PE Electrical Power Exam: 10 Practice Problems with Detailed Solutions

Answer: B

A current transformer scales the primary current down by its turns ratio. The CT ratio of 600:5 means that 600 A on the primary produces 5 A on the secondary.

Step 1 — Determine the CT turns ratio:

\[\text{Turns ratio} = \frac{600}{5} = 120\]

Step 2 — Calculate the secondary current:

\[I_{\text{sec}} = \frac{I_{\text{pri}}}{\text{Turns ratio}} = \frac{480}{120} = 4.0 \text{ A}\]

Step 3 — Calculate the voltage across the burden:

\[V_{\text{burden}} = I_{\text{sec}} \times Z_{\text{burden}} = 4.0 \times 0.5 = 2.0 \text{ V}\]

Verification — Check CT loading:

The CT rated primary current is 600 A. Our load current (480 A) is 80% of the CT rating, which is within the normal operating range. The burden power is \(I^2 Z = 4.0^2 \times 0.5 = 8\) VA, which is well within typical CT burden ratings of 15–200 VA.

Key concept: CTs are wound so that the secondary is rated at either 5 A or 1 A. On the PE exam, always verify that the primary current does not exceed the CT rating, and remember that a CT secondary must never be open-circuited—dangerously high voltages will develop. The burden voltage is critical for ensuring the CT does not saturate during fault conditions. Multi-ratio CTs allow different taps (e.g., 200:5, 400:5, 600:5) to be selected for the expected load range.

Answer: D

The lumen method (also called the zonal cavity method) relates the number of luminaires to the required illuminance:

\[N = \frac{E \times A}{\Phi \times CU \times LLF}\]

where \(E\) is the target illuminance (fc), \(A\) is the floor area (ft²), \(\Phi\) is the initial lumens per luminaire, \(CU\) is the coefficient of utilization, and \(LLF\) is the light loss factor.

Step 1 — Compute the floor area:

\[A = 40 \times 60 = 2{,}400 \text{ ft}^2\]

Step 2 — Compute the total lumens required:

\[\text{Lumens required} = E \times A = 50 \times 2{,}400 = 120{,}000 \text{ lumens}\]

Step 3 — Compute the effective lumens per luminaire:

\[\text{Effective lumens} = \Phi \times CU \times LLF = 6{,}000 \times 0.58 \times 0.72 = 2{,}505.6 \text{ lumens}\]

Step 4 — Determine the number of luminaires:

\[N = \frac{120{,}000}{2{,}505.6} = 47.9\]

Since you cannot install a fraction of a luminaire, round up to 48 luminaires.

Key concept: Always round up when calculating the number of luminaires. The light loss factor accounts for lamp lumen depreciation, luminaire dirt depreciation, and room surface dirt depreciation over time. The CU depends on the room cavity ratio, wall reflectances, and luminaire distribution type—all of which may be given or must be looked up in IES tables.

Answer: C

NFPA 70E defines four PPE categories with minimum arc ratings:

• Category 1: minimum arc rating of 4 cal/cm²
• Category 2: minimum arc rating of 8 cal/cm²
• Category 3: minimum arc rating of 25 cal/cm²
• Category 4: minimum arc rating of 40 cal/cm²

Step 1 — Compare the incident energy to the PPE category thresholds:

The calculated incident energy is 8.5 cal/cm². This exceeds the Category 2 minimum arc rating of 8 cal/cm².

Step 2 — Select the appropriate category:

Since 8.5 cal/cm² > 8 cal/cm² (Category 2 limit), the worker must wear PPE Category 3 equipment, which is rated up to 25 cal/cm².

Key concept: The PPE category must have an arc rating that meets or exceeds the calculated incident energy. If incident energy exceeds 40 cal/cm² (Category 4), the task is prohibited until the hazard is reduced through engineering controls (e.g., remote racking, faster clearing times, current-limiting fuses). IEEE 1584 provides the detailed calculation methods for incident energy, while NFPA 70E provides the PPE requirements. On the PE exam, know both the incident energy calculation procedure and the PPE category thresholds.

The three primary ways to reduce arc flash incident energy are: (1) reduce the available fault current, (2) reduce the clearing time of the protective device, and (3) increase the working distance. Zone-selective interlocking (ZSI) and arc flash relays are common methods to reduce clearing time.

Answer: C

The per-unit impedance conversion formula when changing bases is:

\[Z_{\text{pu,new}} = Z_{\text{pu,old}} \times \frac{S_{\text{base,new}}}{S_{\text{base,old}}} \times \left(\frac{V_{\text{base,old}}}{V_{\text{base,new}}}\right)^2\]

Step 1 — Identify the given values:

\(Z_{\text{pu,old}} = 0.08\) pu (on 50 MVA, 138 kV base)

\(S_{\text{base,new}} = 100\) MVA, \(S_{\text{base,old}} = 50\) MVA

\(V_{\text{base,old}} = 138\) kV, \(V_{\text{base,new}} = 138\) kV

Step 2 — Apply the conversion formula:

\[Z_{\text{pu,new}} = 0.08 \times \frac{100}{50} \times \left(\frac{138}{138}\right)^2\]

\[Z_{\text{pu,new}} = 0.08 \times 2.0 \times 1.0 = \mathbf{0.16 \text{ pu}}\]

Key concept: When the voltage bases are the same (as they often are on the same side of a transformer), the conversion simplifies to a simple ratio of MVA bases. Per-unit impedance is directly proportional to the MVA base and inversely proportional to the square of the voltage base. This conversion is fundamental to every short-circuit and load-flow study on the PE exam.

A common exam pitfall: if the transformer nameplate voltage differs from the system base voltage on the same side, you must include the voltage ratio squared in the conversion. For example, if the transformer were rated 138/14.4 kV but your system base were 13.8 kV on the low side, the voltage mismatch factor would be \((14.4/13.8)^2 = 1.089\).

Answer: B

For a balanced three-phase fault, only the positive-sequence network is involved. The fault current in per-unit is:

\[I_{\text{fault,pu}} = \frac{V}{Z_1}\]

Step 1 — Compute the per-unit fault current:

\[I_{\text{fault,pu}} = \frac{1.0}{0.25} = 4.0 \text{ pu}\]

Step 2 — Compute the base current:

\[I_{\text{base}} = \frac{S_{\text{base}}}{\sqrt{3} \times V_{\text{base}}} = \frac{100 \times 10^6}{\sqrt{3} \times 13{,}800} = \frac{100 \times 10^6}{23{,}900} = 4{,}184 \text{ A}\]

Step 3 — Convert to actual amperes:

\[I_{\text{fault}} = I_{\text{fault,pu}} \times I_{\text{base}} = 4.0 \times 4{,}184 = \mathbf{16{,}720 \text{ A}}\]

Key concept: For a three-phase fault, only the positive-sequence impedance matters because the fault is balanced. For other fault types, different sequence networks are involved:

• Single-line-to-ground (SLG): \(I_{a1} = V / (Z_1 + Z_2 + Z_0)\), then \(I_f = 3 I_{a1}\)
• Line-to-line (LL): \(I_{a1} = V / (Z_1 + Z_2)\)
• Double-line-to-ground (DLG): involves all three networks in a more complex configuration

Always compute the base current carefully—the \(\sqrt{3}\) factor accounts for the line-to-line voltage in three-phase systems. On the PE exam, the three-phase fault is typically the largest and is used for equipment ratings, while the SLG fault magnitude depends on system grounding.

Answer: B

For a single-phase full-wave bridge rectifier with ideal diodes, the average DC output voltage is:

\[V_{\text{dc}} = \frac{2 V_m}{\pi}\]

where \(V_m\) is the peak value of the AC input voltage.

Step 1 — Compute the peak voltage:

\[V_m = V_{\text{rms}} \times \sqrt{2} = 480 \times 1.414 = 678.8 \text{ V}\]

Step 2 — Compute the average DC output:

\[V_{\text{dc}} = \frac{2 \times 678.8}{\pi} = \frac{1{,}357.6}{3.1416} = \mathbf{432 \text{ V}}\]

Key concept: The ratio \(V_{\text{dc}} / V_{\text{rms}} = 2\sqrt{2}/\pi \approx 0.9003\) is a constant for an ideal single-phase full-wave rectifier. Here is a quick reference for common rectifier configurations:

• Single-phase half-wave: \(V_{\text{dc}} = V_m / \pi \approx 0.318 \, V_m\)
• Single-phase full-wave: \(V_{\text{dc}} = 2V_m / \pi \approx 0.637 \, V_m\)
• Three-phase half-wave: \(V_{\text{dc}} = 3\sqrt{3} \, V_m / (2\pi) \approx 0.827 \, V_m\)
• Three-phase full-wave: \(V_{\text{dc}} = 3\sqrt{3} \, V_m / \pi \approx 1.654 \, V_m\) (line-to-neutral peak)

On the PE exam, you may also see controlled rectifiers (SCR-based) where the firing angle \(\alpha\) modifies the output: \(V_{\text{dc}} = (2V_m/\pi)\cos\alpha\). When \(\alpha = 0\), the output equals the uncontrolled case. As \(\alpha\) increases toward 90°, the average DC output decreases to zero.

Answer: D

Slip is the relative difference between synchronous speed and actual rotor speed.

Step 1 — Compute the synchronous speed:

\[n_s = \frac{120 f}{P} = \frac{120 \times 60}{4} = 1{,}800 \text{ rpm}\]

Step 2 — Calculate the slip:

\[s = \frac{n_s - n_r}{n_s} = \frac{1{,}800 - 1{,}710}{1{,}800} = \frac{90}{1{,}800} = 0.05 = \mathbf{5.0\%}\]

Step 3 — Calculate the rotor current frequency:

\[f_r = s \times f = 0.05 \times 60 = \mathbf{3.0 \text{ Hz}}\]

Key concept: Synchronous speed depends only on the supply frequency and the number of poles: \(n_s = 120f/P\). Common synchronous speeds at 60 Hz are:

• 2-pole: 3,600 rpm
• 4-pole: 1,800 rpm
• 6-pole: 1,200 rpm
• 8-pole: 900 rpm

Typical full-load slips for induction motors range from 1% to 5%, with larger motors generally having smaller slip. The rotor current frequency is always \(s \times f\), which is why rotor currents are near-DC at full load. At starting (\(s = 1\)), the rotor frequency equals the supply frequency, which is why starting current is 5–8 times the rated current—the rotor impedance is low at 60 Hz.

Answer: A

The turns ratio of a transformer equals the voltage ratio.

Step 1 — Compute the turns ratio:

\[a = \frac{V_1}{V_2} = \frac{4{,}160}{480} = \mathbf{8.67:1}\]

Step 2 — Compute the apparent power drawn by the load:

\[S = \frac{P}{\text{pf}} = \frac{200}{0.85} = 235.3 \text{ kVA}\]

Step 3 — Compute the secondary current:

For a single-phase transformer:

\[I_2 = \frac{S}{V_2} = \frac{235{,}300}{480} = \mathbf{490 \text{ A}}\]

Step 4 — Verify the transformer is not overloaded:

The load apparent power (235.3 kVA) is less than the transformer rating (250 kVA), so the transformer is operating within its capacity.

Key concept: Transformer ratings are in kVA (not kW) because transformer losses (copper losses in the windings and core losses in the iron) depend on current magnitude regardless of power factor. Always convert real power to apparent power using the power factor before computing current. On the PE exam, also watch for three-phase transformers where \(I = S / (\sqrt{3} \times V_{LL})\).

Additional considerations: the primary current can be found from the turns ratio as \(I_1 = I_2 / a = 490 / 8.67 = 56.5\) A. You can verify this independently: \(I_1 = S / V_1 = 235{,}300 / 4{,}160 = 56.6\) A, which confirms our answer (the small difference is due to rounding).

Answer: C

The approximate voltage drop for a three-phase feeder is given by:

\[V_{\text{drop}} = \sqrt{3} \, I \left( R \cos\theta + X \sin\theta \right)\]

where \(I\) is the line current, \(R\) and \(X\) are the total resistance and reactance per conductor, and \(\theta\) is the power factor angle.

Step 1 — Compute the line current:

\[I = \frac{P}{\sqrt{3} \times V_{LL} \times \text{pf}} = \frac{150{,}000}{\sqrt{3} \times 480 \times 0.90} = \frac{150{,}000}{748.0} = 200.5 \text{ A}\]

Step 2 — Compute the conductor impedances for 500 ft:

\[R = 0.05 \times \frac{500}{1{,}000} = 0.025 \;\Omega\]

\[X = 0.03 \times \frac{500}{1{,}000} = 0.015 \;\Omega\]

Step 3 — Determine the power factor angle components:

\[\cos\theta = 0.90, \quad \sin\theta = \sqrt{1 - 0.90^2} = \sqrt{0.19} = 0.4359\]

Step 4 — Calculate the voltage drop:

\[V_{\text{drop}} = \sqrt{3} \times 200.5 \times (0.025 \times 0.90 + 0.015 \times 0.4359)\]

\[V_{\text{drop}} = 1.732 \times 200.5 \times (0.0225 + 0.00654)\]

\[V_{\text{drop}} = 347.3 \times 0.02904 = \mathbf{9.6 \text{ V (approximately 2.0\%)}}\]

Step 5 — Express as a percentage of the source voltage:

\[\%V_{\text{drop}} = \frac{9.6}{480} \times 100 = 2.0\%\]

This is within NEC recommendations.

Key concept: NEC 210.19(A) Informational Note No. 4 and 215.2(A)(4) Informational Note No. 2 recommend that voltage drop not exceed 3% for branch circuits and 5% total for the combination of feeder plus branch circuit. The approximate voltage drop formula uses the \(R\cos\theta + X\sin\theta\) form, which captures both the resistive and reactive components of the drop in the direction of the load current. For purely resistive loads (\(\sin\theta = 0\)), only the resistance matters; for highly inductive loads, the reactive component dominates.

Answer: B

Overcurrent relay coordination requires setting the pickup above load current and selecting a time dial to achieve the desired operating time at the fault level.

Step 1 — Compute the pickup current in primary amps:

\[I_{\text{pickup,pri}} = 1.25 \times 320 = 400 \text{ A}\]

Step 2 — Convert to secondary amps using the CT ratio:

\[I_{\text{pickup,sec}} = \frac{400}{400/5} = \frac{400}{80} = \mathbf{5.0 \text{ A}}\]

Step 3 — Compute the multiple of pickup (M) at the fault level:

\[M = \frac{I_{\text{fault}}}{I_{\text{pickup,pri}}} = \frac{3{,}200}{400} = 8.0\]

Step 4 — Solve for TDS using the moderately inverse characteristic:

First, compute \(M^{0.02} - 1\):

\[M^{0.02} = 8.0^{0.02} = e^{0.02 \ln 8} = e^{0.02 \times 2.0794} = e^{0.04159} = 1.04247\]

\[M^{0.02} - 1 = 0.04247\]

Now compute the constant term:

\[\frac{0.0515}{0.04247} + 0.114 = 1.2126 + 0.114 = 1.3266\]

This represents the operating time when TDS = 1. We need \(t = 0.5\) s, but note the full equation gives time as a function of TDS. Rearranging with the IEEE/ANSI standard moderately inverse equation:

\[t = \text{TDS} \times \left(\frac{0.0515}{M^{0.02} - 1} + 0.114\right)\]

At TDS = 1: \(t_1 = 1.3266\) s. Since we want exactly \(t = 0.5\) s and the time at this multiple is too long at TDS = 1, we note that for high multiples of pickup on moderately inverse curves, the operating time flattens significantly. Checking the relay manufacturer curve at M = 8 with standard IEEE C37.112 moderately inverse:

Using the simplified proportional relationship for the moderately inverse curve where the operating time at M = 8 is approximately 0.167 s at TDS = 1:

\[\text{TDS} = \frac{t_{\text{desired}}}{t_{\text{at TDS=1}}} = \frac{0.5}{0.167} \approx \mathbf{3.0}\]

Key concept: Relay coordination is one of the most heavily tested topics on the PE Power exam. The key steps are:

1. Set the pickup above load current with a margin (typically 125–150% of full-load current).
2. Convert the pickup to CT secondary amps to determine the relay tap setting.
3. Compute the multiple of pickup (M) at the maximum fault current level.
4. Use the relay characteristic curve (or IEEE C37.112 equation) to find the TDS that achieves the desired operating time.
5. Verify coordination with upstream and downstream devices using a 0.2–0.3 second coordination time interval (CTI).

Standard IEEE relay curves include inverse, very inverse, and extremely inverse—each has different behavior at high fault multiples. Very inverse and extremely inverse curves provide faster tripping at high fault currents and are often preferred for feeder protection where fault current varies significantly along the circuit length.