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PE Electrical Electronics practice questions

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Question 1Electrical and Computer: Electrical and Electronics

A 480 V, three-phase induction motor draws 60 A at a power factor of 0.85 lagging. What is the real power delivered to the motor?

  1. 49.9 kW
  2. 24.5 kW
  3. 42.4 kW Correct answer.
  4. 73.4 kW

Explanation

Real power for a balanced three-phase load is P = √3 · V_L · I_L · cos θ. Substituting gives √3 × 480 × 60 × 0.85 ≈ 42.4 kW. Multiplying by the power factor is essential; the apparent power alone would be about 49.9 kVA.

Question 2Electrical and Computer: Electrical and Electronics

In a balanced three-phase wye-connected system, the relationship between line voltage and phase voltage is:

  1. V_line = √3 · V_phase Correct answer.
  2. V_line = 3 · V_phase
  3. V_line = V_phase
  4. V_line = V_phase / √3

Explanation

For a wye connection, the line voltage is the phasor difference of two phase voltages displaced by 120°, giving a magnitude of √3 times the phase voltage. The line current equals the phase current in a wye connection. Delta connections reverse these relationships.

Question 3Electrical and Computer: Electrical and Electronics

An ideal transformer has 200 primary turns and 50 secondary turns. If the primary is connected to a 240 V source, what is the secondary voltage?

  1. 960 V
  2. 30 V
  3. 120 V
  4. 60 V Correct answer.

Explanation

The voltage ratio of an ideal transformer equals the turns ratio: V_s / V_p = N_s / N_p. Here 50/200 = 0.25, so the secondary voltage is 0.25 × 240 = 60 V. This is a step-down transformer.

Question 4Electrical and Computer: Electrical and Electronics

The primary purpose of a commutator in a DC machine is to:

  1. provide mechanical support to the field winding
  2. reduce core eddy currents
  3. convert the alternating armature EMF into direct current at the terminals Correct answer.
  4. regulate the field current automatically

Explanation

The armature coils of a DC machine inherently generate alternating EMF as they rotate through the field. The commutator and brushes act as a mechanical rectifier, delivering unidirectional current to the external circuit. In a motor, the same action reverses armature current to maintain torque direction.

Question 5Electrical and Computer: Electrical and Electronics

In a linear circuit containing several independent sources, the superposition theorem states that the response in any branch equals:

  1. the average of the responses caused by each source
  2. the response caused by the largest source only
  3. the sum of the responses caused by each source acting alone with others deactivated Correct answer.
  4. the product of the responses caused by each source

Explanation

Superposition applies to linear circuits: each independent source is considered alone while the others are set to zero (voltage sources shorted, current sources opened). The individual branch responses are then added algebraically. The theorem does not apply to power, which is a nonlinear quantity.

Question 6Electrical and Computer: Electrical and Electronics

A silicon pn-junction diode is forward biased. Which condition best describes the steady-state behavior?

  1. The diode behaves as an open circuit until breakdown
  2. Current is constant regardless of forward voltage
  3. The depletion region narrows and current rises exponentially with forward voltage Correct answer.
  4. The depletion region widens and current is negligible

Explanation

Forward bias reduces the potential barrier, narrowing the depletion region and allowing diffusion current to flow. The current follows an exponential relationship with forward voltage above the knee (about 0.7 V for silicon). Reverse bias produces the opposite effect.

Question 7Electrical and Computer: Electrical and Electronics

An ideal operational amplifier used in a voltage-follower configuration has which key property?

  1. Very low input impedance and high voltage gain
  2. Infinite output impedance and zero bandwidth
  3. Unity current gain and inverted output
  4. Very high input impedance and unity voltage gain Correct answer.

Explanation

In a voltage follower the output is fed back directly to the inverting input, forcing the gain to unity with no phase inversion. The input signal drives the non-inverting terminal, so the ideal op-amp presents very high input impedance. This makes it useful as a buffer between stages.

Question 8Electrical and Computer: Electrical and Electronics

The Boolean expression A + A·B simplifies to which result?

  1. B
  2. A + B
  3. A·B
  4. A Correct answer.

Explanation

The absorption law states that A + A·B = A. Factoring gives A(1 + B), and since 1 + B = 1, the expression reduces to A. The term A·B is redundant because whenever A is true the whole expression is already true.

Question 9Electrical and Computer: Electrical and Electronics

A 10 kΩ resistor carries 2 mA. What is the voltage across it?

  1. 0.2 V
  2. 20 V Correct answer.
  3. 5 V
  4. 200 V

Explanation

Ohm's law gives V = I·R = 0.002 A × 10,000 Ω = 20 V. Converting milliamperes to amperes before multiplying avoids a factor-of-1000 error. The resulting power dissipation would be I²R = 40 mW.

Question 10Electrical and Computer: Electrical and Electronics

A series RLC circuit is at resonance. Which statement about its impedance is correct?

  1. Impedance is purely reactive and at its maximum magnitude
  2. Impedance equals the difference between X_L and X_C
  3. Impedance is zero for all frequencies
  4. Impedance is purely resistive and at its minimum magnitude Correct answer.

Explanation

At series resonance the inductive and capacitive reactances are equal in magnitude and cancel, so the net reactance is zero. The impedance reduces to the resistance R, which is the minimum possible magnitude. This is why current peaks at resonance in a series circuit.

Question 11Electrical and Computer: Electrical and Electronics

A 60 Hz sinusoidal voltage has an RMS value of 120 V. What is its peak value?

  1. 170 V Correct answer.
  2. 85 V
  3. 120 V
  4. 240 V

Explanation

For a sinusoid, V_peak = √2 · V_rms = 1.414 × 120 ≈ 170 V. The RMS value is the equivalent heating value, not the peak. Frequency does not affect this amplitude relationship.

Question 12Electrical and Computer: Electrical and Electronics

A separately excited DC motor is running at steady load. If the field current is reduced while the load torque is held constant, what happens?

  1. Speed increases because flux decreases Correct answer.
  2. Speed decreases because flux decreases
  3. The motor stalls immediately
  4. Speed is unchanged because torque is constant

Explanation

Motor torque is proportional to flux times armature current, and back EMF is proportional to flux times speed. With less flux, a higher speed is needed to generate the back EMF that balances the supply, so speed rises. This field-weakening effect is used for speed control above base speed.

Question 13Electrical and Computer: Electrical and Electronics

A single-phase full-wave bridge rectifier is fed from a 120 V RMS AC source and drives a resistive load. Ignoring diode drops, what is the average DC output voltage?

  1. 120 V
  2. 108 V Correct answer.
  3. 170 V
  4. 54 V

Explanation

A full-wave rectifier produces V_dc = 2·V_peak/π. The peak is 120 × √2 ≈ 170 V, so the average is 2 × 170/π ≈ 108 V. A half-wave rectifier would give half that value.

Question 14Electrical and Computer: Electrical and Electronics

A bipolar junction transistor is biased in the active region with β = 100 and base current of 20 µA. What is the collector current?

  1. 2 mA Correct answer.
  2. 0.2 mA
  3. 20 mA
  4. 5 mA

Explanation

In the active region, I_C = β·I_B = 100 × 20 µA = 2000 µA = 2 mA. The emitter current would then be about 2.02 mA. This linear relationship holds as long as the device remains out of saturation.

Question 15Electrical and Computer: Electrical and Electronics

In a CMOS inverter, both transistors are momentarily on during input transitions. The main consequence is:

  1. a permanent DC path that prevents operation
  2. a large increase in input impedance
  3. a short-circuit current that increases dynamic power dissipation Correct answer.
  4. zero power dissipation during switching

Explanation

During the transition through the threshold region, both the p-channel and n-channel devices conduct, creating a brief path from supply to ground. This short-circuit current adds to the dynamic power consumed by charging and discharging load capacitance. It scales with switching frequency and transition time.

Question 16Electrical and Computer: Electrical and Electronics

A 4-bit successive-approximation ADC has a 5 V full-scale reference. What is the nominal resolution (LSB size)?

  1. 0.156 V
  2. 1.25 V
  3. 0.625 V
  4. 0.3125 V Correct answer.

Explanation

Resolution equals full-scale voltage divided by 2^n, where n is the number of bits. For 4 bits, 5 V / 16 = 0.3125 V per LSB. Increasing the bit count halves the LSB size for each added bit.

Question 17Electrical and Computer: Electrical and Electronics

A 50 Hz, 4-pole induction motor operates with 3% slip. What is the rotor speed?

  1. 1410 rpm
  2. 1455 rpm Correct answer.
  3. 1500 rpm
  4. 750 rpm

Explanation

Synchronous speed is n_s = 120f/p = 120 × 50/4 = 1500 rpm. Rotor speed is n_s(1 − s) = 1500 × 0.97 = 1455 rpm. The slip is necessary for torque production, so the rotor always runs below synchronous speed in motoring.

Question 18Electrical and Computer: Electrical and Electronics

The main advantage of a buck-boost converter over a linear regulator in a high-current application is:

  1. simpler control circuitry with no feedback
  2. the ability to operate without an inductor
  3. higher efficiency because the switching device operates in saturation or cutoff Correct answer.
  4. lower output ripple by design

Explanation

A switching converter dissipates little power because the transistor is either fully on or fully off, unlike a linear regulator that drops excess voltage across a pass element. This yields higher efficiency, especially with large input-output differentials. The tradeoff is added switching ripple and electromagnetic interference.

Question 19Electrical and Computer: Electrical and Electronics

A 100 Ω resistor and a 100 mH inductor are connected in series across a 120 V, 60 Hz source. What is the magnitude of the total impedance?

  1. 100 Ω
  2. 118 Ω Correct answer.
  3. 37.7 Ω
  4. 220 Ω

Explanation

Inductive reactance is X_L = 2πfL = 2π × 60 × 0.1 ≈ 37.7 Ω. The impedance magnitude is √(R² + X_L²) = √(100² + 37.7²) ≈ 118 Ω. Impedances add as phasors, not as simple sums.

Question 20Electrical and Computer: Electrical and Electronics

A Zener diode regulator is designed to maintain a constant output. For proper regulation, the Zener must operate:

  1. at exactly its rated power continuously
  2. in reverse breakdown with current between the knee and maximum rated value Correct answer.
  3. below its breakdown voltage with zero current
  4. in forward conduction with current above the knee

Explanation

Regulation occurs in the reverse-breakdown region, where small voltage changes produce large current changes. The series resistor must limit current so it stays above the knee for minimum load and below the maximum rating at full load. Operating outside this window degrades or destroys regulation.

Question 21Electrical and Computer: Electrical and Electronics

A three-phase 60 Hz transmission line has per-phase inductive reactance of 0.5 Ω/mile and carries 200 A. What is the per-phase reactive voltage drop?

  1. 50 V
  2. 400 V
  3. 200 V
  4. 100 V Correct answer.

Explanation

The reactive drop per phase is I·X_L = 200 A × 0.5 Ω = 100 V. This drop is 90° out of phase with the resistive drop and adds as a phasor. Line reactance is a major determinant of voltage regulation on long lines.

Question 22Electrical and Computer: Electrical and Electronics

A control system has an open-loop transfer function with a pole at the origin. What does this imply about the steady-state error for a step input?

  1. A constant nonzero steady-state error
  2. The system is unstable for all gains
  3. Zero steady-state error for a step input Correct answer.
  4. Infinite steady-state error for a step input

Explanation

A pole at the origin makes the system Type 1, giving infinite DC gain in the open loop. For a unity-feedback closed loop, this yields zero steady-state error to a step input. The error to a ramp input would instead be finite and proportional to 1/K_v.

Question 23Electrical and Computer: Electrical and Electronics

A 3-phase, 480 V, 100 kVA generator supplies a load at 0.8 power factor lagging. What is the reactive power delivered?

  1. 60 kvar Correct answer.
  2. 100 kvar
  3. 80 kvar
  4. 20 kvar

Explanation

Apparent power S is 100 kVA and real power P = S·cos θ = 80 kW. Reactive power is Q = S·sin θ, and with cos θ = 0.8, sin θ = 0.6, giving Q = 60 kvar. The power triangle satisfies S² = P² + Q².

Question 24Electrical and Computer: Electrical and Electronics

A second-order system has a damping ratio of 0.5 and a natural frequency of 10 rad/s. What is the approximate peak overshoot to a unit step?

  1. 0%
  2. 16% Correct answer.
  3. 5%
  4. 50%

Explanation

Peak overshoot is given by exp(−πζ/√(1−ζ²)). With ζ = 0.5, the exponent is −π(0.5)/0.866 ≈ −1.81, giving about 0.163, or roughly 16%. Lower damping ratios produce larger overshoot.

Question 25Electrical and Computer: Electrical and Electronics

In a digital system, a setup time violation on a flip-flop most directly indicates that:

  1. the hold time margin is excessive
  2. the flip-flop output is stuck at a constant value
  3. the clock frequency is too low for the logic
  4. data arrived too late relative to the clock edge to be captured reliably Correct answer.

Explanation

Setup time is the minimum interval data must be stable before the active clock edge. A violation means the data path delay plus clock skew leaves insufficient margin, risking metastability. Hold violations, by contrast, concern data changing too soon after the edge.

Question 26Electrical and Computer: Electrical and Electronics

A single-phase transformer has 500 primary turns and 100 secondary turns. If the primary is connected to a 2400 V RMS source, what is the secondary voltage?

  1. 480 V Correct answer.
  2. 12000 V
  3. 240 V
  4. 48 V

Explanation

The turns ratio is N1/N2 = 500/100 = 5, and for an ideal transformer the voltage scales by the same ratio, so V2 = 2400/5 = 480 V. The secondary voltage is lower because the secondary has fewer turns than the primary.

Question 27Electrical and Computer: Electrical and Electronics

The superposition theorem is applied to a linear circuit containing two independent sources. Which procedure is correct?

  1. Compute the response to each source with the others deactivated, then add the individual responses algebraically Correct answer.
  2. Compute the response with both sources at full value and divide by two
  3. Compute the response to each source with the others deactivated, then multiply the two responses
  4. Compute the response using only the larger source and ignore the smaller one

Explanation

Superposition states that in a linear network the total response equals the algebraic sum of the responses caused by each independent source acting alone. Voltage sources not under consideration are replaced by short circuits and current sources by open circuits. The method fails for power, which is nonlinear in voltage and current.

Question 28Electrical and Computer: Electrical and Electronics

A 12 V battery is connected across three resistors of 6 Ω, 3 Ω, and 2 Ω in parallel. What is the total current drawn from the battery?

  1. 6 A
  2. 11 A Correct answer.
  3. 22 A
  4. 2 A

Explanation

Each parallel branch sees the full 12 V, so the branch currents are 2 A, 4 A, and 6 A. Kirchhoff's current law gives a total of 12 A? No: 2 + 4 + 6 = 12 A.

Question 29Electrical and Computer: Electrical and Electronics

A balanced three-phase wye-connected load has a per-phase impedance of 10 Ω and is supplied from a 208 V line-to-line source. What is the line current magnitude?

  1. 6 A
  2. 20.8 A
  3. 36 A
  4. 12 A Correct answer.

Explanation

In a wye connection the phase voltage is the line voltage divided by √3, giving 208/√3 ≈ 120 V. The phase current is 120/10 = 12 A, and in a wye connection the line current equals the phase current.

Question 30Electrical and Computer: Electrical and Electronics

Which quantity is held constant by an ideal current source, regardless of the voltage across its terminals?

  1. Its output voltage
  2. Its internal resistance
  3. Its output current Correct answer.
  4. Its delivered power

Explanation

An ideal current source fixes the current through it at a specified value independent of terminal voltage. The voltage across it is set by the rest of the circuit, and the power it delivers varies with that voltage.

Question 31Electrical and Computer: Electrical and Electronics

A 555 timer is configured as an astable multivibrator. Which combination of external components primarily sets the oscillation frequency?

  1. Two resistors and one capacitor Correct answer.
  2. One resistor and one diode
  3. Two capacitors and one resistor
  4. One inductor and one capacitor

Explanation

In the astable configuration the timing is set by an RC network of two resistors (RA and RB) and a timing capacitor C, with frequency approximately 1.44/((RA + 2RB)C). The capacitor charges through both resistors and discharges through RB only.

Question 32Electrical and Computer: Electrical and Electronics

A logic gate produces a HIGH output only when all of its inputs are HIGH. Which gate is it?

  1. NAND
  2. XOR
  3. AND Correct answer.
  4. OR

Explanation

The AND function outputs 1 only for the all-ones input combination, so a single LOW input forces the output LOW. The OR gate outputs HIGH when any input is HIGH, and NAND is its complement.

Question 33Electrical and Computer: Electrical and Electronics

A conductor carries 5 A and is placed in a uniform magnetic field of 0.4 T, oriented perpendicular to the field. If the active length in the field is 0.5 m, what is the magnitude of the force on the conductor?

  1. 1.0 N Correct answer.
  2. 4.0 N
  3. 0.1 N
  4. 10 N

Explanation

The force on a current-carrying conductor is F = BIL sin θ, and with perpendicular orientation sin θ = 1. Substituting gives 0.4 × 5 × 0.5 = 1.0 N.

Question 34Electrical and Computer: Electrical and Electronics

A DC shunt generator is building voltage with no load. Loss of residual magnetism in the field poles most directly prevents which of the following?

  1. Voltage build-up during self-excitation Correct answer.
  2. Cooling of the field winding
  3. Commutation of the armature current
  4. Mechanical rotation of the armature

Explanation

Self-excited shunt generators depend on residual flux to induce a small initial armature voltage that drives current into the field and reinforces the flux. Without residual magnetism the build-up process cannot start, and the machine must be separately excited or flashed to restore it.

Question 35Electrical and Computer: Electrical and Electronics

A 480 V, 60 Hz, three-phase induction motor has a synchronous speed of 1800 rpm and runs at 1746 rpm at full load. What is the slip?

  1. 5.4%
  2. 2.7%
  3. 54%
  4. 3.0% Correct answer.

Explanation

Slip is (Ns − N)/Ns, so (1800 − 1746)/1800 = 54/1800 = 0.03, or 3.0%. Slip represents the fractional difference between the rotating field speed and the rotor speed.

Question 36Electrical and Computer: Electrical and Electronics

A boost converter operates in continuous conduction mode with a duty cycle D. Which expression gives the ideal conversion ratio Vout/Vin?

  1. D
  2. 1/(1 − D) Correct answer.
  3. D/(1 − D)
  4. 1 − D

Explanation

In a boost converter the inductor charges during the on-time and discharges into the output during the off-time, giving Vout/Vin = 1/(1 − D). Because the ratio exceeds unity for 0 < D < 1, the output is always higher than the input.

Question 37Electrical and Computer: Electrical and Electronics

A common-emitter amplifier has a voltage gain of −50 and an input signal of 10 mV peak. The supply is ±15 V. What is the ideal output peak amplitude?

  1. 500 mV Correct answer.
  2. 1.5 V
  3. 50 mV
  4. 200 mV

Explanation

The gain magnitude times input amplitude gives 50 × 10 mV = 500 mV peak. The negative sign only indicates phase inversion of the output relative to the input. The result is well within the ±15 V supply, so no clipping occurs.

Question 38Electrical and Computer: Electrical and Electronics

In a microprocessor system, a memory-mapped I/O scheme differs from isolated (port-mapped) I/O mainly in which way?

  1. I/O registers are placed in a separate address space requiring special instructions
  2. I/O registers share the same address space and are accessed with ordinary memory instructions Correct answer.
  3. I/O registers are accessed only through dedicated interrupt vectors
  4. I/O devices must use DMA for every transfer

Explanation

Memory-mapped I/O assigns device registers addresses within the processor's main memory space, so normal load and store instructions address them. Isolated I/O instead uses a separate address space with dedicated instructions and control signals.

Question 39Electrical and Computer: Electrical and Electronics

A first-order low-pass RC filter has R = 1 kΩ and C = 0.1 µF. What is its −3 dB cutoff frequency?

  1. Approximately 159 Hz
  2. Approximately 1.6 kHz Correct answer.
  3. Approximately 16 kHz
  4. Approximately 10 kHz

Explanation

The cutoff frequency is fc = 1/(2πRC). With RC = 1000 × 0.1×10⁻⁶ = 1×10⁻⁴ s, fc = 1/(2π × 10⁻⁴) ≈ 1592 Hz, about 1.6 kHz.

Question 40Electrical and Computer: Electrical and Electronics

A signal is sampled at 8 kHz. To avoid aliasing, the analog anti-alias filter passband must be limited to below which frequency?

  1. 4 kHz Correct answer.
  2. 16 kHz
  3. 8 kHz
  4. 2 kHz

Explanation

The Nyquist criterion requires the sampling rate to exceed twice the highest signal frequency, so content above half the sampling rate (4 kHz) must be removed before sampling. Without this filtering, higher-frequency components fold back into the baseband as aliases.

Question 41Electrical and Computer: Electrical and Electronics

A 3-phase, 60 Hz, 4-pole synchronous generator is driven at 1800 rpm. What is the frequency of the generated voltage?

  1. 30 Hz
  2. 90 Hz
  3. 60 Hz Correct answer.
  4. 120 Hz

Explanation

Synchronous machine frequency is f = (P × N)/120, so (4 × 1800)/120 = 60 Hz. This confirms the machine runs at synchronous speed for a 4-pole, 60 Hz design.

Question 42Electrical and Computer: Electrical and Electronics

An ideal transformer has a primary-to-secondary turns ratio of 10:1 and delivers 100 W to a resistive secondary load. Neglecting losses, what power is drawn by the primary?

  1. 1000 W
  2. 10 W
  3. 1 W
  4. 100 W Correct answer.

Explanation

An ideal transformer transfers power without loss, so input power equals output power regardless of the turns ratio. The turns ratio changes voltage and current inversely but leaves the power product unchanged.

Question 43Electrical and Computer: Electrical and Electronics

In a control system, the root-locus plot shows the closed-loop poles as a parameter varies. Which statement describes a branch of the root locus as gain increases from zero?

  1. It always remains on the imaginary axis
  2. It starts at an open-loop zero and ends at an open-loop pole
  3. It starts at an open-loop pole and ends at an open-loop zero or infinity Correct answer.
  4. It starts at the closed-loop pole and terminates at the origin

Explanation

Root-locus branches originate at the open-loop poles when the gain is zero and terminate at open-loop zeros or at infinity as gain grows. The number of branches equals the number of open-loop poles.

Question 44Electrical and Computer: Electrical and Electronics

A two-input XOR gate has inputs A = 1 and B = 1. What is the output?

  1. 1
  2. 0 Correct answer.
  3. B
  4. A

Explanation

The exclusive-OR output is HIGH only when the inputs differ, so identical HIGH inputs give a LOW output. This is the defining distinction between XOR and the inclusive OR gate.

Question 45Electrical and Computer: Electrical and Electronics

A DC series motor is connected to a constant voltage supply with a light mechanical load. Which hazard arises if the load is removed entirely?

  1. The field winding demagnetizes permanently
  2. The armature reverses rotation direction
  3. The motor speed can rise to a destructive value because flux falls with current Correct answer.
  4. The motor stalls and draws excessive locked-rotor current

Explanation

In a series motor the field flux is produced by the armature current, so light load means low current, weak flux, and high speed for a given back EMF. With little load the speed can run away to a dangerous level, which is why series motors are never belt-coupled or run unloaded.

Question 46Electrical and Computer: Electrical and Electronics

A single-phase transformer delivers 10 kVA at 0.8 power factor lagging. What is the real power delivered?

  1. 6 kvar
  2. 10 kW
  3. 12.5 kW
  4. 8 kW Correct answer.

Explanation

Real power equals apparent power times power factor, so 10 kVA × 0.8 = 8 kW. The remaining reactive component is 6 kvar, which does no useful work but loads the source.

Question 47Electrical and Computer: Electrical and Electronics

A Zener diode shunt regulator uses a series resistor from an unregulated supply. The load current increases while the input voltage stays fixed. What happens to the Zener current?

  1. It increases, because the Zener must absorb the extra load current
  2. It remains exactly constant regardless of load
  3. It decreases, because the series resistor carries more current to the load Correct answer.
  4. It drops to zero immediately when the load is connected

Explanation

The series resistor current is fixed by the voltage drop across it, so as load current rises the Zener must conduct less to satisfy Kirchhoff's current law. Regulation holds only while the Zener remains in reverse breakdown with some minimum current. If load current exceeds the available resistor current, regulation is lost.

Question 48Electrical and Computer: Electrical and Electronics

A linear time-invariant system has a transfer function with a right-half-plane zero. What is the practical control implication?

  1. The steady-state error to a step input is always infinite
  2. The phase margin is always greater than 90 degrees
  3. The step response initially moves in the opposite direction to its final value Correct answer.
  4. The system is unstable and cannot be compensated

Explanation

A right-half-plane zero produces inverse response, an initial undershoot before the output heads toward its final value. It also adds phase lag like a pole, limiting achievable bandwidth, but it does not by itself make the system unstable.

Question 49Electrical and Computer: Electrical and Electronics

A CMOS gate drives a capacitive load. Dynamic power dissipation is best reduced by which change, all else equal?

  1. Lowering the supply voltage Correct answer.
  2. Increasing the load capacitance
  3. Raising the logic swing
  4. Increasing the clock frequency

Explanation

Dynamic power is approximately P = C·V²·f, so supply voltage has a squared influence and is the most effective lever. Reducing voltage lowers energy per transition substantially, although it also reduces noise margin and speed.

Question 50Electrical and Computer: Electrical and Electronics

A negative-feedback amplifier has an open-loop gain of 10000 and a feedback factor of 0.01. What is the approximate closed-loop gain?

  1. Approximately 0.01
  2. Approximately 100 Correct answer.
  3. Approximately 10000
  4. Approximately 50

Explanation

Closed-loop gain is A/(1 + Aβ), so 10000/(1 + 100) ≈ 99, about 100. The large loop gain makes the result depend almost entirely on the feedback network, namely 1/β.