The Router That Ignored the Shorter-Looking Route
Two routes matched the destination. The router chose the more specific prefix, not the visually simpler route.
Solve This Question β The full solution unlocks after the investigation.Practice FE-style Electrical & Computer questions through interactive engineering problems with guided reasoning and detailed solutions.
Two routes matched the destination. The router chose the more specific prefix, not the visually simpler route.
Solve This Question β The full solution unlocks after the investigation.Each topic groups questions around a specific FE knowledge area. Topic pages become indexable only after they contain a meaningful practice set.
Choose a question and solve it before reading the final engineering verdict.
Two routes matched the destination. The router chose the more specific prefix, not the visually simpler route.
Serialization took only 12 microseconds. The fiber path added another 500 microseconds.
The subnet looked large enough from the address range. The usable host count said otherwise.
Four 20 kHz channels looked like an 80 kHz jobβuntil the guard bands were counted.
The voice channel bandwidth was 4 kHz. The data link was sized at 32 kbps and immediately overflowed.
The carrier was 5 V. The message amplitude was 6 V. The envelope detector stopped reproducing the message cleanly.
Increasing gain improved responseβuntil the characteristic equation crossed the stability limit.
The low-frequency gain was 40 dB. One decade past the pole, the response was near 20 dB.
The forward gain was 10. The closed-loop output did not become 10 times the reference.
The secondary load was 8 Ξ©. The primary source did not see 8 Ξ©βor even 32 Ξ©.
The source measured 240 V. The load saw only 228 V because the feeder resistance was treated as one-way.
The load was only 30 kW, but the feeder current was much higher than the kW-only calculation predicted.
The gain equation requests -10 V, but the amplifier only has Β±5 V rails.
An NPN switch has VCC = 5 V, Rc = 100 Ξ©, Ξ² = 100, and Ib = 1 mA. Ξ²Ib predicts 100 mA, but with VCE(sat) β 0.2 V the collector circ
The reverse-polarity protection worked, but the 5 V sensor would not start.
A perfect step arrived. The filter answered 2.0 instead of 10.
A 'passband' tone reached the cutoff and lost nearly 30% of its amplitude.
The machine says 900 Hz. The sampled spectrum insists on 300 Hz.
The numerator says 20. The unit-step output settles at 5. The model is working exactly as written.
Both reactive components looked like opposition. At resonance, they canceled and the current doubled the protection limit.
The steady-state current is zero. At switch-on, the circuit violates its 5 mA startup limit.
The divider measured exactly 6 V with no load. Connect one load, and the output falls to 4 V.
Each branch looked safe. The supply still tripped the instant the second load was connected.
A 12 V source is connected across a required 2 Ξ© load. The calculated and measured current is 6 A, but the resistor overheats and fails within seconds. Identify the missing engineering check, prove the root cause, and choose a safe correction.
Choose a topic, solve the problem before seeing the conclusion, and use the final reasoning trail to identify where your original assumption was supported or needed revision. The goal is not only to reach the answer, but to understand the engineering check behind it.