E5 - Electrical Principles

Resonance in an electrical circuit is the frequency at which the capacitive reactance equals the inductive reactance. The current through and the voltage across a series resonant circuit are in phase. Through a parallel resonant circuit they are also are in phase.

At resonance, the magnitude of the impedance of a circuit with a resistor, an inductor and a capacitor all in parallel is approximately equal to circuit resistance. The magnitude of the circulating current within the components of a parallel L-C circuit at resonance is at a maximum. But at the input of a parallel R-L-C circuit it is minimum.

As the frequency goes through resonance the input current of a series R-L-C circuit is a maximum. Resonance can cause the voltage across reactances in series to be larger than the voltage applied to them.

For questions on the exam, the half-power bandwidth of a parallel circuit for the given resonant frequency and Q is in the table below,

Frequency
MHz
Frequency
kHz
Circuit QFormulaHalf-power Bandwidth
1.8     MHz  1800 kHz  95
kHz
Q
18.9 kHz
3.7     MHz  3700 kHz11831.4 kHz
7.1     MHz  7100 kHz15047.3 kHz
14.25 MHz14250 kHz18776.2 kHz

For questions on the exam, the resonant frequency of a series RLC circuit with given L and C values is in the table below (R is not relevant to resonance),

LCFormula*Resonant Frequency
50 microhenrys  40 picofarads
  1000  
 2πLC
3.56 MHz
40 microhenrys200 picofarads1.78 MHz
50 microhenrys  10 picofarads7.12 MHz
25 microhenrys  10 picofarads10.1 MHz
*The factor of 1000 adjusts for directly using microhenry and picofarad values instead of adjusting them to Henrys and Farads, respectively

The term for the time required for the capacitor in an RC circuit to be charged to 63.2% of the supply voltage is one time constant. It is also 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 one time constant.

A capacitor in an RC circuit is discharged to 13.5% of the starting voltage after two time constants. [.368×.368=.135].

There are two questions on the exam asking for computation of discharge time for an RC circuit and one asking for the time constant. All are really asking for the time constant of the circuit since the amount of discharge in the questions is exactly 36.8% -- one time constant. The table below shows the values in the questions and the computed results,

Initial
Charge
Decreased
Charge
Percent
Discharge
Capacitor
C
Resistor
R
Formula1Time (Sec.)
TC
20 V DC7.36 V DC36.8%0.01-microfarad2-megohmTC=R×C0.02 seconds
800 V DC294 V DC36.8% 450-microfarad1-megohm450 seconds
Note2 220-microfarad1-megohm220 seconds
1 All time constant questions appear to have been selected so that if the nominal R and C values are multiplied you get the right answer, without having to move the decimal points or fool with powers of 10.
2 The time constant of a circuit having two 220-microfarad capacitors and two 1-megohm resistors all in parallel is 220 seconds. In theory the two capacitors in parallel double the capacitance and the two resistors halve the resistance. Because of this, If you make the "mistake" of using the values for one capacitor and one resistor, you still get the right answer.

Current through a capacitor leads the voltage across it by 90 degrees. The voltage across an inductor leads the current through it by 90 degrees.

The system often used to display the resistive, inductive, and/or capacitive reactance components of impedance is the rectangular coordinate system. The two numbers that are used to define a point on a graph using rectangular coordinates represent the coordinate values along the horizontal and vertical axes. When using rectangular coordinates to graph the impedance of a circuit, the horizontal axis represents the voltage or current associated with the resistive component. The vertical axis represents the voltage or current associated with the reactive component.

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? It is equivalent to a pure resistance.


There are eight questions (one question appearing twice, is counted as two) in the exam pool that asks, either explicitly or implicitly from the choice of answers, the polar coordinate form of the impedance of a network composed of resistance and capacitive and/or inductive reactance in series. The following table shows how to do those calculations,

Resistance
R
Capacitive
Reactance
XC
Inductive
Reactance
XL
FormulaTotal
Reactance
X
Formula Network
Impedance
I
FormulaPhase
Angle
Θ
100-ohm100-ohm100-ohmX=XLXC      0-ohm
I = R2+X2
100 ohms
Θ = Tan-1(X)
R
    0°
400-ohm300-ohm600-ohm  300-ohm500 ohms  37°
400-ohm 300-ohm  300-ohm500 ohms  37°
     4-ohm      1-ohm     4-ohm      3-ohm    5 ohms  37°
300-ohm400-ohm −400-ohm500 ohms−53°
100-ohm 100-ohm  100-ohm141 ohms  45°
100-ohm*100-ohm* −100-ohm141 ohms−45°
*This problem occurs twice in the question pool, once explicitly specifying 100-ohm value for capacitive reactance, and once simply presenting the impedance as a complex number of the form 100 + j100.

There are two questions in the exam pool that ask the impedance of a network composed of resistance and capacitive or inductive reactance in parallel. The following table shows how to do those calculations,

Resistance
R
Capacitive
Reactance
XC
Inductive
Reactance
XL
Formula*Total
Reactance
X
Formula Network
Impedance
I
FormulaPhase
Angle
Θ
300-ohm 400-ohm X=XLXC  400-ohm
I =   R×X 
R2+X2
240 ohms
Θ = Tan-1( R )
X
  37°
100-ohm100-ohm −100-ohm  71 ohms−45°
*

This is NOT the correct equation for reactances in parallel, but it works as a memory aid to remember the sign as long as either XC or XL is zero. It is used here simply because it's the same formula for reactances in series and makes it easier to remember the sign of the reactance. Nowhere in the exam question pool is there a problem with reactances in parallel.

The correct formula(s) depend on the topography of the network. A few simple rules about reactances in series and parallel are all that are needed to compute the impedance of any complex network. But since it is not required by any question in the Extra examination pool, it is not covered in this summary.

The system often used to display the phase angle of a circuit containing resistance, inductive and/or capacitive reactance is the polar coordinate system.

There are five questions asking for the phase angle from known XC (capacitive reactance), XL (inductive reactance) and R (resistance), all measured in ohms. Computation is unnecessary since all five answers are the same phase angle - 14 degrees, and by checking if XL is greater than XC you know if the voltage leads or lags the current. The table below summarizes the logic steps required,

Resistance
R
Capacitive
Reactance
XC
Inductive
Reactance
XL
FormulaPhase Angle
Θ
Is XL>XC?Voltage
Leads/Lags
Current
100 ohms  25 ohms  50 ohms
Θ = Tan-1(XL-XC)
R
  14°YesLeads
1 kilohm500 ohms250 ohms−14°NoLags
100 ohms100 ohms  75 ohms−14°NoLags
100 ohms  75 ohms  50 ohms−14°NoLags
1 kilohm250 ohms500 ohms  14°YesLeads

There is one question in the exam requiring the computation of inductive reactance based on frequency before expressing the impedance as a complex number. This table shows the calculation,

Resistance
R
Inductance
L
Frequency
f
Formula Inductive
Reactance
XL
FormulaRectangular
Coordinates
40-ohm10-microhenry500 MHz XL=2πfL 31,415-ohm= R + jXL40 + j31,400

There are two questions that state the impedance of a circuit in polar coordinates as an admittance in millisiemens. Admittance is the reciprocal of impedance. Just divide the number of millisiemens into 1000 to get the impedance in ohms and change the sign of the angle from + to , or vice versa. The table below shows this simple conversion,

Admittance
M
FormulaImpedance
I
Phase Angle
ΘA
FormulaPhase Angle
Θ
7.09 millisiemens
I = 1000
M
127 ohm +45° Θ = − ΘA −45°
5      millisiemens200 ohm−30°+30°

In all the questions in Extra pool there is only one that calls for an impedance expressed in polar coordinates to be converted into rectangular coordinates. The conversion is shown below,

Impedance
I
Phase Angle
Θ
FormulaResistance
R
FormulaReactance
X
FormulaRectangular
Coordinates
200-ohm+30°R = I × Cos(Θ)174-ohm X = I × Sin(Θ)100-ohm= R + jX174 + j100

Four questions involve computing the impedance of a series circuit consisting of a resister, capacitor and/or inductor from their nominal values, operating at a given frequency and then locating correct impedance on Figure E5-2, where the horizontal axis represents resistance and the vertical axis reactance. Note that inductive reactance XL is positive and capacitive reactance XC is negative. Resistance is always positive so neither Point 5 nor Point 7 is an acceptable answer. The following table summarizes the computations,

Resistance
R
Frequency
f
Capacitance
picofarads

Capacitance
microfarads
C
Formula Capacitive
Reactance
XC
Inductance
microhenrys
L
Formula Inductive
Reactance
XL
FormulaReactance
X
Point on
Figure E5-2
300-ohm24.9 MHz85.000085
XC   1   
2πfC
75-ohm 0.64 XL=2πfL 100-ohm X=XLXC 25-ohm 300 + j25
Point 8
300-ohm3.505 MHz    18396-ohm396-ohm300 + j396
Point 3
300-ohm21.200 MHz19.000019395-ohm  l −395-ohm300 − j395
Point 1
400-ohm14 MHz38.000039291-ohm   −291-ohm400 − j291
Point 4

The phenomenon that as frequency increases, RF current flows in a thinner layer of the conductor, closer to the surface, is called the skin effect. Because of skin effect the resistance of a conductor is different for RF currents than for direct currents.

A capacitor is a device used to store electrical energy in an electrostatic field. The unit that measures electrical energy stored in an electrostatic field is the Joule.

A magnetic field is the region surrounding a magnet through which a magnetic force acts. The strength of a magnetic field around a conductor is determined by the amount of current. 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.

The term for energy stored in electromagnetic or electrostatic fields is potential energy.

The term for an out-of-phase, nonproductive power associated with inductors and capacitors is reactive power. Reactive power is wattless, nonproductive power. 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.

Given the phase angle between the voltage and current, there are three questions on the exam asking for the power factor of a R-L circuit. The mathematical answer is Trigonometric Cosine of the phase angle. A Math calculator can give you the answer but if you know how to use one, you probably have the values for 30°, 45° and 60° memorized. The table below gives the values,

Circuit TypePhase Angle
Θ
FormulaPower Factor
PF
R-L60°PF = Cos(Θ)0.500
R-L45°0.707
R-L30°0.866

The true power in an AC circuit where the voltage and current are out of phase is determined by multiplying the apparent power times the power factor. There are three questions in the exam asking to compute the power consumed in a circuit with a known power factor. The table below summarizes these calculations,

Voltage
E
Current
I
FormulaPower
P
Power Factor
PF
FormulaPower Consumed
PC
100-V AC4 amperesP=E × I  400 Watts0.2PC=PF × P  80 watts
200-V AC5 amperes1000 Watts0.6600 watts
   500 Watts  0.71355 watts

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