. Above the resonant frequency, the circuit is predominantly inductive. 82

VS

R L C

1.0 kW XC = 5.0 kW

XL = 2.0 kW

The total reactance is 2.0 k 5.0 k 3.0 ktot L CX X X

The total impedance is 2 2 2 21.0 k +3.0 ktot totZ R X 3.16 k

The circuit is capacitive, so I leads V by 71.6o.

The phase angle is 1 1 3.0 ktan tan1.0 k

totXR

71.6o

Example 1 What is the total impedance and phase angle of the series RLC circuit if R= 1.0 k , XL = 2.0 k , and XC = 5.0 k ?

83

R L C

VS

330 H

f = 100 kHz

470 2000 pF

2 2 100 kHz 330 H 207 LX fL

1 1 796 2 2 100 kHz 2000 pFCX

fC

207 796 589 tot L CX X X

2 2 = 470 589 Z 753

Example 2 What is the magnitude of the impedance for the circuit?

84

XL

f

XC

X

XL

XC

Impedance of Series RLC Depending on the frequency, the circuit can appear to be capacitive or inductive. The circuit in the Example-2 was capacitive because XC > XL

85

R L C

VS

330 mH f = 400 kHz

470 W 2000 pF

The circuit is now inductive.

2 2 400 kHz 330 H 829 LX fL1 1 199

2 2 400 kHz 2000 pFCXfC

829 199 630 tot L CX X X

2 2 = 470 630 Z 786

Example 3 What is the total impedance for the circuit when the frequency is increased to 400 Hz?

86

XL

f

XC

X

By changing the frequency, the circuit in Example-3 is now inductive because XL>XC

XL

XC

Impedance of Series RLC

87

Voltages in a series RLC circuits

Notice that VC is out of phase with VL. When they are algebraically added,

0

VC

VL

This example is inductive.

Voltages in Series RLC The voltages across the RLC components must add to the source voltage in accordance with KVL. Because of the opposite phase shift due to L and C, VL and VC effectively subtract.

88

89

Exercise A series RLC circuit has a resistance of 25.0 , a capacitance of 50.0 F, and an inductance of 0.300 H. If the circuit is driven by a 120 V, 60 Hz source, calculate

a)The total impedance of the circuit b)The rms current in the circuit c)The phase angle between the voltage and the

current.

90

64.9 , 1.85 A, 67.3o

Summary 18.1 Alternating currents through a resistor Power, Prms = Vrms Irms R.m.s. value, Irms = 0.707I0, Vrms = 0.707V0 18.2 Alternating currents through an inductor V leads I by /2 radians I = I0sin t, V = V0cos t, P=1/2 I0Vosin(2 t) 18.3 Alternating currents through a capacitor I leads V by /2 radians I = I0cos t, V = V0sin t, P=1/2 I0Vosin(2 t) 18.4 R-C and R-L circuits in series Impedance, 91

MORE ABOUT

92

Half-wave Rectifiers

Forward biased

93

Half-wave Rectifiers

Reverse biased

Output result

94

Half-wave Rectifier Note that the frequency stays the same Strength of the signal is reduced Vavg = Vp(out)/ = 0.318 x Vp(out) [31.8 % of Vp] Vp(out) = Vp(in) VBar For Silicon VBar = 0.7 V

Half-wave Rectifier

Vp(out) Vp(in)

Vavg

2 95

Half-wave Rectifier - Example

Output:

Peak Inverse Voltage (PIV) The peak voltage at which the diode is reverse biased In this example PIV = Vp(in)-

Hence, the diode must be rated for PIV = 100 V

Draw the output signal Vp(out) = Vp(in) 0.7 Vavg = 99.3/What happens to the

frequency?

96

Transformers (Review) Transformer: Two inductors coupled together separated by a dielectric

When the input magnetic field is changing voltage is induced on the second inductor The dot represents the + (voltage direction)

Applications: Step-up/down Isolate sources

Turns ratio (n) n = Sec. turns / Pri. turns = Nsec/Npri

Vsec = n. Vpri depending on value of n : step-up or step-down

Center-tapped transformer Voltage on each side is Vsec/2

97

Half-wave Rectifier - Example Example:

Assume that the input is a sinusoidal signal with Vp=156 V & T = 2 msec; assume Nsec:Npri = 1:2 Draw the signal Find turns ratio; Find Vsec; Find Vout.

n = ½ = 0.5 Vsec = n.Vpri = 78 V Vout = Vsec 0.7 = 77.3 V

78-0.7

98

Full-wave Rectifier Note that the frequency is doubled Vavg = 2Vp(out)/ = 0.637 x Vp(out)

99

Full-wave Rectifier Circuit

Center-tapped full-wave rectifier Each half has a voltage = Vsec/2

Only one diode is forward biased at a time The voltages at different halves are opposite of each other

100

Full-wave Rectifier Circuit Center-tapped full-wave rectifier

Each half has a voltage = Vsec/2

Only one diode is forward biased at a time The voltages at different halves are opposite of each other

101

Full-wave Rectifier Circuit Vout = Vsec /2 0.7 Peak Inverse Voltage (PIV)

PIV = (Vsec/2 0.7)- (-Vsec/2) = Vsec 0.7 Vout = Vsec/2 0.7

Assuming D2 is reverse-biased No current through D2

102

Full-wave Rectifier - Example Assuming a center-tapped transformer

Find the turns ratio Find Vsec Find Vout Find PIV Draw the Vsec and Vout What is the output freq?

Vsec

n=1:2=0.5 Vsec=n*Vpri=25 Vout = Vsec/2 0.7 PIV = Vsec-0.7=24.3 V

103

Full-wave Rectifier - Multisim XFrmr can be virtual or real Use View Grapher to see the details of your results The wire-color can determine the waveform color Make sure the ground is connected to the scope.

104

Bridge Full-wave Rectifier Uses an untapped transformer larger Vsec Four diodes connected creating a bridge

When positive voltage D1 and D2 are forward biased When negative voltage D3 and D4 are forward biased

Two diodes are always in series with the load

Vp(out) = Vp(sec) 1.4V The negative voltage is inverted

The Peak Inverse Voltage (PIV) PIV=Vp(out)+0.7 105

Bridge Full-wave Rectifier - Example Assume 12 Vrms secondary voltage for the standard 120 Vrms across the primary

Find the turns ratio Find Vp(sec) Show the signal

direction when Vin is positive Find PIV rating

n=Vsec/Vpri = 0.1 10:1 Vp(sec) = (0.707)-1 x Vrms = 1.414(12)=17 V Vp(out) = V(sec) (0.7 + 0.7) = 15.6 V through D1&D2 PIV = Vp(out) + 0.7 = 16.3 V

120Vrms

Note: Vp-Vbr ; hence, always convert from rms to Vp 106

Bridge Full-wave Rectifier - Comparison

120Vrm

s

Vp(2)=Peak secondary voltage ; Vp(out) Peak output voltage ; Idc = dc load current

107

Smoothing by capacitors Filters and Regulators

108

Filters and Regulators

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-Ripple voltage depends on voltage variation across the capacitor - Large ripple means less effective filter


peak-to-peak ripple voltage

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Too much ripple is bad! Ripple factor = Vr (pp) / VDC Vr(pp) = (1/ fRLC) x Vp(unfiltered) VDC = (1 1/ fRLC) x Vp(unfiltered)


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Summary

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outline e. electricity and magnetism...

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