chapter 21: alternating currents

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Fisica Generale - Alan Giambattista, Betty McCarty Richardson

Copyright © 2008 – The McGraw-Hill Companies s.r.l.

Chapter 21: Alternating Currents

•Sinusoidal Voltages and Currents

•Capacitors, Resistors, and Inductors in AC Circuits

•Series RLC Circuits

•Resonance

•AC to DC Conversion

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§21.1 Sinusoidal Currents and Voltage

A power supply can be set to give an EMF of form:

tt sin)( 0

This EMF is time dependent, has an amplitude 0, and varies with angular frequency .

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f 2

angular frequency in rads/sec

frequency in cycles/sec or Hz

The current in a resistor is still given by Ohm’s Law:

tItRR

ttI

sinsin)(

)( 00

The current has an amplitude of I0=0/R.

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The instantaneous power dissipated in a resistor will be:

tIttI

tVtIP R

20000 sinsinsin

)()(

The power dissipated depends on t (where in the cycle the current/voltage are).

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What is the average power dissipated by a resistor in one cycle?

The average value sin2t over one cycle is 1/2.

.2

100IPav The average power is

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What are the averages of V(t) and I(t) over one cycle?

The “problem” here is that the average value of sin t over one complete cycle is zero! This is not a useful way to characterize the quantities V(t) and I(t).

To fix this problem we use the root mean square (rms) as the characteristic value over one cycle.

2 and

20

rms0

rms

I

I

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In terms of rms quantities, the power dissipated by a resistor can be written as:

R

222

1

2rms2

rmsrmsrms

0000av

RII

IIP

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Example (text problem 21.4): A circuit breaker trips when the rms current exceeds 20.0 A. How many 100.0 W light bulbs can run on this circuit without tripping the breaker? (The voltage is 120 V rms.)

Each light bulb draws a current given by:

Amps 83.0

V 120 Watts100

rms

rms

rmsrmsav

I

I

IP

If 20 amps is the maximum current, and 0.83 amps is the current drawn per light bulb, then you can run 24 light bulbs without tripping the breaker.

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Example (text problem 21.10): A hair dryer has a power rating of 1200 W at 120 V rms. Assume the hair dryer is the only resistance in the circuit.

(a) What is the resistance of the heating element?

12

V 120 Watts1200

2

rms2

av

RR

RP

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(b) What is the rms current drawn by the hair dryer?

Amps 10

V 120 Watts1200

rms

rms

rmsrmsav

I

I

IP

(c) What is the maximum instantaneous power that the resistance must withstand?

00max2

00 sin IPtIP 002

1 IPav

Pmax = 2Pav = 2400 Watts

Example continued:

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§21.3-4 Capacitors, Resistors and Inductors in AC circuits

)()( tCVtQ CFor a capacitor:

In the circuit:

t

tVC

t

tQtI C )()()(

Slope of the plot V(t) vs. t

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The current in the circuit and the voltage drop across the capacitor are 1/4 cycle out of phase. Here the current leads the voltage by 1/4 cycle.

Here it is true that VCI. The equality is Vc = IXC where XC is called capacitive reactance. (Think Ohm’s Law!)

CXC

1 Reactance has

units of ohms.

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For a resistor in an AC circuit,

.)()( RtItV

The voltage and current will be in phase with each other.

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For an inductor in an AC circuit:

t

tILVL

)(

Also, VL = IXL where the inductive reactance is:

LX L

Slope of an I(t) vs. t plot

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The current in the circuit and the voltage drop across the inductor are 1/4 cycle out of phase. Here the current lags the voltage by 1/4 cycle.

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Plot of I(t), V(t), and P(t) for a capacitor.

The average power over one cycle is zero. An ideal capacitor dissipates no energy.

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A similar result is found for inductors; no energy is dissipated by an ideal inductor.

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§21.5 Series RLC Circuits

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2sinsin

2sin

sin)( 0

tVtVtV

tt

CRL

Applying Kirchhoff’s loop rule:

0)()()()( tVtVtVt CRL

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To find the amplitude (0) and phase () of the total voltage we add VL, VR, and VC together by using phasors.

IZ

XXRI

IXIXIR

VVV

CL

CL

CLR

22

22

220

Z is called impedance.

X

y

VR

VL

VC

0

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The phase angle between the current in the circuit and the input voltage is:

Z

RV

R

XX

V

VV

R

CL

R

CL

0

cos

tan

>0 when XL> XC and the voltage leads the current (shown above).

<0 when XL< XC and the voltage lags the current.

X

y

VR

VL

VC

0

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Example (text problem 21.79): In an RLC circuit these three elements are connected in series: a resistor of 20.0 , a 35.0 mH inductor, and a 50.0 F capacitor. The AC source has an rms voltage of 100.0 V and an angular frequency of 1.0103 rad/sec. Find…

(a) The reactances of the capacitor and the inductor.

0.201

0.35

CX

LX

C

L

(b) The impedance.

0.2522CL XXRZ

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(c) The rms current:

Amps 00.4 25.0

V 0.100

Zrms

rms

rmsrms

I

ZI

(d) The current amplitude:

Amps 66.52

2

rms0

0rms

II

II

Example continued:

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(e) The phase angle:

rads 644.075.0tan

75.020

2035tan

1

R

XX CL

(f) The rms voltages across each circuit element:

V 0.80

V 140

V 0.80

rms,rms

rms,rms

rms,rms

CC

LL

R

XIV

XIV

RIV

(Or 37°)

Example continued:

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(g) Does the current lead or lag the voltage?

(h) Draw a phasor diagram.

Since XL>XC, is a positive angle. The voltage leads the current.

y

XVR

VL

VC

rms

Example continued:

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The power dissipated by a resistor is:

cosrmsrmsrms,rmsav IIP R

where cos is called the power factor (compare to slide 7; Why is there a difference?).

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§21.6 Resonance in RLC Circuits

A plot of I vs. for a series RLC circuit has a peak at = 0.

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The peak occurs at the resonant frequency for the circuit.

22CL XXRZ

I

The current will be a maximum when Z is a minimum. This occurs when XL = XC (or when Z=R).

LC

CL

XX CL

1

1

0

00

This is the resonance frequency for the circuit.

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At resonance:

1cos

0tan

R

RR

XX CL

The phase angle is 0; the voltage and the current are in phase. The current in the circuit is a maximum as is the power dissipated by the resistor.

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§21.7 Converting AC to DC; Filters

A diode is a circuit element that allows current to pass through in one direction, but not the other.

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The plot shows the voltage drop across the resistor. During half a cycle, it is zero.

Putting a capacitor in the circuit “smoothes” out VR, producing a nearly constant voltage drop (a DC voltage).

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A capacitor may be used as a filter.

Low-pass filter. When XC << R ( is large) the output voltage will be small compared to the input voltage.

When XC >> R ( is small), the output voltage will be comparable to the input voltage.

This circuit will allow low frequency signals to pass through while filtering out high frequency signals.

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A high-pass filter. This will allow high frequency signals to pass through while filtering out low frequency signals.

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Summary

•Difference Between Instantaneous, Average, and rms Values

• Power Dissipation by R, L, and C

•Reactance for R, L, and C

•Impedance and Phase Angle

•Resonance in an RLC Circuit

•Diodes

•High- and Low-Pass Filters

chapter 21: alternating currents

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