G5 Electrical Principles

C. Resonance causes impedance to be very low

In a series LC circuit, inductive and capacitive reactance act in opposite directions, so when they are equal (at resonance) they cancel. Only the circuit's resistance is left to oppose current, so the impedance is at its minimum and the current is at its maximum.

• A. Very high impedance at resonance describes a parallel LC circuit, where current circulating between the inductor and capacitor makes the source see a high impedance.
• B. Inductance (in henries) and capacitance (in farads) cannot be averaged into an impedance (in ohms); at series resonance the reactances cancel and the impedance equals the circuit's resistance.
• D. An arithmetic mean of inductance and capacitance mixes henries and farads and is not an impedance; at series resonance the impedance equals the circuit's resistance.

B. Opposition to the flow of alternating current caused by capacitance or inductance

Reactance is the opposition that inductors and capacitors present to alternating current. Unlike resistance, it stores energy in a magnetic or electric field and returns it to the circuit instead of turning it into heat, and its value changes with frequency.

• A. Opposition to direct current caused by resistance is resistance itself; reactance comes from inductance and capacitance and applies to AC.
• C. Resistance never reinforces current; it always opposes it, and reactance is a separate property that applies to AC.
• D. Inductance and capacitance oppose changes in current and voltage; they do not reinforce the flow of alternating current.

D. Reactance

An inductor opposes alternating current through inductive reactance, measured in ohms. It rises with frequency: XL = 2 x π x f x L.

• A. Conductance is the reciprocal of resistance, measured in siemens; it describes how easily current flows, not opposition to AC.
• B. Reluctance is the opposition to magnetic flux in a magnetic circuit, such as a transformer core, not opposition to current.
• C. Admittance is the reciprocal of impedance; it describes how easily AC flows rather than how strongly it is opposed.

C. Reactance

A capacitor opposes alternating current through capacitive reactance, measured in ohms. It falls as frequency rises: XC = 1 / (2 x π x f x C).

• A. Conductance is the reciprocal of resistance, measured in siemens; it describes how easily current flows, not opposition to AC.
• B. Reluctance is the opposition to magnetic flux in a magnetic circuit, such as a transformer core, not opposition to current.
• D. Admittance is the reciprocal of impedance; it describes how easily AC flows rather than how strongly it is opposed.

D. As the frequency of the applied AC increases, the reactance increases

Inductive reactance is XL = 2 x π x f x L, so it rises in proportion to frequency. Doubling the frequency doubles an inductor's reactance.

• A. Reactance that falls as frequency rises describes a capacitor, not an inductor.
• B. An inductor's reactance depends on frequency and inductance, not on the amplitude of the applied AC.
• C. Changing the amplitude of the AC does not change an inductor's reactance; only frequency and inductance appear in XL = 2 x π x f x L.

A. As the frequency of the applied AC increases, the reactance decreases

Capacitive reactance is XC = 1 / (2 x π x f x C), so it falls as frequency rises. Doubling the frequency halves a capacitor's reactance.

• B. Reactance that rises with frequency describes an inductor, not a capacitor.
• C. A capacitor's reactance depends on frequency and capacitance, not on the amplitude of the applied AC.
• D. Changing the amplitude of the AC does not change a capacitor's reactance; only frequency and capacitance appear in XC = 1 / (2 x π x f x C).

D. Admittance

Admittance (symbol Y) is the reciprocal of impedance: Y = 1 / Z. Impedance measures how strongly a circuit opposes AC; admittance measures how readily it passes AC, in siemens.

• A. Conductance is the reciprocal of resistance alone, not of the full impedance.
• B. Susceptance is the reciprocal of reactance alone, not of the full impedance.
• C. Reluctance is the opposition to magnetic flux in a magnetic circuit and has nothing to do with the reciprocal of impedance.

C. The ratio of voltage to current

Impedance (Z) is the total opposition to alternating current, combining resistance and reactance. Like resistance in Ohm's law, it equals voltage divided by current: Z = E / I, in ohms.

• A. The ratio of current to voltage is the reciprocal of impedance, which is admittance.
• B. The product of current and voltage is power (watts or volt-amperes), not impedance.
• D. The product of current and reactance gives the voltage across that reactance, not impedance.

B. Ohm

Reactance is opposition to current, like resistance, so it is measured in the same unit: the ohm. A coil might have, for example, 50 ohms of reactance at a given frequency.

• A. The farad is the unit of capacitance, the property that produces capacitive reactance, not the unit of reactance itself.
• C. The ampere is the unit of current.
• D. The siemens is the unit of conductance, susceptance and admittance, which are reciprocals of opposition quantities.

D. All these choices are correct

Several devices can match impedances at RF: a transformer (its impedance ratio is the square of its turns ratio), a Pi-network of capacitors and an inductor, and a section of transmission line such as a quarter-wave matching section. All three are used.

• A. A transformer does match impedances, but it is only one of the listed devices that can.
• B. A Pi-network does match impedances and is common in transmitter outputs and antenna tuners, but it is only one of the listed devices.
• C. A length of transmission line, such as a quarter-wave section, can transform impedance, but it is only one of the listed devices.

B. X

Reactance is represented by the letter X, with subscripts for the two kinds: XL for inductive reactance and XC for capacitive reactance.

• A. The letter Z represents impedance, the combination of resistance and reactance.
• C. The letter B represents susceptance, the reciprocal of reactance.
• D. The letter Y represents admittance, the reciprocal of impedance.

D. Inductive reactance and capacitive reactance cancel

Resonance occurs at the frequency where inductive reactance equals capacitive reactance. Because the two act in opposite directions, they cancel, and only resistance remains to oppose current.

• A. Current and voltage are measured in different units and cannot be equal; at resonance the remaining opposition is resistive, so they are in phase.
• B. Resistance is not cancelled at resonance; the reactances cancel, leaving resistance as the only opposition.
• C. An LC circuit is not an antenna; at resonance its energy moves back and forth between the inductor and capacitor, and losses are mostly heat.

B. Approximately 3 dB

A power ratio in decibels is 10 x log(P2 / P1). For a factor of two, 10 x log(2) = 3.01 dB, so doubling power is about +3 dB and halving it is about -3 dB.

• A. 2 dB is a power ratio of about 1.6, not 2.
• C. 6 dB is a power ratio of about 4 (two doublings); it is also the change for doubling voltage, not power.
• D. 9 dB is a power ratio of about 8 (three doublings).

C. It equals the sum of the currents through each branch

In a parallel circuit, current divides among the branches, and the total current from the source equals the sum of the branch currents (Kirchhoff's current law). Branches drawing 1 A, 2 A and 3 A draw 6 A in total.

• A. The total is the sum of the branch currents, not their average; with more than one branch, the average is smaller than the total.
• B. Adding parallel branches lowers the total resistance, so the total current increases rather than decreases.
• D. Every branch of a parallel circuit has the same voltage across it; the total current is the sum of branch currents, not a sum of reciprocal voltages.

B. 200 watts

Power from voltage and resistance is P = E² / R. Here P = 400² / 800 = 160,000 / 800 = 200 watts.

• A. 0.5 is the current in amperes (I = E / R = 400 / 800), not the power.
• C. 400 watts at 400 volts would require a 400-ohm load (400² / 400), not 800 ohms.
• D. 3200 watts at 400 volts would require a 50-ohm load (400² / 50), not 800 ohms.

A. 2.4 watts

Power equals voltage times current: P = E x I = 12 V x 0.2 A = 2.4 watts.

• B. 24 watts is ten times too large; at 12 volts it would require a current of 2 A.
• C. 6 watts at 12 volts would require a current of 0.5 A, not 0.2 A.
• D. 60 is the bulb's resistance in ohms (R = E / I = 12 / 0.2), not its power.

A. Approximately 61 milliwatts

With current and resistance known, P = I² x R. Converting 7.0 mA to 0.007 A gives P = 0.007² x 1250 = 0.06125 W, about 61 milliwatts.

• B. 61 watts has the right digits but the wrong prefix; the result is 0.061 W, which is 61 milliwatts.
• C. 11 milliwatts does not follow from P = I² x R; the actual result, 61 mW, is more than five times larger.
• D. 11 watts has neither the right digits nor the right size; P = I² x R gives 0.061 W.

B. 100 watts

Halve the peak-to-peak voltage to get the peak: 200 / 2 = 100 V. Multiply by 0.707 for RMS: 70.7 V. Then PEP = E² / R = 70.7² / 50 = 100 watts.

• A. 1.4 is the RMS current in amperes (70.7 V / 50 ohms), not the power.
• C. 353.5 watts does not follow from the conversion; peak-to-peak to peak to RMS, then E² / R, gives 100 watts.
• D. 400 watts results from treating 200 volts as the peak voltage instead of the peak-to-peak voltage, which doubles the RMS voltage and quadruples the power.

C. The RMS value

The RMS (root mean square) value of an AC voltage is defined as the DC voltage that produces the same heating in a resistor. For a sine wave, RMS = 0.707 x peak; the 120 VAC of household power is an RMS value.

• A. The peak-to-peak value of a sine wave is 2.828 times its RMS value, so it greatly overstates the heating effect.
• B. The peak value is reached only for an instant in each half cycle; it is 1.414 times the RMS value and overstates the heating effect.
• D. The reciprocal of the RMS value is not a voltage at all, so it cannot be compared with a DC voltage.

D. 339.4 volts

For a sine wave, peak = RMS x 1.414 and peak-to-peak = 2 x peak. So 120 x 1.414 = 169.7 V peak, and 2 x 169.7 = 339.4 V peak-to-peak.

• A. 84.8 volts is 120 x 0.707, the peak-to-RMS conversion applied in the wrong direction.
• B. 169.7 volts is the peak voltage; the peak-to-peak voltage is twice that.
• C. 240 volts doubles the RMS value without first converting it to peak.

B. 12 volts

For a sine wave, RMS = peak x 0.707. Here 17 x 0.707 = 12.0 volts.

• A. 8.5 volts is half the peak value; the RMS conversion factor is 0.707, not 0.5.
• C. 24 volts is 17 x 1.414, the RMS-to-peak conversion applied in the wrong direction.
• D. 34 volts is the peak-to-peak value, twice the 17-volt peak.

C. 20.6 percent

Working backward from dB = 10 x log(Pout / Pin), a loss of 1 dB means Pout / Pin = 0.794. The output is 79.4 percent of the input, so 20.6 percent of the power is lost.

• A. 10.9 percent is the drop in voltage for a 1 dB loss (voltage ratio 0.891); power drops more because it depends on the square of voltage.
• B. 12.2 percent is the rise in voltage for a 1 dB gain (voltage ratio 1.122), not a power loss.
• D. 25.9 percent is the rise in power for a 1 dB gain (power ratio 1.259), not the loss for 1 dB.

B. 1.00

PEP is the average power during one RF cycle at the highest point of the modulation envelope. An unmodulated carrier has a constant envelope, so every cycle is the same and PEP equals average power: a ratio of 1.00.

• A. 0.707 is the ratio of RMS to peak voltage for a sine wave, not a power ratio for a carrier.
• C. 1.414 is the ratio of peak to RMS voltage for a sine wave, not a power ratio for a carrier.
• D. A PEP to average ratio of 2 applies to a two-tone SSB test signal, whose envelope varies; an unmodulated carrier's envelope does not.

B. 245 volts

From P = E² / R, the voltage is E = square root of (P x R) = square root of (1200 x 50) = square root of 60,000 = 245 volts RMS.

• A. 173 volts is 245 x 0.707; the formula already gives an RMS voltage, so no further conversion is needed.
• C. 346 volts is the peak voltage (245 x 1.414), not the RMS voltage.
• D. 692 volts is the peak-to-peak voltage, twice the 346-volt peak.

B. 1060 watts

An unmodulated carrier has a constant envelope, so its peak envelope power equals its average power. With 1060 watts average, the PEP is also 1060 watts.

• A. 530 watts is half the average power; PEP is never less than average power.
• C. 1500 watts is 1060 x 1.414; that factor relates peak and RMS voltage, and for a steady carrier the PEP to average power ratio is 1.
• D. 2120 watts doubles the average power; a ratio of 2 applies to a two-tone test signal, not a steady carrier.

B. 625 watts

Peak voltage is 500 / 2 = 250 V, and RMS voltage is 250 x 0.707 = 176.8 V. Then PEP = E² / R = 176.8² / 50 = 625 watts.

• A. 8.75 watts is far too low; it would correspond to only about 21 volts RMS across 50 ohms.
• C. 2500 watts results from treating 500 volts as the peak voltage instead of the peak-to-peak voltage.
• D. 5000 watts comes from squaring the peak-to-peak voltage directly (500² / 50) without converting to RMS.

C. Mutual inductance

AC in the primary winding creates a changing magnetic field in the core. That field passes through the secondary winding and induces a voltage in it; this magnetic coupling between windings is called mutual inductance.

• A. Capacitive coupling transfers energy through the electric field between conductors; a transformer works through its shared magnetic field.
• B. Displacement current is the changing electric field that lets AC pass through a capacitor; it is not how a transformer couples its windings.
• D. Mutual capacitance is the capacitance between two conductors; some exists between windings, but it is not what produces the secondary voltage.

A. The input voltage is multiplied by 4

A transformer works in either direction. Driving the secondary of a 4:1 step-down transformer turns it into a 1:4 step-up transformer, so the output voltage is the input voltage multiplied by 4.

• B. Dividing by 4 is the normal step-down result when the signal goes into the primary, not the secondary.
• C. No series resistance is needed; reversing the transformer only changes it from step-down to step-up.
• D. No parallel resistance is needed; driven from the secondary, the transformer steps the voltage up by 4.

A. 5.9 ohms

For resistors in parallel, R = 1 / (1/R1 + 1/R2 + 1/R3). Here 1/10 + 1/20 + 1/50 = 0.1 + 0.05 + 0.02 = 0.17, and 1 / 0.17 = 5.9 ohms.

• B. 0.17 is the sum of the reciprocals; the last step, 1 / 0.17, is missing.
• C. 17 ohms is larger than the smallest resistor (10 ohms), which is impossible for resistors in parallel.
• D. 80 ohms is the series total (10 + 20 + 50), not the parallel total.

D. 67 ohms

For two resistors in parallel, R = (R1 x R2) / (R1 + R2) = (100 x 200) / (100 + 200) = 20,000 / 300 = 66.7 ohms, about 67 ohms.

• A. 300 ohms is the series total (100 + 200), not the parallel total.
• B. 150 ohms is the average of the two values; the parallel total is always less than the smallest resistor.
• C. 75 ohms does not follow from the product-over-sum formula, which gives 66.7 ohms.

B. To accommodate the higher current of the primary

A transformer delivers about the same power as it takes in. In a step-up transformer, the secondary has the higher voltage and therefore the lower current, so the primary carries the higher current and needs heavier wire.

• A. Coupling between windings depends on the core and the arrangement of the windings, not on wire size.
• C. Parasitic oscillations come from unwanted feedback in amplifiers; transformer wire size is chosen for current, not to prevent them.
• D. There is no requirement for windings of equal volume; wire size is set by the current each winding carries.

A. 360 volts

Transformer voltage scales with the turns ratio: Es = Ep x (Ns / Np) = 120 x (1500 / 500) = 120 x 3 = 360 volts.

• B. 120 volts would require equal turns on both windings; this secondary has three times as many turns as the primary.
• C. 40 volts inverts the turns ratio (120 x 500 / 1500); with more turns on the secondary, the voltage steps up, not down.
• D. 25.5 volts does not follow from the turns ratio; a 1:3 step-up transformer triples the voltage.

A. 3.5 to 1

A transformer's impedance ratio is the square of its turns ratio, so the turns ratio is the square root of the impedance ratio: square root of (600 / 50) = square root of 12 = 3.46, about 3.5 to 1.

• B. 12 to 1 is the impedance ratio; the turns ratio is its square root.
• C. 24 to 1 doubles the impedance ratio; the turns ratio is the square root of the impedance ratio, not a multiple of it.
• D. 144 to 1 squares the impedance ratio instead of taking its square root.

D. 10.750 nanofarads

Capacitors in parallel add directly. Converting 750 pF to 0.75 nF: 5.0 + 5.0 + 0.75 = 10.75 nanofarads (10,750 pF).

• A. 576.9 is the value of the three capacitors in series, and in picofarads rather than nanofarads; parallel capacitors add instead.
• B. 1,733 matches the digits of the sum of reciprocals in the series formula (1/5000 + 1/5000 + 1/750 = 0.001733 per pF); it is not the parallel total.
• C. 3,583 picofarads is the average of the three values (10,750 / 3); parallel capacitors add rather than average.

C. 33.3 microfarads

Capacitors in series combine like resistors in parallel: C = 1 / (1/C1 + 1/C2 + 1/C3). For three equal 100 µF capacitors, that is 100 / 3 = 33.3 microfarads.

• A. 0.33 microfarads is 100 times too small; one third of 100 µF is 33.3 µF.
• B. 3.0 microfarads does not follow from the series formula; three equal capacitors in series give one third of one capacitor's value, 33.3 µF.
• D. 300 microfarads is the parallel total; capacitors in series give less capacitance than any one of them.

C. 3.3 millihenries

Inductors in parallel combine like resistors in parallel. For three equal 10 mH inductors, the total is 10 / 3 = 3.3 millihenries.

• A. 0.30 henries is 300 mH, ten times even the series total of 30 mH.
• B. 3.3 henries has the right digits but the wrong unit; it is 1000 times 3.3 mH.
• D. 30 millihenries is the series total (10 + 10 + 10); inductors in parallel give less inductance than any one of them.

C. 70 millihenries

Inductors in series add like resistors in series: 20 mH + 50 mH = 70 millihenries (assuming no magnetic coupling between them).

• A. 7 millihenries is ten times too small; the series total is 20 + 50 = 70 mH.
• B. 14.3 millihenries is the parallel combination (20 x 50 / 70); inductors in series add.
• D. 1,000 is the product of the two values (20 x 50), which is only the numerator of the parallel formula.

B. 14.3 microfarads

Capacitors in series combine like resistors in parallel: C = (C1 x C2) / (C1 + C2) = (20 x 50) / (20 + 50) = 1000 / 70 = 14.3 microfarads.

• A. 0.07 microfarads is far too small; two capacitors in series give a total between half the smaller value and the smaller value, here 14.3 µF.
• C. 70 microfarads is the sum of the two values, which is how capacitors combine in parallel, not in series.
• D. 1,000 is the product of the two values (20 x 50), only the numerator of the series formula; it must be divided by the sum, 70.

C. A capacitor in parallel

Capacitors in parallel add, because together they act like one capacitor with a larger plate area. Adding a capacitor in parallel therefore increases the total capacitance.

• A. An inductor in series with a capacitor forms a series resonant circuit; it adds inductive reactance, not capacitance.
• B. An inductor in parallel with a capacitor forms a parallel resonant (tank) circuit; it does not add capacitance.
• D. A capacitor in series reduces the total capacitance to less than the smaller of the two.

D. An inductor in series

Inductors in series add, like resistors in series, so adding an inductor in series increases the total inductance (assuming the two are not magnetically coupled).

• A. A capacitor in series with an inductor forms a series resonant circuit; it does not add inductance.
• B. A capacitor in parallel with an inductor forms a parallel resonant (tank) circuit; it does not add inductance.
• C. An inductor in parallel reduces the total inductance to less than the smaller of the two.