amateur extra licensing class lake area radio klub spring 2012 circuits & resonance for all!

Amateur Extra Licensing Class

Lake Area Radio KlubSpring 2012

Circuits & Resonance for All!

Amateur Radio Extra ClassElement 4 Course Presentation

ELEMENT 4 Groupings

• Rules & Regs• Skywaves & Contesting• Outer Space Comms• Visuals & Video Modes• Digital Excitement with Computers & Radios• Modulate Your Transmitters• Amps & Power Supplies• Receivers with Great Filters

Amateur Radio Extra ClassElement 4 Course Presentation

ELEMENT 4 Groupings

• Oscillate & Synthesize This!• Circuits & Resonance for All!• Components in Your New Rig• Logically Speaking of Counters• Optos & OpAmps Plus Solar• Test Gear, Testing, Testing 1,2,3• Antennas• Feedlines & Safety

Amateur Radio Extra Class


• E5A01… Resonance can cause the voltage across reactances in series to be larger than the voltage applied to them.

Let’s go through the math step by step………

Amateur Radio Extra ClassCircuits & Resonance for All!

FLxL 2FC

XC 21

Cf 2L2

12

Resonance occurs in a circuit when XL is equal to XC.

Therefore…..What we do to the left side of the equation, we must do to the right side, and what we do to the numerator we must do to the denominator, to maintain equality

LCf

212

2

FCFL

π2

12

LCf

π2

1o

This is the resonant frequency formula.

LCf

212

2




Circuits & Resonance for All!• E5A02… Resonance in an electrical circuit is the frequency at

which the capacitive reactance equals the inductive reactance.

• E5A03… The magnitude of the impedance of a series R-L-C circuit at resonance is approximately equal to circuit resistance.

• E5A04… The magnitude of the impedance of a circuit with a resistor, an inductor and a capacitor all in parallel, at resonance is approximately equal to circuit resistance.

At resonance, a series resonant circuit L and C present a low impedance so the circuit resistance is set by the resistor.

At resonance, a parallel resonant circuit presents a very high impedance across the resistor.


Circuits & Resonance for All!• E5A05…The magnitude of the current at the input of a series R-L-C

circuit as the frequency goes through resonance is Maximum.

At resonance a series circuit presents a low impedance and current would be limited by the resistor Tuning to either side of resonance would cause additional reactive resistance and therefore lower current flow in the circuit.

Series and Parallel Resonant Circuits.


Circuits & Resonance for All!• E5A06… The magnitude of the circulating current within the

components of a parallel L-C circuit at resonance is at a maximum.

Variation of Inductive and capacitive reactance with frequency(graph not to exact log-log scale)


Circuits & Resonance for All!• E5A07… The magnitude of the current at the input of a parallel R-

L-C circuit at resonance is at a Minimum.

• E5A08… The voltage and the current through and the voltage across a series resonant circuit are in phase.

• E5A09… The current through and the voltage across a parallel resonant circuit are in phase. (also true for a series circuit at

resonance)



• E5B07… The phase angle between the voltage across and the current through a series R-L-C circuit if XC is 500 ohms, R is 1 kilohm, and XL is 250 ohms is 14.0 degrees with the voltage lagging the current.

Series RLC Circuits for Phase angle Calculations

Tangent of θ = Y / X Tangent of θ = 250/1000 Tangent of θ = .25 θ = 14.04°


Circuits & Resonance for All!• E5B08… The phase angle between the voltage across and the

current through a series R-L-C circuit if XC is 100 ohms, R is 100 ohms, and XL is 75 ohms is 14 degrees with the voltage lagging the current.

Rules for calculating impedances and phase angles1) Impedances in series add together2) Admittance is the reciprocal of impedance3) Admittances in parallel add together4) Inductive and capacitive reactance in series cancel5) 1/j=-j

Tangent of θ = Y / X Tangent of θ = (75-100)/100

Tangent of θ = -.25

θ = -14.04°

Complex number axis diagram.



Vector Addition

• Here is more detail. In an ac circuit, when calculating the impedance of the circuit, the reactance and resistance must be added vectorially rather than algebraically. This vector addition can be understood best by looking at the following diagram:


Circuits & Resonance for All!• E5B09… The relationship between the current through and the

voltage across a capacitor is that the current leads the voltage by 90 degrees.

• E5B10… The relationship between the current through an inductor and the voltage across an inductor is that the voltage leads current by 90 degrees.

• E5B11… The phase angle between the voltage across and the current through a series RLC circuit if XC is 25 ohms, R is 100 ohms, and XL is 50 ohms is 14 degrees with the voltage leading the current.

Tangent of θ = Y / X Tangent of θ = (50-25)/100

Tangent of θ = .25

θ = 14.04°

j 50

- j 25

100 Ω

Remember: ELI the ICE man….



current through a series RLC circuit if XC is 75 ohms, R is 100 ohms, and XL is 50 ohms is 14 degrees with the voltage lagging the current.Tangent of θ = Y / X Tangent of θ =

(50-75)/100Tangent of θ = -.25

θ = -14.04°

j50

- j 75

100 Ω



current through a series RLC circuit if XC is 250 ohms, R is 1 kilohm, and XL is 500 ohms is 14.04 degrees with the voltage leading the current.Tangent of θ = Y / X Tangent of θ =

(500-250)/1000Tangent of θ = 0.25

θ = 14.04°

j 500

- j 250

1000 Ω


Circuits & Resonance for All!• E5C13… The Rectangular coordinate system is often used to

display the resistive, inductive, and/or capacitive reactance components of impedance.

• E5C09… When using rectangular coordinates to graph the impedance of a circuit, the horizontal axis represents the voltage or current associated with the resistive component.



Rectangular Coordinates



Polar Coordinates


Circuits & Resonance for All!• E5C22… In rectangular coordinates, what is the impedance of a

network comprised of a 10-microhenry inductor in series with a 40-ohm resistor at 500 MHz?

R = 40Ω

Remember Inductive Reactance is positive so the answer is:40 + j 31,400

• E5C17… In rectangular coordinates, the impedance of a circuit that has an admittance of 5 millisiemens at -30 degrees is 173 + j100 ohms.

XL= (2 π FL)

XL = (6.28 x 500 x 10+6 x 10 x 10-6) XL = 31,416 Ω

Polar Impedance (Z) =1/admittance

Z= 1/.005 Z= 200ΩPolar angle = 1/admittance

angle= 1/- 30°

= +30°Cos θ= resistance (R) / Impedance

(Z)R = 200Ω x Cosine 30°R = 200Ω x .866173.2 Ω

Sin θ= reactance (j) / Impedance (Z)

j = 200 x Sine 30°

j = 200Ω x .50 j100Ω

Don’t be tempted to use a “-j” in front of the reactance with the admittance given initially with -30o angle.


Circuits & Resonance for All!• E5C10… When using rectangular coordinates to graph the

impedance of a circuit, the vertical axis represents the voltage or current associated with the reactive component.

• E5C11… The two numbers used to define

• E5C12… If you plot the impedance of a circuit using the rectangular coordinate system and find the impedance point falls on the right side of the graph on the horizontal line, you know the circuit is equivalent to a pure resistance.

a point on a graph using rectangular coordinates represent the coordinate values along the horizontal and vertical axes.


Circuits & Resonance for All!• E5C19… In Figure E5-2, point 4

best represents that impedance of a series circuit consisting of a 400 ohm resistor and a 38 picofarad capacitor at 14 MHz.

R = 400 Ω

XC = 1/ (2 π FC)

XC= 1/(6.28 x 14 x .000038) XC = -300 Ω

Remember:Capacitive reactance is negative.

Figure E5-2


Circuits & Resonance for All!• E5C20… In Figure E5-2, Point 3

best represents the impedance of a series circuit consisting of a 300 ohm resistor and an 18 microhenry inductor at 3.505 MHz.

R =300 ΩXL = (2 π FL)

XL = (6.28 x 3.505 x 18)

XL = 396.4 Ω

Remember:Inductive reactance is positive

Answer is 300 Ω + j 395 Ω Figure E5-2


Circuits & Resonance for All!• E5C21… In Figure E5-2, Point 1

best represents the impedance of a series circuit consisting of a 300 ohm resistor and a 19 picofarad capacitor at 21.200 MHz.

R = 300 ΩXC = 1/ (2 π FC)

XC = 1/(6.28 x 21.2 x .000019)

XC = -395.1 Ω

Remember:Capacitive reactance is negative

Answer is 300 Ω –j 395 Figure E5-2


Circuits & Resonance for All!• E5C23… On Figure E5-2, Point 8

best represents the impedance of a series circuit consisting of a 300-ohm resistor, a 0.64-microhenry inductor and an 85-picofarad capacitor at 24.900 MHz.

R = 300 ΩXC = 1/ (2 π FC) XC = 1/(6.28 x 24.9 x .000085) XC = -75.19 Ω (XC is negative)XL = (2 π FL) XL = (6.28 x 24.9 x .64) XL = 100.12 Ω (XL is positive)Net reactance is the sum of XC and XL

-75.19 + 100.12 = +24.9 Answer is 300 Ω + j 24.9

Figure E5-2


Circuits & Resonance for All!• E5C14… The Polar coordinate system is often used to display the

phase angle of a circuit containing resistance, inductive and/or capacitive reactance.

• E5C04… In polar coordinates, the impedance of a network consisting of a 400-ohm-reactance capacitor in series with a 300-ohm resistor is 500 ohms at an angle of -53.1 degrees.

300Ω

-j400

Z= √(X² + (XL –XC)²)

Z= √( 300² + (0-400)²) Z= √(250,000)Z= 500 Ω θ = arc tan (reactance/resistance)arc tan (-400/300) arc tan (- 1.33) -53.13°



Complex Numbers (Real and Imaginary and Operator j



j Operator as Vector Rotator


Circuits & Resonance for All!• E5C01… In polar coordinates, the impedance of a network consisting

of a 100-ohm-reactance inductor in series with a 100-ohm resistor is 141 ohms at an angle of 45°.

Z= √(X² + Y²)

Z= √(100² + 100²)

Z= √( 20,000)

Z= 141.42 Ω

θ = arc tan (reactance/resistance)

arc tan 100/100

arc tan 1 or 45°

j 100

100 Ω


Circuits & Resonance for All!• E5C05… In polar coordinates, the impedance of a network

consisting of a 400-ohm-reactance inductor in parallel with a 300-ohm resistor is 240 ohms at an angle of 36.9 degrees.

Impedance = = = 120,000 / 500 = 240 Ω

θ = arctan 1/ (Reactance/Resistance)

θ= arctan 1/ (400 / 300)

θ= arctan 1/ 1.333

arctan =.750

θ =36.87°



of a 100-ohm-reactance inductor, a 100-ohm-reactance capacitor, and a 100-ohm resistor, all connected in series is 100 ohms at an angle of 0 degrees. Z= √( R² + (XL –XC)²)

Z= √( 100² + (100-100)²)Z= √( 10,000)Z= 100

Ωθ = arc tan (reactance/resistance)

arc tan 0/100arc tan 0

0°

100 Ω

- j 100

j 100

Note- the Y side is the vector sum of the inductive reactance and capacitive reactance or (XL –Xc) j 100-j 100100 Ω



of a 300-ohm-reactance capacitor, a 600-ohm-reactance inductor, and a 400-ohm resistor, all connected in series is 500 ohms at an angle of 37 degrees.

Z= √(R² + (XL –Xc)²)

Z= √( 400² + (600-300)²)

Z= √( 250,000)

Z= 500 Ω

θ = arc tan (reactance/resistance)

300/400

arc tan .75

36.9°

400Ω

j 600

- j300


Circuits & Resonance for All!• E5C06… In polar coordinates, the impedance of a network

consisting of a 100-ohm-reactance capacitor in series with a 100-ohm resistor is 141 ohms at an angle of -45 degrees.

• E5C07… In polar coordinates, the impedance of a network comprised of a 100-ohm-reactance capacitor in parallel with a 100-ohm resistor is 71 ohms at an angle of -45 degrees.

Z= √(X² + (XL –Xc)²)

Z= √( 100² + (-100)²) Z= √(20,000) Z= 141.4 Ω

Angle is arctan 1/ (reactance/resistance) arctan 1/ (100/100) arc tan (- 1) -45°

Admittance = 1/100 +( -j/100)0.01 + j 0.01 Angle = arc tan .01/.01 45° (-45° in polar coordinates)

Impedance = 1/ (√ ( (.01)² x (.01)² ))

1/(.0141)

70.71 Ώ


Circuits & Resonance for All!• E5C07… In polar coordinates, the impedance of a network comprised

of a 100-ohm-reactance capacitor in parallel with a 100-ohm resistor is 71 ohms at an angle of -45 degrees.

• E5C08… In polar coordinates, the impedance of a network comprised of a 300-ohm-reactance inductor in series with a 400-ohm resistor is 500 ohms at an angle of 37 degrees.

Admittance = 1/100 +(-j/100)0.01 +j 0.01 Angle =arc tan .01/.01 45° (-45° in polar coordinates)

Impedance = 1/ (√ ( (.01)² x (.01)² ))1/(.0141)

70.71 Ώ

Z= √(X² + (XL –XC)²)

Z= √( 400² + (0-300)²)Z= √(250,000) Z= 500 Ω

Angle is arc tan (X/R)arc tan (300/400)arc tan (.75) 36.86° 400Ω

j300


Circuits & Resonance for All!• E5C15… In polar coordinates, the impedance of a circuit of 100 -j100

ohms impedance is 141 ohms at an angle of -45 degrees.

• E5C16… In polar coordinates, the impedance of a circuit that has an admittance of 7.09 milli-siemens at 45 degrees is 141 ohms at an angle of -45 degrees.

Z= √(X² + (XL –XC)²)

Z= √( 100² + ( -100)²)

Z= √(20,000) Z= 141.42 Ω

Angle is arc tan (X/R)

arc tan (-100/100)

arc tan (-1) -45°

Polar Impedance (Z)

=1/admittance

Z= 1/.00709

Z= 141.04Ω

Polar angle = 1 / j (admittance angle)

1/j(45°)

-j45°


Circuits & Resonance for All!• E5C18… In polar coordinates, the impedance of a series circuit

consisting of a resistance of 4 ohms, an inductive reactance of 4 ohms, and a capacitive reactance of 1 ohm is 5 ohms at an angle of 37 degrees.

Z= √(X² + (XL –Xc)²)

Z= √( 4² + (4-1)²)

Z= √(25) or Z= 5 Ω

Angle is arc tan (X/R)

arc tan (3/4)

arc tan (.75)

36.86°

- j 1

+ j 4

4Ω



• E4B17… The bandwidth of the circuit's frequency response can be used as a relative measurement of the Q for a series-tuned circuit.

The Narrower the bandwidth the higher the Q of the circuit.

A large loading coil on a mobile whip helps antennas

achieve high Q resonance.


Circuits & Resonance for All!• E5A10… The half-power bandwidth of a parallel resonant circuit

that has a resonant frequency of 1.8 MHz and a Q of 95 is 18.9 kHz.B/W = Frequency/Q 1,800 KHz/95 18.94 KHz

For tuned circuits the quality factor, Q, is:

For tuned circuits with Q greater than 10:


Circuits & Resonance for All!• E5A11… The half-power bandwidth of a parallel resonant circuit that

has a resonant frequency of 7.1 MHz and a Q of 150 is 47.3 kHz.B/W = Frequency/Q 7,100

KHz/15047.3 KHz

An amplifier’s voltage gain will vary with frequency. At the cutoff frequencies, the voltage gain drops to 0.070 of what is in the mid-band. These frequencies f1 and f2 are called the half-power frequencies.

If the output voltage is 10 volts across a 100-ohm load when the gain is A at the mid-band, then the power output, PO at mid-band is:



• E5A12… The half-power bandwidth of a parallel resonant circuit that has a resonant frequency of 3.7 MHz and a Q of 118 is 31.4 kHz.

The power output at the cutoff frequency points is one-half the mid-band power. The half-power bandwidth is between the frequencies f1 and f2.

At the cutoff frequency, the output voltage will be 0.707 of what is at the mid-band; therefore, 7.07 volts. The power output is:

The power output at the 0.707

frequencies is 3 dB down from the mid-

band power.

Power Ratio in dB

B/W = Frequency/Q 3,700 KHz/118

31.36 KHz


Circuits & Resonance for All!•

E5A13… The half-power bandwidth of a parallel resonant circuit that has a resonant frequency of 14.25 MHz and a Q of 187 is 76.2 kHz.

• E5A14… The resonant frequency of a series RLC circuit with R of 22 ohms, L of 50 microhenrys and C of 40 picofarads is 3.56 MHz.

The equation can be solved with: L in Henries or Micro Henries and C in Farads or Micro Farads

B/W = Frequency/Q 14,250 KHz/187

76.20 KHz

2 π L x C

1

6.28 x 50x10^-6 x 40 x10^-12

1

=FR = = 3.56 MHz

40 pf

50 mh

22 ohms


Circuits & Resonance for All!• E5A15… The resonant frequency of a series RLC circuit with R

of 56 ohms, L of 40 microhenrys and C of 200 picofarads is 1.78 MHz.

• E5A16… The resonant frequency of a parallel RLC circuit with R of 33 ohms, L of 50 microhenrys and C of 10 picofarads is 7.12 MHz.

• E5A17… The resonant frequency of a parallel RLC circuit with R of 47 ohms, L of 25 microhenrys and C of 10 picofarads is 10.1 MHz.

2 π L x C

1 1

=FR = = 1.78 MHz6.28 x 40x10^-6 x 200x10^-12

2 π L x C

1 1

=FR = = 7.121 MHz6.28 x 50x10^-6 x 10x10^-12

2 π L x C

1 1

=FR = = 10.1 MHz6.28 x 25x10^-6 x 10x10^-12



• E5B01… One time constant is the term for the time required for the capacitor in an RC circuit to be charged to 63.2% of the supply voltage.

Schematics Curves


Circuits & Resonance for All!• E5B02… One time constant is the time it takes for a charged

capacitor in an RC circuit to discharge to 36.8% of its initial value of stored charge.

Conversely a time constant is the time it takes a discharged capacitor to reach 63.2% of the applied voltage.

Time Constants Charge % of applied voltage Discharge % of starting voltage

1 63.20% 36.80%

2 86.50% 13.50%

3 95.00% 5.00%

4 98.20% 1.80%

5 99.30% 0.70%


Circuits & Resonance for All!• E5B03… The capacitor in an RC circuit is discharged to 13.5%

percentage of the starting voltage after two time constants.

%= (100-((100 x .632)) – (100 – (100 x.632) x .632))100+(- 63.2 – 23.25) 13.54%

• E5B04… The time constant of a circuit having two 220-microfarad capacitors and two 1-megohm resistors all in parallel is 220 seconds.

TC (seconds) = R (megohms) x C (microfarads)

TC =(1/2) x (220 x 2)

0.5 x 440

220 seconds

Remember that capacitors in parallel add and resistors of equal value in parallel are equal to one resistor divided by the number of resistors.


Circuits & Resonance for All!•

E5B05… It will take .020 seconds (or 20 milliseconds) for an initial charge of 20 V DC to decrease to 7.36 V DC in a 0.01-microfarad capacitor when a 2-megohm resistor is connected across it.

To discharge to 7.36 VDC would take one time constant20V – (.632 x 20V)

7.36 Volts

• E5B06… It takes 450 seconds for an initial charge of 800 V DC to decrease to 294 V DC in a 450-microfarad capacitor when a 1-megohm resistor is connected across it.

To discharge to 294 VDC would take one time constant 800V – (.632 x 800V) = 294.4V

0.02 seconds

20 millisecondsTC = 2 x .01

TC = 1 x 450 450 seconds


Circuits & Resonance for All!• E5D01… As frequency increases, RF current flows in a thinner layer of

the conductor, closer to the surface this is called skin effect.

• E5D02… The resistance of a conductor is different for RF currents than for direct currents because of skin effect.

• E5D03… A capacitor is a device that is used to store electrical energy in an electrostatic field.

• E5D04… The Joule is the unit of electrical energy stored in an electrostatic field. A Joule is defined as a quantity of energy equal to one Newton of force acting over 1

meter

Current flow in cross section of a conductor

UHFRFDC


Circuits & Resonance for All!• E5D05… The region surrounding a magnet through which a magnetic

force acts is a magnetic field.• E5D06… The direction of the magnetic field oriented about a

conductor in relation to the direction of electron flow is in a direction determined by the left-hand rule.

Left-Hand Rule

Direction of Magnetic Field

Magnetic Field surrounding wire

Wire or Conductor with current through it


Circuits & Resonance for All!• E5D07… The amount of current determines the strength of a

magnetic field around a conductor. • E5D08… Potential energy is the term for energy that is stored in an

electromagnetic or electrostatic field.• E5D09… Reactive power is the term for an out-of-phase,

nonproductive power associated with inductors and capacitors.• E5D10… In a circuit that has both inductors and capacitors the

reactive power is repeatedly exchanged between the associated magnetic and electric fields, but is not dissipated.

• E5D11… The true power can be determined in an AC circuit where the voltage and current are out of phase by multiplying the apparent power times the power factor.

Apparent power is the voltage times the current into the circuitTrue Power is the apparent power times the power factor The only time true power and apparent power are the same is if the power

factor is 1.00 … (the phase angle is zero)

(assuming perfect lossless components)



Apparent and True Power


Circuits & Resonance for All!• E5D12… The power factor (PF) of an R-L circuit having a 60 degree

phase angle between the voltage and the current is 0.5.PF is the cosine function of the voltage to current angle

►PF =cosine of 60°

PF= 0.5

• E5D13… 80 watts are consumed in a circuit having a power factor of 0.2 if the input is 100-V AC at 4 amperes.

Power Consumed = V x I x PF

100 x 4 x .2

80 watts



• E5D14… The power is consumed in a circuit consisting of a 100 ohm resistor in series with a 100 ohm inductive reactance drawing 1 ampere is 100 Watts.

Power (real) = I² x R Power (real) = (1)² x 100

100 watts. (Only the circuit resistance consumes power)

• E5D15… Wattless, nonproductive power is reactive power.

• E5D16… The power factor of an RL circuit having a 45 degree phase angle between the voltage and the current is 0.707.

PF = Cosine of 45°

PF = 0.707


Circuits & Resonance for All!• E5D17… The power factor of an RL circuit having a 30 degree

phase angle between the voltage and the current is 0.866.PF Cosine of 30°

PF = 0.866 • E5D18… 600 watts are consumed in a circuit having a power

factor of 0.6 if the input is 200V AC at 5 amperes.Power Consumed = V x I x PF200 x 5 x .6

600 watts

• E5D19… The power consumed in a circuit having a power factor of 0.71 if the apparent power is 500 watts is 355 W.

Power Consumed = Apparent power x PF

500 x .71

355 watts


Circuits & Resonance for All!• E4E04… Conducted and radiated noise caused by an automobile

alternator can be suppressed by connecting the radio's power leads directly to the battery and by installing Feed Through capacitors in line with the alternator leads.

• E4E05… Noise from an electric motor can be suppressed by installing a brute-force AC-line filter in series with the motor leads.

• E6D08… Core permeability (for a given size core) is the property that determines the inductance of a toroidal inductor with a 10-turn winding.


Circuits & Resonance for All!• E6D09… The usable frequency range of inductors that use toroidal

cores, assuming a correct selection of core material for the frequency being used is from less than 20 Hz to approximately 300 MHz.

• E6D10… One important reason for using powdered-iron toroids rather than ferrite toroids in an inductor is that powdered-iron toroids generally have better temperature stability.

Applications for powdered Iron toroids would be oscillator and filter circuits where inductance stability with temperature is important.

• E6D12… A primary advantage of using a toroidal core instead of a solenoidal core in an inductor is that toroidal cores contain most of the magnetic field within the core material.


Circuits & Resonance for All!• E6D13… Forty three turns of wire will be required to produce a 1-mH

inductor using a ferrite toroidal core that has an inductance index (AL) value of 523 millihenrys/1000 turns.

N turns = 1000 x (√ (L / AL))

N turns = 1000 x (√ (1 / 523))

43.7 turns • E6D14… Thirty five turns of wire will be required to produce a 5-

microhenry inductor using a powdered-iron toroidal core that has an inductance index (A L) value of 40 microhenrys/100 turns.

N turns = 100 x (√ (L / AL))N turns = 100 x (√ (5 / 40))

35.35 turns• E6D18… One reason for using ferrite toroids rather than powdered-

iron toroids in an inductor is that Ferrite toroids generally require fewer turns to produce a given inductance value.

Element 4 Extra Class Question Pool

Valid July 1, 2008

Through

June 30, 2012


E5A01 What can cause the voltage across reactances in series to be larger than the voltage applied to them?

A. ResonanceB. CapacitanceC. ConductanceD. Resistance

E5A02 What is resonance in an electrical circuit?

A. The highest frequency that will pass current

B. The lowest frequency that will pass current

C. The frequency at which the capacitive reactance equals the inductive reactance

D. The frequency at which the reactive impedance equals the resistive impedance

E5A03 What is the magnitude of the impedance of a series R-L-C circuit at resonance?

A. High, as compared to the circuit resistance

B. Approximately equal to capacitive reactance

C. Approximately equal to inductive reactance

D. Approximately equal to circuit resistance

E5A04 What is the magnitude of the impedance of a circuit with a resistor, an inductor and a capacitor all in parallel, at resonance?

A. Approximately equal to circuit resistance

B. Approximately equal to inductive reactance

C. Low, as compared to the circuit resistance

D. Approximately equal to capacitive reactance

E5A05 What is the magnitude of the current at the input of a series R-L-C circuit as the frequency goes through resonance?

A. MinimumB. MaximumC. R/LD.L/R

E5A06 What is the magnitude of the circulating current within the components of a parallel L-C circuit at resonance?

A. It is at a minimumB. It is at a maximumC. It equals 1 divided by the quantity [ 2

multiplied by Pi, multiplied by the square root of ( inductance "L" multiplied by capacitance "C" ) ]

D. It equals 2 multiplied by Pi, multiplied by frequency "F", multiplied by inductance "L"

E5A07 What is the magnitude of the current at the input of a parallel R-L-C circuit at resonance?

A.MinimumB.MaximumC.R/LD.L/R

E5A08 What is the phase relationship between the current through and the voltage across a series resonant circuit?

A. The voltage leads the current by 90 degrees

B. The current leads the voltage by 90 degrees

C. The voltage and current are in phase

D. The voltage and current are 180 degrees out of phase

E5A09 What is the phase relationship between the current through and the voltage across a parallel resonant circuit?

A. The voltage leads the current by 90 degrees

B. The current leads the voltage by 90 degrees

C. The voltage and current are in phase

D. The voltage and current are 180 degrees out of phase

E5B07 What is the phase angle between the voltage across and the current through a series R-L-C circuit if XC is 500 ohms, R is 1 kilohm, and XL is 250 ohms?

A. 68.2 degrees with the voltage leading the current

B. 14.0 degrees with the voltage leading the current

C. 14.0 degrees with the voltage lagging the current

D. 68.2 degrees with the voltage lagging the current

E5B08 What is the phase angle between the voltage across and the current through a series R-L-C circuit if XC is 100 ohms, R is 100 ohms, and XL is 75 ohms?

A. 14 degrees with the voltage lagging the current

B. 14 degrees with the voltage leading the current

C. 76 degrees with the voltage leading the current

D. 76 degrees with the voltage lagging the current

E5B09 What is the relationship between the current through and the voltage across a capacitor?

A.Voltage and current are in phaseB.Voltage and current are 180

degrees out of phaseC.Voltage leads current by 90

degreesD.Current leads voltage by 90

degrees

E5B10 What is the relationship between the current through an inductor and the voltage across an inductor?

A. Voltage leads current by 90 degrees

B. Current leads voltage by 90 degrees

C. Voltage and current are 180 degrees out of phase

D. Voltage and current are in phase

E5B11 What is the phase angle between the voltage across and the current through a series RLC circuit if XC is 25 ohms, R is 100 ohms, and XL is 50 ohms?

A. 14 degrees with the voltage lagging the current

B. 14 degrees with the voltage leading the current

C. 76 degrees with the voltage lagging the current

D. 76 degrees with the voltage leading the current

E5B12 What is the phase angle between the voltage across and the current through a series RLC circuit if XC is 75 ohms, R is 100 ohms, and XL is 50 ohms?

A.76 degrees with the voltage lagging the current

B.14 degrees with the voltage leading the current

C.14 degrees with the voltage lagging the current

D.76 degrees with the voltage leading the current

E5B13 What is the phase angle between the voltage across and the current through a series RLC circuit if XC is 250 ohms, R is 1 kilohm, and XL is 500 ohms?

A. 81.47 degrees with the voltage lagging the current

B. 81.47 degrees with the voltage leading the current

C. 14.04 degrees with the voltage lagging the current

D. 14.04 degrees with the voltage leading the current

E5C13 What coordinate system is often used to display the resistive, inductive, and/or capacitive reactance components of an impedance?

A. Maidenhead gridB. Faraday gridC. Elliptical coordinates D. Rectangular coordinates

E5C09 When using rectangular coordinates to graph the impedance of a circuit, what does the horizontal axis represent?

A. The voltage or current associated with the resistive component

B. The voltage or current associated with the reactive component

C. The sum of the reactive and resistive components

D. The difference between the resistive and reactive components

E5C22 In rectangular coordinates, what is the impedance of a network comprised of a 10-microhenry inductor in series with a 40-ohm resistor at 500 MHz?

A. 40 + j31,400B. 40 - j31,400C. 31,400 + j40D.31,400 - j40

E5C17 In rectangular coordinates, what is the impedance of a circuit that has an admittance of 5 millisiemens at -30 degrees?

A. 173 - j100 ohmsB. 200 + j100 ohmsC. 173 + j100 ohmsD.200 - j100 ohms

E5C10 When using rectangular coordinates to graph the impedance of a circuit, what does the vertical axis represent?

A. The voltage or current associated with the resistive component

B. The voltage or current associated with the reactive component

C. The sum of the reactive and resistive components

D. The difference between the resistive and reactive components

E5C11 What do the two numbers represent that are used to define a point on a graph using rectangular coordinates?

A. The magnitude and phase of the point

B. The sine and cosine valuesC. The coordinate values along

the horizontal and vertical axes

D. The tangent and cotangent values

E5C12 If you plot the impedance of a circuit using the rectangular coordinate system and find the impedance point falls on the right side of the graph on the horizontal line, what do you know about the circuit?

A. It has to be a direct current circuitB. It contains resistance and

capacitive reactanceC. It contains resistance and

inductive reactanceD. It is equivalent to a pure

resistance

E5C19 Which point on Figure E5-2 best represents that impedance of a series circuit consisting of a 400 ohm resistor and a 38 picofarad capacitor at 14 MHz?

A. Point 2B. Point 4C. Point 5D.Point 6

Figure E5-2

Point 3

Point 2

Point 6Point 8

Point 5

Point 7 Point 1

Point 4

E5C20 Which point in Figure E5-2 best represents the impedance of a series circuit consisting of a 300 ohm resistor and an 18 microhenry inductor at 3.505 MHz?


Figure E5-2

Point 3

Point 2

Point 6Point 8

Point 5

Point 7 Point 1

Point 4

E5C21 Which point on Figure E5-2 best represents the impedance of a series circuit consisting of a 300 ohm resistor and a 19 picofarad capacitor at 21.200 MHz?

A. Point 1B. Point 3C. Point 7D. Point 8

Figure E5-2

Point 3

Point 2

Point 6Point 8

Point 5

Point 7 Point 1

Point 4

E5C23 Which point on Figure E5-2 best represents the impedance of a series circuit consisting of a 300-ohm resistor, a 0.64-micro-henry inductor and an 85-picofarad capacitor at 24.900 MHz?


Figure E5-2

Point 3

Point 2

Point 6Point 8

Point 5

Point 7 Point 1

Point 4

E5C14 What coordinate system is often used to display the phase angle of a circuit containing resistance, inductive and/or capacitive reactance?

A. Maidenhead gridB. Faraday gridC. Elliptical coordinatesD.Polar coordinates

E5C04 In polar coordinates, what is the impedance of a network consisting of a 400-ohm-reactance capacitor in series with a 300-ohm resistor?

A. 240 ohms at an angle of 36.9 degrees

B. 240 ohms at an angle of -36.9 degrees

C. 500 ohms at an angle of 53.1 degrees

D. 500 ohms at an angle of -53.1 degrees

E5C01 In polar coordinates, what is the impedance of a network consisting of a 100-ohm-reactance inductor in series with a 100-ohm resistor?

A. 121 ohms at an angle of 35 degrees

B. 141 ohms at an angle of 45 degrees

C. 161 ohms at an angle of 55 degrees

D. 181 ohms at an angle of 65 degrees

E5C05 In polar coordinates, what is the impedance of a network consisting of a 400-ohm-reactance inductor in parallel with a 300-ohm resistor?

A. 240 ohms at an angle of 36.9 degreesB. 240 ohms at an angle of -36.9

degreesC. 500 ohms at an angle of 53.1 degreesD. 500 ohms at an angle of -53.1

degrees

E5C02 In polar coordinates, what is the impedance of a net-work consisting of a 100-ohm-reactance inductor, a 100-ohm-reactance capacitor, and a 100-ohm resistor, all connected in series?

A.100 ohms at an angle of 90 degrees

B.10 ohms at an angle of 0 degreesC.10 ohms at an angle of 90

degreesD.100 ohms at an angle of 0

degrees

E5C03 In polar coordinates, what is the impedance of a net-work consisting of a 300-ohm-reactance capacitor, a 600-ohm-reactance inductor, and a 400-ohm resistor, all connected in series?

A. 500 ohms at an angle of 37 degreesB. 900 ohms at an angle of 53 degreesC. 400 ohms at an angle of 0 degreesD. 1300 ohms at an angle of 180

degrees

E5C06 In polar coordinates, what is the impedance of a network consisting of a 100-ohm-reactance capacitor in series with a 100-ohm resistor?

A.121 ohms at an angle of -25 degrees

B.191 ohms at an angle of -85 degrees

C.161 ohms at an angle of -65 degrees

D.141 ohms at an angle of -45 degrees

E5C07 In polar coordinates, what is the impedance of a network comprised of a 100-ohm-reactance capacitor in parallel with a 100-ohm resistor?

A. 31 ohms at an angle of -15 degrees

B. 51 ohms at an angle of -25 degrees

C. 71 ohms at an angle of -45 degrees

D. 91 ohms at an angle of -65 degrees

E5C08 In polar coordinates, what is the impedance of a network comprised of a 300-ohm-reactance inductor in series with a 400-ohm resistor?

A. 400 ohms at an angle of 27 degrees




E5C15 In polar coordinates, what is the impedance of a circuit of 100 -j100 ohms impedance?

A. 141 ohms at an angle of -45 degrees


C. 100 ohms at an angle of -45 degrees


E5C16 In polar coordinates, what is the impedance of a circuit that has an admittance of 7.09 millisiemens at 45 degrees?

A. 5.03 x 10 –E05 ohms at an angle of 45 degrees

B. 141 ohms at an angle of -45 degrees

C. 19,900 ohms at an angle of -45 degrees


E5C18 In polar coordinates, what is the impedance of a series circuit consisting of a resistance of 4 ohms, an inductive reactance of 4 ohms, and a capacitive reactance of 1 ohm?

A. 6.4 ohms at an angle of 53 degrees



D. 10 ohms at an angle of -51 degrees

E4B17 Which of the following can be used as a relative measurement of the Q for a series-tuned circuit?

A. The inductance to capacitance ratio

B. The frequency shift

C. The bandwidth of the circuit's frequency response

D. The resonant frequency of the circuit

E5A10 What is the half-power bandwidth of a parallel resonant circuit that has a resonant frequency of 1.8 MHz and a Q of 95?

A. 18.9 kHzB. 1.89 kHzC. 94.5 kHzD.9.45 kHz


A.157.8 HzB.315.6 HzC.47.3 kHzD.23.67 kHz


A. 436.6 kHzB. 218.3 kHzC. 31.4 kHzD. 15.7 kHz


A. 38.1 kHzB. 76.2 kHzC. 1.332 kHzD.2.665 kHz

E5A14 What is the resonant frequency of a series RLC circuit if R is 22 ohms, L is 50 microhenrys and C is 40 picofarads?

A. 44.72 MHzB. 22.36 MHzC. 3.56 MHzD.1.78 MHz

E5A15 What is the resonant frequency of a series RLC circuit if R is 56 ohms, L is 40 microhenrys and C is 200 picofarads?

A. 3.76 MHzB. 1.78 MHzC. 11.18 MHzD. 22.36 MHz

E5A16 What is the resonant frequency of a parallel RLC circuit if R is 33 ohms, L is 50 microhenrys and C is 10 picofarads?

A. 23.5 MHzB. 23.5 kHzC. 7.12 kHzD.7.12 MHz

E5A17 What is the resonant frequency of a parallel RLC circuit if R is 47 ohms, L is 25 microhenrys and C is 10 picofarads?

A. 10.1 MHzB. 63.2 MHzC. 10.1 kHzD.63.2 kHz

E5B01 What is the term for the time required for the capacitor in an RC circuit to be charged to 63.2% of the supply voltage?

A. An exponential rate of oneB. One time constantC. One exponential periodD.A time factor of one

E5B02 What is the term for the time it takes for a charged capacitor in an RC circuit to discharge to 36.8% of its initial value of stored charge?

A. One discharge periodB. An exponential discharge rate of oneC. A discharge factor of oneD. One time constant

E5B03 The capacitor in an RC circuit is discharged to what percentage of the starting voltage after two time constants?

A. 86.5%B. 63.2%C. 36.8%D.13.5%

E5B04 What is the time constant of a circuit having two 220-microfarad capacitors and two 1-megohm resistors all in parallel?

A. 55 secondsB. 110 secondsC. 440 secondsD.220 seconds

E5B05 How long does it take for an initial charge of 20 V DC to decrease to 7.36 V DC in a 0.01-microfarad capacitor when a 2-megohm resistor is connected across it?

A.0.02 secondsB.0.04 secondsC.20 secondsD.40 seconds

E5B06 How long does it take for an initial charge of 800 V DC to decrease to 294 V DC in a 450-microfarad capacitor when a 1-megohm resistor is connected across it?

A. 4.50 secondsB. 9 secondsC. 450 secondsD.900 seconds

E5D01 What is the result of skin effect?

A. As frequency increases, RF current flows in a thinner layer of the conductor, closer to the surface

B. As frequency decreases, RF current flows in a thinner layer of the conductor, closer to the surface

C. Thermal effects on the surface of the conductor increase the impedance

D. Thermal effects on the surface of the conductor decrease the impedance

E5D02 Why is the resistance of a conductor different for RF currents than for direct currents?

A. Because the insulation conducts current at high frequencies

B. Because of the Heisenburg Effect

C. Because of skin effectD.Because conductors are non-

linear devices

E5D03 What device is used to store electrical energy in an electrostatic field?

A. A batteryB. A transformerC. A capacitorD.An inductor

E5D04 What unit measures electrical energy stored in an electrostatic field?

A. CoulombB. JouleC. WattD.Volt

E5D05 What is a magnetic field?

A. Electric current through the space around a permanent magnet

B. The region surrounding a magnet through which a magnetic force acts

C. The space between the plates of a charged capacitor, through which a magnetic force acts

D. The force that drives current through a resistor

E5D06 In what direction is the magnetic field oriented about a conductor in relation to the direction of electron flow?

A. In the same direction as the current

B. In a direction opposite to the current

C. In all directions; omnidirectionalD. In a direction determined by the

left-hand rule

E5D07 What determines the strength of a magnetic field around a conductor?

A. The resistance divided by the current

B. The ratio of the current to the resistance

C. The diameter of the conductorD.The amount of current

E5D08 What is the term for energy that is stored in an electromagnetic or electrostatic field?

A.Amperes-joulesB.Potential energyC.Joules-coulombsD.Kinetic energy

E5D09 What is the term for an out-of-phase, non-productive power associated with inductors and capacitors?

A. Effective powerB. True powerC. Peak envelope powerD.Reactive power

E5D10 In a circuit that has both inductors and capacitors, what happens to reactive power?

A. It is dissipated as heat in the circuit

B. It is repeatedly exchanged between the associated magnetic and electric fields, but is not dissipated

C. It is dissipated as kinetic energy in the circuit

D. It is dissipated in the formation of inductive and capacitive fields

E5D11 How can the true power be determined in an AC circuit where the voltage and current are out of phase?

A. By multiplying the apparent power times the power factor

B. By dividing the reactive power by the power factor

C. By dividing the apparent power by the power factor

D. By multiplying the reactive power times the power factor

E5D12 What is the power factor of an R-L circuit having a 60 degree phase angle between the voltage and the current?

A. 1.414B. 0.866C. 0.5D. 1.73

E5D13 How many watts are consumed in a circuit having a power factor of 0.2 if the input is 100-V AC at 4 amperes?

A. 400 wattsB. 80 wattsC. 2000 wattsD.50 watts

E5D14 How much power is consumed in a circuit consisting of a 100 ohm resistor in series with a 100 ohm inductive reactance drawing 1 ampere?

A. 70.7 WattsB. 100 WattsC. 141.4 WattsD. 200 Watts

E5D15 What is reactive power?

A. Wattless, nonproductive power

B. Power consumed in wire resistance in an inductor

C. Power lost because of capacitor leakage

D. Power consumed in circuit Q

E5D16 What is the power factor of an RL circuit having a 45 degree phase angle between the voltage and the current?

A. 0.866B. 1.0C. 0.5D.0.707

E5D17 What is the power factor of an RL circuit having a 30 degree phase angle between the voltage and the current?

A. 1.73B. 0.5C. 0.866D. 0.577

E5D18 How many watts are consumed in a circuit having a power factor of 0.6 if the input is 200V AC at 5 amperes?

A. 200 wattsB. 1000 wattsC. 1600 wattsD. 600 watts

E5D19 How many watts are consumed in a circuit having a power factor of 0.71 if the apparent power is 500 watts?

A. 704 WB. 355 WC. 252 WD. 1.42 mW

E4E04 How can conducted and radiated noise caused by an automobile alternator be suppressed?

A. By installing filter capacitors in series with the DC power lead and by installing a blocking capacitor in the field lead

B. By connecting the radio to the battery by the longest possible path and installing a blocking capacitor in both leads

C. By installing a high-pass filter in series with the radio's power lead and a low-pass filter in parallel with the field lead

D. By connecting the radio's power leads directly to the battery and by installing coaxial capacitors in line with the alternator leads

E4E05 How can noise from an electric motor be suppressed?

A. By installing a ferrite bead on the AC line used to power the motor

B. By installing a brute-force AC-line filter in series with the motor leads

C. By installing a bypass capacitor in series with the motor leads

D. By using a ground-fault current interrupter in the circuit used to power the motor

E6D08 What material property determines the inductance of a toroidal inductor with a 10-turn winding?

A. Core load currentB. Core resistanceC. Core reactivityD. Core permeability

E6D09 What is the usable frequency range of inductors that use toroidal cores, assuming a correct selection of core material for the frequency being used?

A. From a few kHz to no more than 30 MHz

B. From less than 20 Hz to approximately 300 MHz

C. From approximately 1000 Hz to no more than 3000 kHz

D. From about 100 kHz to at least 1000 GHz

E6D10 What is one important reason for using powdered-iron toroids rather than ferrite toroids in an inductor?

A. Powdered-iron toroids generally have greater initial permeabilities

B. Powdered-iron toroids generally have better temperature stability

C. Powdered-iron toroids generally require fewer turns to produce a given inductance value

D. Powdered-iron toroids have the highest power handling capacity

E6D12 What is a primary advantage of using a toroidal core instead of a solenoidal core in an inductor?

A. Toroidal cores contain most of the magnetic field within the core material

B. Toroidal cores make it easier to couple the magnetic energy into other components

C. Toroidal cores exhibit greater hysteresis

D. Toroidal cores have lower Q characteristics

E6D13 How many turns will be required to produce a 1-mH inductor using a ferrite toroidal core that has an inductance index (A L) value of 523 millihenrys/1000 turns?

A. 2 turnsB. 4 turnsC. 43 turnsD.229 turns

E6D14 How many turns will be required to produce a 5-microhenry inductor using a powdered-iron toroidal core that has an inductance index (A L) value of 40 microhenrys/100 turns?

A. 35 turnsB. 13 turnsC. 79 turnsD.141 turns

E6D18 What is one reason for using ferrite toroids rather than powdered-iron toroids in an inductor?

A. Ferrite toroids generally have lower initial permeabilities

B. Ferrite toroids generally have better temperature stability

C. Ferrite toroids generally require fewer turns to produce a given inductance value

D. Ferrite toroids are easier to use with surface mount technology

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