FE Electrical and Computer Practice Problems: 19 Free Questions with Solutions

Phasor addition requires rectangular components, since the two branch currents differ in phase by 65 degrees.

Component-wise addition leaves a resultant almost aligned with the real axis, the imaginary parts nearly cancelling.

Magnitude of the resultant follows from its rectangular components.

Supplier share and supplier-specific defect rate combine into the unconditional defect probability that serves as the denominator of the reverse conditional.

Bayes' theorem rescales the single branch belonging to supplier X against that total.

Concurrent-use terms cap simultaneous users at the number of seats held, so the entitlement must cover the logged peak; sizing to the average of 8 leaves the firm in breach on every occasion demand climbs above the seat count.

The seats already licensed remain in force, so only the deficit between the required count and the existing entitlement is purchased.

The per-seat charge is a recurring subscription rate, which makes the compliance cost an ongoing yearly commitment rather than a one-time purchase.

Costs arriving on different schedules become comparable only when the first cost, the annual series, and the end-of-life salvage are all referred to time zero over the same 15-year horizon.

Unit A carries the heavier purchase price but recovers part of it at retirement while committing to the smaller maintenance stream.

Unit B earns no salvage credit, so its cost stream is purchase price plus the discounted maintenance alone.

Fifteen years of the USD 2,300 per year maintenance penalty on Unit B swamp its USD 15,000 lower first cost, leaving Unit A ahead by about USD 7,200.

Compensated material carries a majority density equal to the net dopant excess, since each acceptor consumes one donor electron.

Mass action then fixes the minority population, which in heavily doped silicon sits many orders below the intrinsic level.

Reduction starts at the far end, where the 12 Ω and 18 Ω resistors carry a common current and combine before the parallel step.

With 30 Ω of total loop resistance, the source drives 4 A, and the 10 Ω resistor absorbs part of the supply voltage before the parallel section.

Splitting that section voltage between the two series resistors of the composite branch leaves the larger share on the 18 Ω element.

Matching the denominator against the standard second-order form fixes both parameters at once.

Overshoot of an underdamped step response is set by the damping ratio alone, since $\omega_n$ only rescales time.

With $\zeta = 0.3$ the exponent falls just below unity, leaving a peak more than one-third above the final value.

Causality admits only right-sided sequences, whose z-transforms converge in the exterior of a circle passing through the pole of largest magnitude.

Bounded-input bounded-output stability additionally demands that this exterior region contain the unit circle, which is impossible with a pole at radius 1.25.

Series-resistor current is set entirely by the 18 V source and the clamped 9.1 V output, regardless of how that current later divides between diode and load.

Load current is whatever the Zener does not take, so the load is maximized at the instant the diode carries only its minimum breakdown current.

Hydraulic power available at the runner is the weight flow rate carried through the net head.

Converting at $550\ \text{ft}\cdot\text{lbf/s}$ per horsepower and charging the single combined efficiency against the shaft leaves the station output.

Charge on the inner conductor produces a purely radial field, and integrating that field from inner to outer radius yields the logarithmic potential difference behind the coaxial capacitance expression.

Only the ratio of the radii enters, so the millimetre units require no conversion.

Polyethylene multiplies the air-filled value by its relative permittivity, and the 30 m length scales it linearly.

Integral action inserts a pole at the origin, making the loop type 1, for which a ramp command produces a finite error set by the velocity error constant.

Only the integral gain survives the limit, so the proportional term contributes nothing to $K_v$.

Error scales with the commanded rate, so the ramp slope must be carried through rather than assumed unity.

Total power in a tone-modulated full-carrier signal divides between an unmodulated carrier and two sidebands, so the carrier level is recovered from the total once the index is known.

Each sideband carries one quarter of the carrier power scaled by the square of the index, the pair together making up the modulation-dependent remainder.

Sideband power is thus only 12 percent of the transmitted total even at 80 percent modulation, the well-known power penalty of retaining the carrier.

Two addresses in every IPv4 subnet are unusable, the all-zeros network identifier and the all-ones broadcast address, so the host field must exceed the required host count by that margin.

Eight host bits fall short at 254 usable addresses, so nine bits is the smallest qualifying field and it fixes the subnet prefix.

The /19 block supplies 13 host bits, four of which become the subnet field, so sixteen equal /23 subnets fit and each carries the same capacity.

Both words carry a clear sign bit, so each represents a positive quantity, and their true sum climbs past the $+127$ ceiling that eight-bit two's complement can hold.

Column-by-column addition propagates a carry into the sign column but produces no carry out of it, the defining signature of two's complement overflow, so the hardware discards nothing yet stores a pattern that no longer matches the arithmetic sum.

Read back under the same convention, the leading one carries weight $-128$ while the remaining set bits contribute 19, so the register reports a negative quantity even though both inputs were positive.

Multilevel caches compose serially: every reference pays the L1 hit time, and only the fraction that misses reaches the next level.

Quoted for the L2 alone, the 25% miss rate is a local rate and therefore weights only the traffic that reaches L2.

Weighting that 29 ns penalty by the 6% of references that miss in L1 gives the latency averaged over the whole reference stream.

Runtime for a comparison sort is proportional to $n \log_2 n$, so the unknown machine constant cancels when two problem sizes are compared as a ratio.

The record count rises by a factor of eight while the logarithmic factor rises only from $\log_2(100{,}000) = 16.610$ to $\log_2(800{,}000) = 19.610$.

Scaling the measured 0.42 s by that factor gives the projected batch time.