PE Electrical Power common mistakes
PE Electrical Power · Understand the traps and how to avoid them.
Computes three-phase power as 3·V_line·I_line, or uses √3 with a phase voltage, and carries the wrong result into the rest of the problem.
The student memorizes 'S = √3·V·I' without tracking whether the V in hand is line-to-line or line-to-neutral, and formula sheets group the two forms closely together.
What to do instead
Before substituting, label every voltage as L-L or L-N and write the power relation twice: S = √3·V_LL·I_L and S = 3·V_LN·I_L. In a wye source V_LN = V_LL/√3; in a delta load V_phase = V_LL while I_line = √3·I_phase.
Reads a wattmeter, ammeter, or loss figure as the actual primary quantity without applying the CT and PT ratios, or ignores meter burden when judging whether a CT will saturate.
Instrument transformers are often described only by their ratio class, so the secondary reading looks like a final answer; burden calculations feel like a separate topic and get skipped.
What to do instead
Always convert secondary to primary with I_pri = I_sec × (CT ratio) and V_pri = V_sec × (PT ratio), noting CT ratio is rated primary over rated secondary. Check saturation by summing relay, lead, and meter burden in ohms against the CT secondary excitation curve.
Equates rotor speed with synchronous speed when computing slip and then uses the wrong frequency for rotor copper loss, or treats air-gap power as equal to shaft output.
The per-unit slip is small (a few percent), so a slip error barely changes speed and is easy to miss, while the power flow chain P_ag → P_conv → P_out → P_shaft has several subtraction steps that get collapsed.
What to do instead
Write the chain explicitly each time: n_s = 120f/p, s = (n_s − n_r)/n_s, P_rotor Cu = s·P_ag, P_conv = (1 − s)·P_ag, then subtract friction and windage. Rotor frequency is s·f, not f.
Applies a transformer's line-to-line voltage or line current directly as a per-phase winding quantity in a delta or wye bank, or changes a per-unit impedance to a new base using the wrong kVA or voltage base.
Nameplate values are line quantities, but the equivalent circuit is per phase; and the base-change formula has two squared ratios that are easy to invert.
What to do instead
Identify the connection first, then convert to winding values (delta: V_w = V_LL, I_w = I_L/√3; wye: V_w = V_LL/√3, I_w = I_L); note the kVA rating is unchanged by connection. For base changes use Z_pu,new = Z_pu,old × (S_new/S_old) × (V_old/V_new)².
Uses a single voltage-drop formula for every circuit - typically V_drop = I·R for three-phase feeders - and omits reactance on overhead lines.
The 2·I·R form is the one most often seen in training material, and low-voltage intuition says reactance is negligible.
What to do instead
Select by circuit: single phase V_d = 2·I·(R cosθ + X sinθ); three phase V_d = √3·I·(R cosθ + X sinθ) using line current and per-phase impedance. On overhead feeders X is comparable to or larger than R, so include it.
Uses conductor resistance at 20 °C for an ampacity, loss, or sag calculation on a conductor operating near 75–90 °C, or neglects the √3 when finding current from a three-phase MVA rating.
Tables list R at a reference temperature, and the temperature coefficient correction is an extra step not present in the formula as memorized.
What to do instead
Correct resistance with R_2 = R_1[1 + α(T_2 − T_1)], using α ≈ 0.0039 /°C for copper and 0.0040 /°C for aluminum at 20 °C reference. Get current from I = S/(√3·V_LL) before entering any loss or sag expression.
Connects the zero-sequence network with the grounding impedance used at face value, or models a line-to-ground fault with only the positive-sequence impedance.
Sequence networks are drawn separately and the fault-type connection rule is easy to misremember under time pressure.
What to do instead
For a line-to-ground fault at a solidly grounded bus, I_a1 = V_f/(Z_1 + Z_2 + Z_0) and the ground path carries 3·I_a0 through 3·Z_n, so the grounding impedance appears as 3·Z_n in the zero-sequence branch. Draw the three networks and the interconnection before substituting numbers.
Builds the bus admittance matrix with self-admittances that omit the shunt terms, or carries mixed impedance and admittance values into the load-flow equations.
Ybus assembly is mechanical, and one-line diagrams show impedances while the equations need admittances.
What to do instead
Convert every branch to admittance first (y = 1/z). Then Y_ii = sum of all admittances connected to bus i including shunts, and Y_ij = negative of the branch admittance between buses i and j. Verify each row sums to the shunt admittance only.
Compares a fault current in primary amperes directly against a relay pickup stated in secondary amperes, or sets an instantaneous element without checking transformer inrush and CT saturation.
Relay data sheets state settings on the relay base (1 A or 5 A), while fault studies produce primary currents, and the two are separated by the CT ratio.
What to do instead
Refer the primary fault current into the relay with I_sec = I_pri/(CT ratio), then compare with pickup and apply the time-current curve. Check the instantaneous setting sits above maximum inrush (roughly 8–12 times rated) and confirm the CT will not saturate at the maximum through-fault.
Uses the half-wave average value (V_m/π) for a full-wave rectifier, or reports the fundamental displacement factor as the true power factor of a phase-controlled converter feeding a nonlinear load.
The two rectifier forms sit adjacent in memory, and power factor is assumed to equal cos α without accounting for current distortion.
What to do instead
Recall V_dc = V_m/π for a half-wave and 2V_m/π for a full-wave bridge, and subtract commutation drop where given. True power factor is (V₁I₁ cos α)/(V_rms I_rms) = displacement factor × distortion factor, and for a sinusoidal source feeding a converter the input current is nonsinusoidal.