General G5B: Power math, RMS and decibels
These are the 14 official questions in group G5B of the 2023–2027 General question pool, with the correct answer marked and a plain-English explanation of why it's right. The real exam asks one question from this group.
- G5B01Question 1
What dB change represents a factor of two increase or decrease in power?
- AApproximately 2 dB
- BApproximately 3 dB Correct
- CApproximately 6 dB
- DApproximately 9 dB
Why: Doubling or halving power is about 3 dB.
- G5B02Question 2
How does the total current relate to the individual currents in a circuit of parallel resistors?
- AIt equals the average of the branch currents
- BIt decreases as more parallel branches are added to the circuit
- CIt equals the sum of the currents through each branch Correct
- DIt is the sum of the reciprocal of each individual voltage drop
Why: Total current in a parallel circuit is the sum of the branch currents.
- G5B03Question 3
How many watts of electrical power are consumed if 400 VDC is supplied to an 800-ohm load?
- A0.5 watts
- B200 watts Correct
- C400 watts
- D3200 watts
Why: P = E² ÷ R = 400² ÷ 800 = 200 W.
- G5B04Question 4
How many watts of electrical power are consumed by a 12 VDC light bulb that draws 0.2 amperes?
- A2.4 watts Correct
- B24 watts
- C6 watts
- D60 watts
Why: P = E × I = 12 × 0.2 = 2.4 W.
- G5B05Question 5
How many watts are consumed when a current of 7.0 milliamperes flows through a 1,250-ohm resistance?
- AApproximately 61 milliwatts Correct
- BApproximately 61 watts
- CApproximately 11 milliwatts
- DApproximately 11 watts
Why: P = I² × R = 0.007² × 1,250 ≈ 0.061 W, about 61 mW.
- G5B06Question 6
What is the PEP produced by 200 volts peak-to-peak across a 50-ohm dummy load?
- A1.4 watts
- B100 watts Correct
- C353.5 watts
- D400 watts
Why: 200 Vpp = 100 V peak = 70.7 V RMS; 70.7² ÷ 50 = 100 W.
- G5B07Question 7
What value of an AC signal produces the same power dissipation in a resistor as a DC voltage of the same value?
- AThe peak-to-peak value
- BThe peak value
- CThe RMS value Correct
- DThe reciprocal of the RMS value
Why: The RMS value of AC heats a resistor the same as that much DC.
- G5B08Question 8
What is the peak-to-peak voltage of a sine wave with an RMS voltage of 120 volts?
- A84.8 volts
- B169.7 volts
- C240.0 volts
- D339.4 volts Correct
Why: 120 V RMS × 1.414 = 169.7 V peak, × 2 = 339.4 V peak-to-peak.
- G5B09Question 9
What is the RMS voltage of a sine wave with a value of 17 volts peak?
- A8.5 volts
- B12 volts Correct
- C24 volts
- D34 volts
Why: 17 V peak × 0.707 ≈ 12 V RMS.
- G5B10Question 10
What percentage of power loss is equivalent to a loss of 1 dB?
- A10.9 percent
- B12.2 percent
- C20.6 percent Correct
- D25.9 percent
Why: 1 dB down is 79.4% of the power, a 20.6% loss.
- G5B11Question 11
What is the ratio of PEP to average power for an unmodulated carrier?
- A0.707
- B1.00 Correct
- C1.414
- D2.00
Why: An unmodulated carrier's PEP equals its average power: a ratio of 1.00.
- G5B12Question 12
What is the RMS voltage across a 50-ohm dummy load dissipating 1200 watts?
- A173 volts
- B245 volts Correct
- C346 volts
- D692 volts
Why: E = √(P × R) = √(1,200 × 50) ≈ 245 V RMS.
- G5B13Question 13
What is the output PEP of an unmodulated carrier if the average power is 1060 watts?
- A530 watts
- B1060 watts Correct
- C1500 watts
- D2120 watts
Why: For an unmodulated carrier, PEP equals average power: 1,060 W.
- G5B14Question 14
What is the output PEP of 500 volts peak-to-peak across a 50-ohm load?
- A8.75 watts
- B625 watts Correct
- C2500 watts
- D5000 watts
Why: 500 Vpp = 250 V peak = 176.8 V RMS; 176.8² ÷ 50 ≈ 625 W.