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Sample SHGM Part-66 B2 Practice Questions
Try these sample questions to review concepts for the SHGM Part-66 B2 exam. Each question includes a detailed explanation. Start the interactive quiz above for the full 28+ question experience with AI tutoring.
1A 28 V supply feeds series resistors of 3 Ω and 4 Ω. What voltage appears across the 4 Ω resistor?
A.7 V
B.12 V
C.16 V
D.28 V
Explanation: Series current is 28/(3 + 4) = 4 A. The 4 Ω resistor drops 4 × 4 = 16 V.
2A capacitor's reactance is Xc = 1/(2πfC). If frequency doubles with capacitance unchanged, what happens to reactance?
A.It doubles
B.It halves
C.It stays constant
D.It becomes four times greater
Explanation: Frequency is in the denominator. Doubling frequency therefore halves capacitive reactance.
3Using stored energy E = ½CV², what energy is stored in a 0.01 F capacitor charged to 20 V?
A.0.1 J
B.1 J
C.2 J
D.4 J
Explanation: Energy is 0.5 × 0.01 × 20² = 2 J. Voltage is squared, so an apparently modest voltage change can strongly affect stored energy.
4In steady, straight, level unaccelerated flight, how does total vertical aerodynamic lift compare with aircraft weight?
A.Lift is greater whenever airspeed increases
B.Lift balances weight
C.Lift is always half the weight
D.Lift must be zero if pitch attitude is zero
Explanation: Zero vertical acceleration requires balanced vertical forces. Airspeed and angle of attack can change while the required total vertical lift remains equal to weight in this condition.
5What is the basic purpose of a rectifier in an aircraft electrical system?
A.Convert AC to DC
B.Convert DC to AC
C.Change AC voltage by transformer action alone
D.Store electrical energy for later discharge
Explanation: Rectification converts alternating current to unidirectional current. Filtering and regulation may then be needed to obtain the specified DC supply quality.
6A single-phase AC load takes 5 A at 115 V RMS with power factor 0.8. What is its real power?
A.92 W
B.460 W
C.575 W
D.718.75 W
Explanation: Real power is voltage × current × power factor. Thus 115 × 5 × 0.8 = 460 W; apparent power is 575 VA.
7Why does a normal transformer require changing magnetic flux for sustained secondary voltage?
A.A steady DC current always gives the rated AC output
B.Secondary voltage is induced by changing flux linking its winding
C.The secondary must be directly connected to the primary conductor
D.The core must rotate mechanically
Explanation: Electromagnetic induction depends on a change of linked magnetic flux. Applying steady DC to an ordinary AC transformer does not provide its normal sustained secondary output and can overheat it.
8Why is a current-limiting device selected to protect the circuit wiring as well as the load?
A.An oversized protector can allow wiring to overheat before opening
B.Only the load's purchase price determines protection
C.Any fuse rating is safe if it fits the holder
D.A circuit breaker increases wire insulation temperature limits
Explanation: Overcurrent protection must coordinate with the circuit's approved design. Physical fit does not establish the correct current or time characteristic.
9For a two-input NOR gate, which input combination produces a HIGH output?
A.LOW, LOW
B.LOW, HIGH
C.HIGH, LOW
D.HIGH, HIGH
Explanation: NOR is the inversion of OR. OR is LOW only when both inputs are LOW, so NOR is HIGH in that case.
10An ideal 8-bit converter divides a 0–10 V input range into 256 equal quantization intervals. What is one interval approximately?
A.3.9 mV
B.39 mV
C.125 mV
D.256 mV
Explanation: The stated interval width is 10/256 V = 0.03906 V. This is approximately 39 mV under the exercise's explicit convention.
About the SHGM Part-66 B2 Practice Questions
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