Free PE Power practice exam

The published characteristic for this effect is not a single derating number. It is a family of curves of percent of rated capacity against discharge current, one curve per temperature, and reading its SHAPE is what settles the question.

1 — Which way does temperature move capacity? The curves are ordered by temperature, with normal room temperature sitting in the middle of the family and the higher temperatures lying above it. A cell held below room temperature therefore delivers a smaller fraction of its rated capacity than the same cell in a conditioned room. Cold slows the electrochemistry and raises the internal resistance, so less of the stored charge can be got out before the cell reaches its end voltage.

2 — Is the loss a fixed percentage? No, and this is the part the shape decides. At a very light load every curve in the family starts near 100 % of rating and the curves lie close together: a battery discharged slowly gives up nearly its full capacity at any of these temperatures. As the discharge current rises, each curve turns down, and the colder ones turn down sooner and more steeply, so the curves fan apart. The penalty for cold is small at a trickle load and large at a heavy one.

3 — What this means for the substation. The same battery that comfortably carries the station's continuous panel load in an outdoor cabinet may fall well short on a high-current duty such as a switchgear closing operation or a motor-operated disconnect. Sizing a station battery for the coldest expected cabinet temperature at the HIGHEST current the duty draws — not at its average — is what the characteristic is telling the designer to do.

The correct statement is therefore that the deliverable fraction of rated capacity falls in the cold, and that the shortfall grows with discharge current.

One pole of the open breaker has a single bus phase on one side of it and the corresponding generator phase on the other. What stands across it is the phasor difference of those two line-to-neutral voltages — not the difference of two line-to-line values, and certainly not the arithmetic difference of the magnitudes, which is zero here.

1 — Line-to-neutral magnitude on each side. Both sides are 13.8 kV line to line, and for a wye reference V_L = √3 · V_LN, so

2 — Place the two phasors. Take the bus phase-a voltage as the reference:

3 — Subtract.

4 — The closed form, as a check. Two phasors of equal magnitude V separated by δ differ by

which agrees exactly, because the two phasors and their difference form an isosceles triangle with apex angle δ.

What the number means. 2.77 kV is 20 % of nominal standing across an open pole, and it is why a synchro-check relay is set to permit closing only within a few degrees: closing at 20° drives a large transient current through the machine and a shock torque through its shaft. Note also that the two components of the difference are very unequal — 481 V along the reference against 2,725 V at right angles to it. For a small angular separation, the difference between two equal phasors is almost entirely perpendicular to them.

The familiar ρ = 2πaR is not a law; it is the equally spaced four-pin relation after the burial depth has been thrown away, and it is only allowed when the pins are driven to l ≤ 0.1a. Here a = 2.5 m, so that limit is 0.25 m and the pins are at 1.0 m — four times too deep. The general form has to be used.

1 — Assemble the depth-dependent denominator.

a = 2.5 m l = 1.0 m

a² + 4l² = 6.25 + 4.00 = 10.25 √ = 3.2016 2a/3.2016 = 1.5617

a² + l² = 6.25 + 1.00 = 7.25 √ = 2.6926 a/2.6926 = 0.9285

1 + 1.5617 − 0.9285 = 1.6333

2 — Apply the general equally spaced four-pin relation.

ρ = 4πaR / [1 + 2a/√(a² + 4l²) − a/√(a² + l²)]

4πaR = 4π(2.5)(4.50) = 141.37 Ω·m

ρ = 141.37/1.6333 = 86.6 Ω·m

Why the shortcut fails by so much. The simplified form ρ = 2πaR returns 70.7 Ω·m — 18% low. That is not a rounding difference; it is the whole content of the l ≤ 0.1a condition. Driving the pins deep pulls more of the injected current through the shallow soil close to each rod, so the same resistivity registers a lower resistance, and reading it with the shallow-pin form under-reports the earth. On a real traverse the fix is to drive the pins shallower or to widen the spacing until l/a falls back under a tenth; where site conditions rule that out, as they do here, the depth-corrected form is the only defensible reduction.

The point method is an inverse-square law with a projection correction, and both parts have to be built from the geometry. The luminaire is not above the desk, so the distance the light travels and the plane the light lands on are two different things.

1 — The distance the light actually travels. The desk is offset, so the ray is the hypotenuse, not the mounting height:

2 — The angle from nadir. θ is measured at the luminaire, between straight down and the ray to the desk, so its cosine is the vertical leg over the hypotenuse:

3 — Illuminance normal to the ray.

4 — Project onto the horizontal work plane.

Intensity in candelas divided by a distance in meters squared gives lux directly, since one candela is one lumen per steradian and one lux is one lumen per square meter.

The same result follows from the form E = I·cos³θ/D², where D is the mounting height: 2,400 × 0.800³/3.6² = 2,400 × 0.512/12.96 = 94.8 lx. That version folds the slant distance into the cosine — it is the identical relation, not a competing convention, and it is a useful check that the geometry has been handled once and only once.

A 3.6 m mount reaching 2.7 m off-axis is already at 37° from nadir, where the two corrections together cut the on-axis value nearly in half. That is why an accent luminaire aimed at a specific object has to be located, not just specified.

A capacitor bank changes the reactive power a load draws from the system; it does not change the real power. So the quantity that stays fixed through the correction is P, and the bank supplies the difference between the reactive power the load needs and the reactive power the system will still deliver.

1 — Real power, which is what the relation is built on.

2 — The two power angles.

3 — The change in reactive power (Handbook §5.1.3):

A 3.8 MVAR bank is therefore required.

Two checks a designer makes without redoing the arithmetic. Before correction the load draws Q₁ = 9.36 × 0.8023 = 7.51 MVAR; after correction it draws Q₂ = 9.36 × 0.3952 = 3.70 MVAR from the system, and 7.51 − 3.70 = 3.81 MVAR is what the bank must make locally. And the corrected apparent power falls to 9.36/0.93 = 10.06 MVA, a 16% reduction in the current the substation transformer and the incoming line carry — which is the reason the bank is installed.

Three independent features of the trace each rule out one of the wrong arrangements, and only one arrangement survives all three.

1 — Count the pulses against the dashed supply trace. The load carries one pulse in the first half cycle and another in the second, two per supply cycle. Both half cycles of the supply therefore reach the load, so the circuit is full-wave. A single-device half-wave circuit delivers one pulse per cycle and would leave a whole half cycle blank.

2 — Look at where each pulse begins and where it ends. Every pulse starts flat on zero, jumps vertically onto the supply sinusoid 60° after the supply's zero crossing, and then decays smoothly back to zero at the next crossing. A diode conducts the instant it becomes forward-biased, which is the zero crossing; the only way to hold conduction off for part of a half cycle is to gate the device. The dead interval at the start of each half cycle is therefore the firing delay of a controlled device, not diode behavior. The smooth ending at the zero crossing is the resistive load turning the device current off by itself.

3 — Check the polarity of every pulse. All four pulses are positive. A converter that merely chops the supply and passes it through leaves the negative half cycles negative; an all-positive output means the supply has been rectified.

Full-wave, gated, rectified: a phase-controlled thyristor bridge.

One honest qualification. On a purely resistive load, a fully controlled bridge with four thyristors and a half-controlled bridge with two thyristors and two diodes produce the same output waveform — a resistive load cannot distinguish them, because the load current falls to zero at the crossing either way and neither bridge can drive the output negative. The trace identifies the family, not the device count, which is why the answer names the arrangement rather than the number of thyristors in it. Put an inductive load on the same two bridges and they separate immediately: the fully controlled bridge can push its output negative and invert, the half-controlled bridge cannot.

Step 1 — get the resistance of the rod bed on its own. The rods are not independent electrodes: each one drives current into earth that the others also have to push against, so the group is worse than n identical rods in parallel. The printed expression for the resistance of all ground rods carries that coupling in its last term (Handbook §2.2.6.5):

with ρ_a = 250 Ω·m, n = 12, l₂ = 9.0 m, d₂ = 0.019 m, A = 30 × 30 = 900 m², and K₁ = 1.37 given.

Step 2 — combine the grid and the rod bed, mutual resistance included. Two electrodes that share the same volume of earth are not in parallel, because the current from each raises the potential of the other. The printed combination is (Handbook §2.2.6.5):

Answer: most nearly 3.08 Ω.

Two checks the result has to pass. First, adding electrodes can only lower the resistance, so R_g must be below the grid's own 3.8 Ω — it is, by 19%. Second, R_g must also be below the rod bed's own 3.41 Ω, and it is. But it is nowhere near the 1.80 Ω a parallel combination would give, and that gap is the physics of the problem: the rods are driven inside the grid's own footprint, into earth the grid has already raised in potential, so most of what each electrode could do on its own is already being done by the other. That is what R₁₂ = 2.6 Ω measures, and it is large — two thirds of the grid's own resistance.

Why the coupling term inside R₂ matters too. Drop it and the twelve rods behave as twelve independent electrodes in parallel: a single 9.0 m rod in 250 Ω·m soil is (250/2π × 9.0)(8.240 − 1) = 32.0 Ω, and 32.0/12 = 2.67 Ω. That is 22% below the coupled value, and it propagates straight into the combination. Coupling appears twice in this problem — once among the rods and once between the rods and the grid — and each has its own term.

Per-unit work is base bookkeeping. The base apparent power is common to the whole network, but the base voltage changes at every transformer — and it changes by the transformer's NAMEPLATE TURNS RATIO, not by whatever the buses are nominally called.

1 — Base voltage in the MCC zone. The base is declared as 12.47 kV in the utility zone. T1 is wound 13.8 kV to 480 V, so it steps by 0.48/13.8:

It is not 0.48 kV. A 13.8 kV-rated transformer energized from a 12.47 kV system delivers proportionally less on the secondary, and the base has to follow the winding, not the label on the bus.

2 — Base current in the MCC zone. For a three-phase system,

3 — Per-unit current.

Check from the other end. The same answer must come out of an apparent-power ratio. The load is √3 × 0.4337 kV × 640 A = 481 kVA, and 481/1,000 = 0.481 per unit. Per-unit current and per-unit apparent power are equal when the per-unit voltage is 1.0, which is what the measured current implies here.

Plausibility. 481 kVA through a 1,500 kVA transformer is about 32% loading — a normal daytime figure for a plant MCC, and comfortably inside the unit's rating.

The requirement is a TORQUE, the answer is a CURRENT, and the two do not scale with terminal voltage the same way. Work the torque relation backwards to a voltage first, then carry that voltage across to the current.

1 — Torque the motor must develop at standstill.

2 — Terminal voltage that produces it. In the induced-torque expression the terminal voltage appears squared and every other term is a machine constant at a fixed slip, so at standstill T ∝ V₁². At rated voltage the machine develops 1.60 of full-load torque, so

Anything lower fails the requirement, so this is the lowest usable tap.

3 — Current at that voltage. The reactor is in SERIES with the stator, so one current flows through both — the line current IS the motor current. At standstill the motor impedance is fixed, so that current follows the motor terminal voltage directly, not its square:

Why the starter type is the whole item. A series reactor cuts current in proportion to voltage while cutting torque as its square. An autotransformer starter set for the SAME 65.6% motor voltage would produce the same torque but draw only 0.4297 × 1,170 = 503 A from the line, because it transforms the current a second time. That is choice B, and it is a real starter — just not the one the stem describes. The reactor's compensations are lower cost, a genuinely smooth ramp as the motor's own impedance rises with speed, and a closed transition; the price is a supply current about half again as large for the same starting torque.

Plausibility. 767 A on a 460 V feeder against a 180 A full-load motor is 4.3 per unit at standstill, a normal reduced-voltage starting draw.

Step 1 - refuse the shortcut, and know why. The reflex is to scale the fault current by the voltage ratio: 9,000 x 13.8/138 = 900 A. That is correct for a BALANCED fault, where all three windings carry equal currents and everything scales together. A single-line-to-ground fault is not balanced. One secondary winding carries the whole fault current and the other two carry none, so the transformer's connection - not just its ratio - decides what appears on the 138 kV lines. Work through the windings.

Step 2 - the secondary windings. The secondary is wye, so each winding is in series with its own line conductor and carries that line's current:

Step 3 - cross to the primary windings. The turns ratio is set by the voltage each WINDING sees, not by the two line voltages. A delta winding sees the full 138 kV; a wye winding sees 13.8/sqrt(3) = 7.967 kV:

By the ideal-transformer relation I_primary/I_secondary = N_secondary/N_primary = 1/a, the one energized primary winding carries

and the other two primary windings carry nothing.

Step 4 - cross from the primary windings to the primary lines. In a delta, each line current is the DIFFERENCE of the two winding currents that meet at that corner. With only one winding energized at 519.6 A, the two corners it touches each carry 519.6 A and the third corner carries zero:

So the largest current in the 138 kV line conductors is 519.6 A, most nearly 520 A - about 58 percent of what the voltage ratio alone would suggest.

Step 5 - check it a second way. In sequence terms the secondary carries I_0 = I_1 = I_2 = 9,000/3 = 3,000 A. The delta primary blocks zero sequence, so only the positive- and negative-sequence currents reach the lines, each 300 A after the ratio, and the delta-wye connection shifts them by +30 and -30 degrees. Adding them: 300 angle 30 + 300 angle -30 = 2 x 300 x cos 30 = 519.6 A. The two derivations agree.

Why this matters for coordination. A primary fuse or relay set from the balanced referral will see only 58 percent of the current the setting assumed, and a device that looked coordinated on a three-phase fault can fail to operate for a secondary ground fault.