ISM chapter33 tủ tài liệu training

This charge, and hence the voltage ΔV C, is a maximum when the current has zero value and is in the process of reversing direction after having been flowing in one direction for a half c

33.7 Resonance in a Series RLC Circuit

33.8 The Transformer and Power Transmission

33.9 Rectifiers and Filters

* An asterisk indicates a question or problem new to this edition

ANSWERS TO OBJECTIVE QUESTIONS

OQ33.1 Answer (c) When a power source, AC or DC, is first connected to a

RL combination, the presence of the inductor impedes the buildup of

a current in the circuit The value of the current starts at zero and increases as the back emf induced across the inductor decreases in magnitude

OQ33.2 (i) Answer (e) Inductive reactance, X L=ωL , doubles when the

frequency doubles, so the rms current is halved

I( rms= ΔVrms X L)

(ii) Answer (b) Capacitive reactance, X C = 1ωC , is cut in half

when frequency doubles, so the rms current doubles

I( rms= ΔVrms X C)

(iii) Answer (d) The resistance remains unchanged I( rms= ΔVrms R)

OQ33.3 Answer (a) The voltage across the capacitor is proportional to the

stored charge This charge, and hence the voltage ΔV C, is a maximum when the current has zero value and is in the process of reversing direction after having been flowing in one direction for a half cycle Thus, the voltage across the capacitor lags behind the

current by 90° The capacitive reactance, X C = 1 ωC , decreases as

frequency increases, causing the impedance to decrease and the current to increase

OQ33.4 (i) Answer (d)

OQ33.5 Answer (d) If the voltage is a maximum when the current is zero, the

voltage is either leading or lagging the current by 90° (or a quarter

cycle) in phase Thus, the element could be either an inductor or a

capacitor It could not be a resistor since the voltage across a resistor

is always in phase with the current If the current and voltage were out of phase by 180°, one would be a maximum in one direction when the other was a maximum value in the opposite direction

OQ33.6 (i) Answer (e) The voltage varies between +170 V and –170 V

(ii) Answer (c) The average of a sine waveform is zero

OQ33.8 Answer (e) is false In an RLC circuit, the instantaneous voltages Δv R,

Δv L , and Δv C (across the resistance, inductance, and capacitance respectively) are not in phase with each other The instantaneous

voltage Δv R is in phase with the current, ∆v L leads the current by 90°,

while Δv C lags behind the current by 90° The instantaneous values

of these three voltages do add algebraically to give the instantaneous

voltages across the RLC combination, but the rms voltages across

these components do not add algebraically The rms voltages across the three components must be added as vectors (or phasors) to obtain

the correct rms voltage across the RLC combination

OQ33.9 Answer (c) At resonance the inductive reactance and capacitive

reactance cancel out

OQ33.10 Answer (c) At resonance the inductive reactance and capacitive

reactance add to zero: φ = tan–1

[(X L – X C )/R] = 0

OQ33.11 The ranking is (a) > (d) > (b) > (c) > (e) At the resonance frequency

f0 = 1 000 Hz both X L and X C are equal: call their mutual value X A high-Q value means the resonance has a small width, so X L and X C are also much larger than R at f0 Inductive reactance X L is

proportional to frequency, and capacitive reactance X C is inversely

proportional to frequency In terms of X, the choices have the values: (a) f = f0 /2, so X C = 2X (b) f = 3f0 /2, so X C = 2X/3 (c) f = f0 /2, so X L = X/2 (d) f = 3f0 /2, so X L = 3X/2 (e) R is independent of frequency, and

R is less than X Thus, we have (a) 2X > (d) 3X/2 > (b) 2X/3 > (c) X/2 >

(e) less than X

OQ33.12 Answer (e) The battery produces a constant current in the primary

coil, which generates a constant flux through the secondary coil With no change in flux through the secondary coil, there is no induced voltage across the secondary coil

OQ33.13 Answer (c) AC ammeters and voltmeters read rms values With an

oscilloscope you can read a maximum voltage, or test whether the average is zero

ANSWERS TO CONCEPTUAL QUESTIONS

CQ33.1 (a) The Q factor determines the selectivity of the radio receiver For

example, a receiver with a very low Q factor will respond to a

wide range of frequencies and might pick up several adjacent radio stations at the same time To discriminate between

102.5 MHz and 102.7 MHz requires a high-Q circuit

(b) Typically, lowering the resistance in the circuit is the way to get

a higher quality resonance

CQ33.2 (a) The second letter in each word stands for the circuit element

For an inductor L, the emf ε leads the current I—thus ELI For a capacitor C, the current leads the voltage across the device In a

circuit in which the capacitive reactance is larger than the inductive reactance, the current leads the source emf—thus ICE (b) CIVIL – in a capacitor C the current (I) leads voltage

(represented by V), voltage leads current in an inductor L

CQ33.3 The voltages are not added in a scalar form, but in a vector form, as

shown in the phasor diagrams throughout the chapter Kirchhoff’s loop rule is true at any instant, but the voltages across different circuit elements are not simultaneously at their maximum values Do not forget that an inductor can induce an emf in itself and that the

voltage across it is 90° ahead of the current in the circuit in phase

CQ33.4 (a) In an RLC series circuit, the phase angle depends on the source

frequency At very low frequency, the capacitor dominates the impedance and the phase angle is near –90° At very high frequencies, the inductor dominates the impedance and the phase angle is near –90°

(b) When the inductive reactance equals the capacitive reactance, the frequency is the resonance frequency; the phase angle is zero

CQ33.5 In 1881, an assassin shot President James Garfield The bullet was lost

in his body Alexander Graham Bell invented the metal detector in an effort to save the President’s life The coil is preserved in the

Smithsonian Institution The detector was thrown off by metal springs in Garfield’s mattress, a new invention itself Surgeons went hunting for the bullet in the wrong place Garfield died

CQ33.6 (a) The person is doing work at a rate of P = Fv cosθ

(b) Compare the previous equation to P = ΔVrmsIrmscosφ One can consider the emf as the “force” that moves the charges through the circuit, and the current as the “speed” of the moving

charges The cosθ factor measures the effectiveness of the cause

in producing the effect Theta is an angle in real space for the vacuum cleaner and phi is the analogous angle of phase difference between the emf and the current in the circuit

CQ33.7 The circuit can be considered an RLC series circuit

(a) Yes The circuit is in resonance because the inductive reactance

and capcitive reactance are equal, so the total impedance Z = R

(b) Total power output by the emf Pemf = I2Rtotal, where Rtotal = 10 Ω (source resistance) + 10 Ω (load resistance) = 20 Ω Power

delivered to the load Pload = I2R L , where R L = 10 Ω Fraction of average power delivered to the load to average power delivered

by the source of emf:

fraction of the emf’s power go to the load Then the emf would

put out a lot less power and less power would reach the load

CQ33.8 No A voltage is only induced in the secondary coil if the flux

through the core changes in time No changing current, no changing flux, no induced voltage

CQ33.9 (a) The capacitive reactance is proportional to the inverse of the

frequency At higher and higher frequencies, the capacitive reactance approaches zero, making a capacitor behave like a wire

(b) As the frequency goes to zero, the capacitive reactance approaches infinity—the resistance of an open circuit

CQ33.10 The ratio of turns indicates the ratio of voltages: N1/N2 = ΔV1/ΔV2,

where ΔV2 = 120 V; therefore, ΔV1 = 12 kV In its intended use, the transformer takes in energy by electric transmission at 12 kV and puts out nearly the same energy by electric transmission at 120 V With the small generator putting energy into the secondary side of the transformer at 120 V, the primary side has 12 kV induced across

it It is deadly dangerous for the repairman

SOLUTIONS TO END-OF-CHAPTER PROBLEMS

P33.1 (a) The rms voltage across the resistor is given by

ΔV R,rms = IrmsR= 8.00 A( ) (12.0 Ω)= 96.0 V (b) From Equation 33.5,

ΔV R,max = 2 ΔV( R,rms)= 2 96.0 V( )= 136 V

ΔV R,max = 2 ΔV( R,rms)= 2 120 V( )= 170 V (b) From the definition of power,

(c)

Because Pavg =(ΔVrms)2

R     →    R =(ΔVrms)2

Pavg , a 100-W bulb has less

resistance than a 60.0-W bulb

P33.5 The current as a function of time is

⎠⎟sinωt Given the

value of t, we want to identify a point with

again For this occurrence, ω t = π − 0.253 rad = 2.89 rad (to

understand why this is true, recall the identity sin π −θ( )= sinθ from trigonometry) Thus,

t= 2.89 rad25.3 rad s= 0.114 s

P33.7 We are given ΔVmax = 15.0 V and Rtotal = 8.20 Ω + 10.4 Ω = 18.6 Ω The

maximum current in the circuit is

P33.8 All lamps are connected in parallel with the voltage source, so

ΔVrms= 120 V for each lamp Also, the current is Irms = Pavg ΔVrms and the resistance is R = ΔVrms Irms

(a) For the 150-W bulbs,

Irms = 150 W

120 V =1.25 A For the 100-W bulb,

Irms = 100 W

120 V =0.833 A The rms current in each 150-W bulb is 1.25 A The rms current in the 100-W bulb is 0.833 A

(b) The resistance in bulbs 1 and 2 is

R1 = R2 = 120 V

1.25 A = 96.0 Ω and the resistance in bulb 3 is

R3 = 120 V0.833 A = 144 Ω (c) The bulbs are in parallel, so

196.0 Ω +

1

144 Ω

Req = 36.0 Ω

Section 33.3 Inductors in an AC Circuit

P33.9 Inductive reactance is proportional to frequency

(b) From

Imax = ΔVmax

X L = ΔVmax

ωL , we see that is current inversely

proportional to angular frequency:

Imax = ΔVmax

X L = 80.0 V

14.3 Ω = 5.60 AEquation 33.7 lets us evaluate the current:

i = −Imaxcosωt= − 5.60 A( )cos 65.0⎡⎣( π s−1) (0.015 5 s)⎤⎦

= − 5.60 A( )cos 3.17 rad( )= +5.60 A

P33.12 The relationship between current, inductance, and maximum voltage

L> ΔV L,max

2 2( )π f(2.00× 10−3 A) = 4.00 V

P33.14 In the inductor, because

U B = 1

2Li L2 = 1

2(0.020 0 H) (19.5 A)2 = 3.80 J

P33.15 The flux Φ B through each turn of the inductor is related to the

inductance by

L= NΦB i

Irms= ΔV L,rms

X L = 84.9 V

47.1 Ω= 1.80 A (e) Imax = 2 Irms = 2 1.80 A( )= 2.55 A

P33.17 Current leads voltage by 90° in a capacitor, and because charge is

proportional to voltage, current leads charge by 90° If Δv C=

ΔVmaxsinω t , then q = C ΔV( max)sinωt so that the stored energy is

U C= q2

2C = 0 when t = 0 Therefore, the current is given by

i C = Imaxsin(ω t + 90°)= ΔVmax

X C sin(ω t + 90°)

The capacitive reactance is

X C = 1

ω C =

12π 60.0 s( ) (1.00× 10−3 C V)= 2.65 Ω

and the current at

12π 60.0 Hz( ) (12.0× 10−6 F) = 221 Ω

(d) No Current leads voltage, and thus charge, by 90° in a

capacitor The current reaches its maximum value one-quarter cycle before the voltage reaches its maximum value From the definition of capacitance, the capacitor reaches its maximum charge when the voltage across it is also a maximum

Consequently, the maximum charge and the maximum current

do not occur at the same time

P33.19 (a) We require

X C = 12π f C < 175 Ω , or

12π f 22.0 × 10( −6 F)< 175 Ω

Solving,

12π 22.0 × 10( −6 F) (175 Ω)< f

or

f > 41.3 Hz

X C = 12π fC =

Imax = 48.0 V( ) ( )2π (90.0 Hz) (3.70× 10−6 F)= 100 mA

P33.22 The maximum charge is given by

Qmax = C ΔV( max)= C 2 ΔV⎡⎣ ( rms)⎤⎦ = 2C ΔV( rms)

P33.23 The maximum current in the capacitor is given by

Imax = 2 240 V( )2π 50.0 s( ) (2.20× 10−6 F)= 235 mA

P33.24 We first determine the reactances of the circuit The capacitive

the inductive reactance is,

X L =ω L = 2π 50.0( ) (185× 10−3 H)= 58.1 Ωand the impedance Z of the circuit is

Imax = ΔVmax

41.0 Ω= 3.66 A

ANS FIG P33.24

(a) The maximum voltage between points a and b is the potential

drop across the resistor:

ΔV R = ImaxR= 3.66 A( ) (40.0 Ω)= 146 V

(b) The maximum voltage between points b and c is the potential

drop across the coil:

ΔV L = ImaxX L = 3.66 A( ) (58.1 Ω)= 212.5 V = 212 V

(c) The maximum voltage between points c and d is the potential

drop across the capacitor:

ω = 100 s−1 The resistance of the circuit is R = 68.0 Ω, the

inductive reactance of the circuit is

X L =ω L = 100 s( −1) (0.160 H)= 16.0 Ω

The capacitive reactance of the circuit is

X C = 1

ω C = 2π 50.0 s⎡⎣ ( −1) (5.00× 10−6 F)⎤⎦−1

= 637 Ω (c) The impedance of the circuit is

Z= ΔVmax

Imax = 240 V

0.100 A = 2.40 × 103 Ω = 2.40 kΩ (d) From the definition of impedance,

⎛

⎝⎜ ⎞⎠⎟ = −14.2°

P33.28 From the definitions of inductive and capacitive reactance, X L =ω L

X C = 1

ωC =

12π fC =

12π 60.0 s( –1) (21.0 × 10–6 F)= 126 Ω

(b) Since X L > X C, φ is positive, so voltage leads the current This

means that the power-supply or total voltage goes through each maximum, zero-crossing, and minimum earlier in time than the current does

P33.30 The Phasors for the three cases are shown in ANS FIG P33.30

(a) 25.0 sin ωt at ωt = 90.0°

(b) 30.0 sin ωt at ωt = 60.0°

(c) 18.0 sin ωt at ωt = 300°

ANS FIG P33.30

P33.31 (a) We first find the impedances of the inductor and the capacitor:

X L =ω L = 2π 50.0( ) (400× 10−3)= 126 Ω and

X C = 1

ω C =

12π 50.0( ) (4.43× 10−6)= 719 Ω

We then compute the impedance of the circuit:

X C = 12π f C =

12π 60.0 Hz( ) (30.0× 10−6 F)= 88.4 Ω

(b) The impedance of the circuit is

(e)

Adding an inductor will change the impedance, and hence the current in the circuit The current could be larger or smaller, depending on the inductance added

The largest current would result when the inductive reactance equals the capacitive reactance, the impedance has its minimum value, equal to 60.0 Ω, and the

current in the circuit is

Imax= ΔVmax

Z = ΔVmax

R =1.20× 102 V

60.0 Ω = 2.00 A

P33.33 Let X C represent the initial capacitive reactance Moving the plates to

half their original separation doubles the capacitance (

P33.34 The power factor for a series RLC circuit is given by

In order for the power factor to be equal to 1.00, we would have to

have X L = 0, which would require either L or f to be zero Because this

is not the case, the situation is impossible

P33.35 From the definition of impedance,

P33.37 The rms current in the circuit is

P33.38 Given v = ΔVmaxsin( )ωt = 90.0 V( )sin 350t( ), observe that

ΔVmax = 90.0 V and ω = 350 rad/s Also, the net reactance is

only occur if X L = X C, or

9.42 Ω = 1

2π 60.0 s( −1)C → C = 281 µF (d) For the power to equal that before the capacitor was installed, or

P b = P d, we require

(ΔVrms)b( )Irms bcosφb =(ΔVrms)d2

R

Solving for the rms voltage gives

P33.41 One-half of the time, the left side of the

generator is positive, the top diode conducts, and the bottom diode switches off The power supply sees resistance

P33.42 We are given L = 0.020 0 H, C = 100 × 10–9 F, R = 20.0 Ω, and

ΔVmax = 100 V

(a) The resonant frequency for a series RLC circuit is

f = ω02π =

12π

1

LC = 3.56 kHz

ANS FIG P33.41

(d) At resonance, the amplitude of the voltage across the inductor is

ΔV L, max = X L Imax =ω0LImax = 2.24 kV

P33.43 The circuit is to be in resonance when

L C

The impedance of the circuit is

L C

The impedance of the circuit is

The power delivered to the circuit is

X L = 2π f L = 2π 1.00 × 10( 10 Hz) (400× 10−12 H)= 25.1 Ω

Section 33.8 The Transformer and Power Transmission

P33.48 (a) The output voltage is found from Equation 33.41,

Δv2= N2

N1Δv1 Therefore,

Irms = P

ΔVrms =

5.00× 106 W5.00× 105 V = 10.0 AThe power loss during transmission is

Ploss = Irms2 R= 10.0 A( )2(290 Ω)= 29.0 kW(b) The fraction of input power lost is

Ploss

P = 2.90× 104W5.00× 106W = 5.80 × 10−3

(c) It is impossible to transmit so much power at such low voltage Maximum power transfer occurs when load resistance equals the line resistance of 290 Ω, and is (4.50× 103 V)2

2⋅2 290 Ω( ) = 17.5 kW, far

below the required 5 000 kW