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G5B

ELECTRICAL PRINCIPLES

- The decibel; current and voltage dividers; electrical power calculations; sine wave root-mean-square (RMS) values; PEP calculations

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G5B011 of 14

What dB change represents a factor of two increase or decrease in power?

Why Power ratios in decibels use dB = 10 x log10(P2/P1). For a ratio of 2, log10(2) = 0.301, so the change is 10 x 0.301 = about 3 dB (positive for doubling, negative for halving). This is why 3 dB is called the half-power point on filter and antenna response curves.
Watch out The choice near 6 dB is the value you get for doubling voltage or current (20 x log10(2)), or for quadrupling power, not for doubling power.
Double the power = 3 dB. Double the voltage = 6 dB. Ten times power = 10 dB.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
G5B022 of 14

How does the total current relate to the individual currents in a circuit of parallel resistors?

Why In a parallel circuit every branch sees the same voltage, and each branch draws its own current given by I = V/R. Kirchhoff's current law says charge cannot pile up at a node, so whatever flows into the junction must flow out: the source current is the arithmetic sum of all the branch currents. Adding another branch gives the current one more path, so the total current goes up while the total resistance goes down.
Watch out The idea that total current decreases as branches are added gets it backward; it is the total resistance that decreases. Averaging branch currents would violate conservation of charge, and summing reciprocals of voltage drops is a garbled version of the parallel resistance formula.
Parallel = more paths = more current. Currents add in parallel, voltages add in series.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
G5B033 of 14

How many watts of electrical power are consumed if 400 VDC is supplied to an 800-ohm load?

Why Power can be found from voltage and resistance with P = E squared divided by R. Here 400 V squared is 160,000, and 160,000 divided by 800 ohms gives 200 watts. Equivalently, the current is 400/800 = 0.5 A, and P = E x I = 400 x 0.5 = 200 watts.
Watch out The choice of 0.5 is the current in amperes (400 V / 800 ohms), not the power; you still have to multiply that current by the voltage.
P = E squared / R. Square the volts first, then divide by ohms.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
G5B044 of 14

How many watts of electrical power are consumed by a 12 VDC light bulb that draws 0.2 amperes?

Why DC power is simply voltage times current, P = E x I. Here 12 volts multiplied by 0.2 amperes gives 2.4 watts. No other formula is needed because both voltage and current are given directly.
Watch out The 24 watt choice is the answer you get by slipping a decimal place, treating the 0.2 A as 2 A; the 60 watt figure comes from dividing 12 by 0.2, which gives resistance in ohms, not power.
P = E x I. Watch the decimal: 0.2 A is one fifth of an amp, so the answer is small.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
G5B055 of 14

How many watts are consumed when a current of 7.0 milliamperes flows through a 1,250-ohm resistance?

Why Use P = I squared times R when you know current and resistance. Convert first: 7.0 mA = 0.0070 A, and 0.0070 squared = 0.000049 A squared. Multiply by 1,250 ohms and you get 0.061 W, which is about 61 milliwatts.
Watch out The choice reading 61 watts has the right arithmetic but the wrong scale: a few milliamps through a kilohm is a tiny amount of power, so the answer has to land in the milliwatt range.
P = I²R. Convert mA to amps first, or you will be off by a million.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
G5B066 of 14

What is the PEP produced by 200 volts peak-to-peak across a 50-ohm dummy load?

Why Peak-to-peak voltage is twice the peak, so 200 V p-p is 100 V peak. For a sine wave the RMS value is 0.707 times the peak, giving 70.7 V RMS. Power into a resistive load is V(RMS) squared divided by R: 70.7 squared is about 5000, divided by 50 ohms gives 100 watts. PEP into a dummy load with a steady sine wave is simply this average power.
Watch out The 400 watt choice comes from using the full 200 V peak-to-peak as if it were RMS, and 353.5 watts comes from squaring 100 V peak without converting to RMS first (and mixing up the arithmetic); both skip one of the two required conversions.
Two steps: halve p-p to get peak, times 0.707 for RMS, then V(RMS)^2/R. 200 p-p into 50 ohms = 100 W.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
G5B077 of 14

What value of an AC signal produces the same power dissipation in a resistor as a DC voltage of the same value?

Why RMS stands for root-mean-square, and it is defined precisely so that an AC waveform delivers the same average heating power to a resistor as a DC voltage of that numeric value. If 120 V RMS AC and 120 V DC are each applied to the same resistor, both dissipate identical average power, since P = V(RMS)^2 / R. For a sine wave, RMS is about 0.707 times the peak value, or peak divided by the square root of 2.
Watch out The peak value is the highest instantaneous voltage of the waveform, and it is about 1.414 times the RMS value, so a sine wave with a peak equal to a DC voltage delivers only half the power. Peak-to-peak is double the peak and describes the total swing, not the heating value.
RMS = the 'DC equivalent' value. Heating power always works in RMS; peak and peak-to-peak just describe the waveform's size.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
G5B088 of 14

What is the peak-to-peak voltage of a sine wave with an RMS voltage of 120 volts?

Why For a sine wave, peak voltage equals RMS times the square root of 2: 120 x 1.414 = 169.7 V peak. Peak-to-peak spans from the negative peak to the positive peak, so it is twice the peak value: 169.7 x 2 = 339.4 V. A handy shortcut is that peak-to-peak equals RMS x 2.828.
Watch out The choice of 169.7 volts is the peak (not peak-to-peak) value, the classic trap of stopping one step early; 84.8 volts is RMS divided by 1.414, the reverse operation.
RMS to peak-to-peak: multiply by 2.828. 120 V wall outlet actually swings about 339 V tip to tip.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
G5B099 of 14

What is the RMS voltage of a sine wave with a value of 17 volts peak?

Why For a sine wave, the RMS (effective) value is the peak value divided by the square root of 2, or equivalently peak times 0.707. So 17 V peak x 0.707 = about 12 V RMS. RMS matters because it is the AC voltage that produces the same heating power as that many volts of DC.
Watch out The 24 volt answer comes from multiplying by 1.414 instead of dividing, which gives peak from RMS, not the other way around. Half the peak, 8.5 V, is not a meaningful sine wave quantity here, and 34 V is the peak-to-peak value.
RMS = peak x 0.707 (smaller). Peak = RMS x 1.414 (bigger). RMS is always less than peak.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
G5B1010 of 14

What percentage of power loss is equivalent to a loss of 1 dB?

Why A decibel change relates to a power ratio by ratio = 10^(dB/10). For a 1 dB loss the remaining power fraction is 10^(-0.1) = 0.794, so about 79.4 percent gets through and roughly 20.6 percent is lost. This is a handy benchmark alongside the familiar 3 dB = half power (50 percent loss).
Watch out The 10.9 percent figure is what a 0.5 dB loss costs (10^-0.05 = 0.891), and 25.9 percent corresponds to about 1.3 dB, so both come from the right formula with the wrong dB value.
1 dB gone = about 20 percent of the power gone; 3 dB gone = half.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
G5B1111 of 14

What is the ratio of PEP to average power for an unmodulated carrier?

Why PEP is the average power during one RF cycle at the crest of the modulation envelope. An unmodulated carrier is a steady sine wave whose envelope never changes, so the peak envelope cycle is the same as every other cycle and PEP equals the average power. The ratio is therefore exactly 1.00. Only when modulation makes the envelope rise and fall (as in SSB voice) does PEP exceed average power.
Watch out The 0.707 and 1.414 values are the RMS-to-peak conversion factors for instantaneous voltage or current on a sine wave, not a power ratio, and they do not apply to comparing PEP with average power.
Flat envelope, no peaks: unmodulated carrier means PEP = average, ratio 1.00.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
G5B1212 of 14

What is the RMS voltage across a 50-ohm dummy load dissipating 1200 watts?

Why Power in a resistive load is P = V^2/R with V in RMS volts, so rearranged V = sqrt(P x R). Here sqrt(1200 W x 50 ohms) = sqrt(60,000) = about 245 volts RMS. A dummy load is purely resistive, so no reactance or power factor correction is needed.
Watch out The 346 volt choice is the peak voltage (245 x 1.414), and 692 volts is the peak-to-peak value; the question asks for RMS. The 173 volt figure is what you get if you wrongly divide the RMS answer by 1.414.
V(RMS) = sqrt(P x R). 1200 x 50 = 60,000, and sqrt(60,000) is about 245.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
G5B1313 of 14

What is the output PEP of an unmodulated carrier if the average power is 1060 watts?

Why Peak envelope power is the average power during one RF cycle at the crest of the modulation envelope. An unmodulated carrier has a constant envelope, so every cycle is the crest cycle and PEP equals the average power, 1060 watts. This is why steady-carrier modes like unmodulated CW key-down make PEP measurements easy.
Watch out Doubling to 2120 watts or halving to 530 watts comes from confusing this with peak versus RMS voltage relationships or with modulated AM, where a fully modulated carrier's envelope peak reaches four times carrier power; none of that applies to a flat, unmodulated envelope.
Flat envelope, flat answer: no modulation means PEP = average power, same number in, same number out.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
G5B1414 of 14

What is the output PEP of 500 volts peak-to-peak across a 50-ohm load?

Why For a sine wave, PEP equals the RMS voltage squared divided by the load resistance. Half of 500 V peak-to-peak is 250 V peak, and RMS is peak divided by the square root of 2, about 176.8 V. Then 176.8 squared divided by 50 ohms gives 625 watts. The one-step shortcut is P = Vpp squared divided by 8R: 250,000 / (8 x 50) = 625 W.
Watch out The choice that says 2500 watts comes from plugging the full peak-to-peak voltage into P = V squared / (2R) as if it were peak, skipping the conversion to RMS; 5000 watts treats 500 V p-p as if it were RMS.
PEP = Vpp squared / 8R. Divide peak-to-peak by 2 for peak, by 2.828 for RMS.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
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