Free PE ECC Practice Questions

Step 1: Apply Ohm's Law

I = V / R

Step 2: Substitute values

I = 12 V / 4 Ω = 3 A

Reference: PE Reference Handbook — Circuit Analysis, Ohm's Law.

Step 1: Recall the definition

L{u(t)} = ∫₀^∞ u(t) × e^(−st) dt = ∫₀^∞ e^(−st) dt

Step 2: Evaluate the integral

= [−1/s × e^(−st)]₀^∞ = 0 − (−1/s) = 1/s (for s > 0)

This is one of the most fundamental Laplace transform pairs and should be memorized.

Reference: PE Reference Handbook — Laplace Transform Table.

Step 1: Recall wye (Y) connection relationships

In a balanced wye system:

• V_L = √3 × V_ph (line voltage is √3 times phase voltage)

• I_L = I_ph (line current equals phase current)

Step 2: Why √3?

Line voltage is the phasor difference of two phase voltages that are 120° apart. The magnitude of this difference is √3 × V_ph.

Why not the others? V_L = V_ph applies to delta connections. V_L = 3 × V_ph and V_L = V_ph/√3 are not valid for any standard configuration.

Step 1: Recall the Bode magnitude plot rules

Each pole contributes −20 dB/decade of roll-off beyond its break frequency.

A first-order low-pass filter has one pole, so: −20 dB/decade (or equivalently −6 dB/octave).

Why not the others?

• −40 dB/decade → second-order system (two poles)

• −6 dB/decade → this is per octave, not per decade (−6 dB/octave = −20 dB/decade)

• −10 dB/decade → not a standard roll-off rate for any integer-order system

Step 1: Define the modulation index

For AM: m = A_message / A_carrier

The modulated signal is: s(t) = A_c[1 + m × cos(ω_m t)] × cos(ω_c t)

Step 2: Interpret m = 0.5

The carrier amplitude varies between A_c(1 + 0.5) and A_c(1 − 0.5), i.e., the envelope varies by ±50% of the unmodulated carrier amplitude.

Why not the others? m relates to amplitude variation, not frequency ratio (B), bandwidth (C), or power ratio (D).

Step 1: Identify the power formula

P = I² × R (power dissipated in a resistor)

Step 2: Substitute values

P = (3 A)² × 10 Ω = 9 × 10 = 90 W

Alternative check: V = IR = 30 V, then P = V²/R = 900/10 = 90 W ✓

Step 1: Recall Thevenin's theorem

R_th = V_oc / I_sc (Thevenin resistance = open-circuit voltage ÷ short-circuit current)

Step 2: Substitute values

R_th = 20 V / 5 A = 4 Ω

The Thevenin equivalent is a 20 V source in series with a 4 Ω resistor.

Reference: PE Reference Handbook — Circuit Analysis, Thevenin & Norton Equivalents.

Step 1: Recall the Nyquist-Shannon sampling theorem

To avoid aliasing, the sampling rate must satisfy: f_s ≥ 2 × f_max

Step 2: Calculate the minimum sampling rate

f_s,min = 2 × 4 kHz = 8 kHz

Sampling below 8 kHz causes frequency components above f_s/2 to fold back (alias) into the baseband, corrupting the signal.

Reference: PE Reference Handbook — Signal Processing, Sampling Theorem.

Step 1: Recall ideal transformer relationships

V_p/V_s = N_p/N_s and I_p/I_s = N_s/N_p (current is inversely proportional to turns ratio)

Step 2: Calculate primary current

Turns ratio N_p/N_s = 10/1. Therefore I_p/I_s = 1/10.

I_p = I_s × (N_s/N_p) = 100 A × (1/10) = 10 A

Verification: Power balance: V_p × I_p = 4,800 × 10 = 48,000 W. V_s = 4,800/10 = 480 V. V_s × I_s = 480 × 100 = 48,000 W ✓

Step 1: Recall root locus construction rules

Rule 1: The number of branches equals the number of open-loop poles.

Rule 2: Branches start at open-loop poles (K = 0) and end at open-loop zeros or infinity (K → ∞).

Step 2: Answer

The number of branches = number of open-loop poles = order of the characteristic equation.

Why not the others?

• B (zeros) — Zeros are endpoints, not branch count determinants

• C (numerator order) — The numerator gives the zeros, not the branch count

• D (gain margin) — Gain margin is a stability metric, unrelated to branch count

Step 1: Convert SNR from dB to linear

SNR (dB) = 10 × log₁₀(SNR_linear) → 30 = 10 × log₁₀(SNR) → SNR = 10³ = 1,000

Step 2: Apply the Shannon-Hartley theorem

C = B × log₂(1 + SNR)

C = 3,000 Hz × log₂(1 + 1,000) = 3,000 × log₂(1,001)

Step 3: Calculate

log₂(1,001) = ln(1,001) / ln(2) ≈ 6.909 / 0.693 ≈ 9.97

C = 3,000 × 9.97 ≈ 29,900 bps ≈ 30 kbps

Reference: PE Reference Handbook — Communications, Shannon Capacity.

At series RLC resonance (ω₀ = 1/√(LC)):

• A ✅ — X_L = X_C, so they cancel and Z = R (purely resistive)

• B ✅ — Since Z is minimized (Z = R), current I = V/R is at its maximum

• C ✅ — V_L = IX_L and V_C = IX_C; since X_L = X_C, the voltages are equal (though 180° out of phase)

• D ❌ — Power factor = cos(0°) = 1, not zero. PF = 0 would mean purely reactive.

• E ✅ — This is the defining condition of resonance: X_L = X_C (reactances cancel)

Correct answers: A, B, C, E.

Step 1: Apply the z-transform definition

X(z) = Σ_{n=0}^{∞} x[n] z⁻ⁿ = Σ_{n=0}^{∞} aⁿ z⁻ⁿ = Σ_{n=0}^{∞} (a/z)ⁿ

Step 2: Recognize as a geometric series

For |a/z| < 1 (i.e., |z| > |a|): Σ (a/z)ⁿ = 1 / (1 − a/z)

Step 3: Simplify

1 / (1 − a/z) = z / (z − a)

Therefore: X(z) = z / (z − a), valid for |z| > |a|.

This is a fundamental z-transform pair that should be memorized for the PE exam.

Step 1: Recall the per-unit base conversion formula

X_new = X_old × (MVA_new / MVA_old) × (kV_old / kV_new)²

Step 2: Substitute values

X_new = 0.20 × (100 / 50) × (13.8 / 13.8)²

X_new = 0.20 × 2.0 × 1.0 = 0.40 pu

Key insight: Since the voltage bases are the same, only the MVA ratio matters. Doubling the MVA base doubles the per-unit reactance.

Reference: PE Reference Handbook — Power Systems, Per-Unit Calculations.

Butterworth → Maximally flat magnitude response in the passband. No ripple, but slower roll-off than other types of the same order.

Chebyshev Type I → Equiripple (equal ripple) in the passband with a monotonically decreasing stopband. Sharper transition than Butterworth for the same order.

Bessel → Maximally flat group delay, providing the best phase linearity (minimal signal distortion). Slowest roll-off of all four types.

Elliptic (Cauer) → Equiripple in both passband and stopband. Achieves the sharpest transition band for a given filter order, at the cost of ripple everywhere.