Set 36 — Switchgear and Circuit Breakers
Twenty worked problems on what happens inside the breaker. Set 33 computed the fault current at a bus and read a nameplate against it, and that arithmetic is not repeated here. This set asks the next question: given that the contacts have parted and the current has just passed through zero, what voltage does the network rebuild across the gap, how fast, and what can be done about it. Two quantities carry almost the whole subject — \(\sqrt{LC}\), a time, which fixes the frequency, the time to peak and the rate of rise of the restriking voltage; and \(\sqrt{L/C}\), an impedance, which fixes the chopping overvoltage and the critical resistance. Compute both before touching any specific question.
The second half turns to the nameplate itself — symmetrical and asymmetrical breaking current, making current, short-time \(I^2t\), the fictitious MVA rating — and to the one device that beats a breaker at its own game, the fuse.
The arc is deliberate, not accidental. The network stores \(\tfrac12Li^2\) and \(v_L = L\,di/dt\) forbids an instantaneous stop. The arc is the channel through which the energy is surrendered at a survivable rate, so a breaker is designed to make an arc and then destroy it on schedule.
An arc is a negative resistance. Ayrton's \(V_{arc} = A + B\ell + (C + D\ell)/I\) falls as \(I\) rises, so an arc cannot be starved by lowering the voltage across it. Either raise \(R_{arc}\) until the source cannot sustain it, or wait for the current zero the supply provides twice a cycle.
The restriking voltage of a lumped terminal fault. With the source at its crest \(E_m\) at the current zero and the arc having held \(v(0)=0\), the loop equation \(LC\,\ddot v + v = E_m\) gives \(v = E_m(1-\cos\omega_n t)\) with \(\omega_n = 1/\sqrt{LC}\). Peak \(2E_m\) at \(t = \pi\sqrt{LC}\).
Two slopes, and they differ by \(2/\pi\). \((dv/dt)_{\max} = E_m/\sqrt{LC}\) occurs at \(\omega_n t = \pi/2\); the average is peak over time-to-peak, \(2E_m/(\pi\sqrt{LC}) = 0.637\,(dv/dt)_{\max}\). A specification quoted as one must never be compared with a test result quoted as the other.
Two multipliers turn the idealisation into a standard. The amplitude factor \(k \approx 1.3\text{–}1.7\) accounts for damping; the first-pole-to-clear factor \(k_{pp}\) is \(1.5\) for an isolated neutral and \(1.3\) for an effectively earthed system. Then \(u_c = k_{pp}k\,E_m\) and \((dv/dt)_{\max} = k_{pp}E_m/\sqrt{LC}\).
Chopping is an energy balance. A powerful interrupter forces a small current to zero early; \(\tfrac12Li_{ch}^2 = \tfrac12CV^2\) gives \(V = i_{ch}\sqrt{L/C}\), and with a standing \(v_0\) on the capacitance, \(V_{\max} = \sqrt{v_0^2 + i_{ch}^2 L/C}\). The danger is entirely in \(\sqrt{L/C}\), which is kilohms for an unloaded transformer and hundreds of ohms for a feeder.
Resistance switching damps the same oscillation. A shunt \(R\) adds a term \(1/RC\) to the damped equation; critical damping needs \(R_c = \tfrac12\sqrt{L/C}\). Then \(v = E_m[1-(1+\omega_n t)e^{-\omega_n t}]\), the peak falls from \(2E_m\) to \(E_m\) and the maximum RRRV falls by \(e = 2.718\). Smaller \(R\) means more damping, because the resistor is across the capacitance and not in the loop.
The nameplate is five independent tests. \(S_{br} = \sqrt3\,V_rI_{sc}\); \(I_{asym} = I_{ac}\sqrt{1+2\delta^2}\); \(i_p = 2.5I_{sc}\) at 50 Hz; \(I_1^2t_1 = I_2^2t_2\) for the short-time rating. Passing one says nothing about the others — Set 33 works the selection through in detail.
A fuse acts before the first peak arrives, which no mechanical device can. The fusing factor is minimum fusing current over rated current; the cut-off current is the peak actually let through; the let-through \(I^2t\) is the thermal stress passed downstream, and it must satisfy \(I^2t \le k^2S^2\) for the cable behind it.
A 220 kV, 50 Hz system has an effective inductance of 12 mH per phase between the source and the breaker, and the total capacitance to earth at the breaker terminal is 0.015 µF. A three-phase terminal fault is cleared. Find
- the natural frequency of the restriking voltage;
- its peak value and the time taken to reach it;
- the maximum and the average RRRV;
- the fault level the same inductance implies, as a check that the data is realistic.
Build the two governing quantities first. The transient is driven by the crest of the phase voltage, because a current limited by \(L\) lags the source by very nearly 90° and so passes through zero when the source is at its peak:
Everything in parts (i) to (iii) is a rearrangement of those two numbers. The companion quantity \(\sqrt{L/C} = 894\ \Omega\) is not needed here but will be in Problem 8.
(i) The natural frequency. The circuit after interruption is an undamped \(L\!-\!C\) loop:
Two hundred and thirty-seven times the power frequency, which justifies the assumption that \(e(t)\) is frozen at \(E_m\) throughout the transient.
(ii) Peak and time to peak. The solution of \(LC\,\ddot v + v = E_m\) with \(v(0)=0\) and \(\dot v(0)=i(0)/C=0\) is \(v = E_m(1-\cos\omega_n t)\), so the capacitance overshoots the source by exactly as much as it started below it:
(iii) The two slopes. Differentiating, \(dv/dt = E_m\omega_n\sin\omega_n t\), whose largest value is at \(\omega_n t = \pi/2\):
The ratio \(2/\pi = 0.637\) is exact and independent of the data — it is the mean of a half sine over its own half period.
(iv) The sanity check. The same 12 mH fixes the fault current, so the data cannot be chosen freely:
A strong 220 kV bus, within the 40 kA breakers built for that class. So the inductance is realistic and the RRRV computed from it is a defensible upper bound.
Why it is an upper bound and not a specification. A single \(L\!-\!C\) pair has one natural frequency; a real 220 kV bus carries several lines and transformers whose capacitance is in parallel with the 0.015 µF and whose inductance is in parallel with the 12 mH. Both effects slow the rise. A published TRV envelope for this class specifies about 2 kV/µs, so the lumped estimate is some six times conservative — useful for showing which parameter matters, not for a purchase order.
An oscillogram taken during a short-circuit test on a 132 kV breaker shows a peak restriking voltage of 216 kV and a maximum RRRV of 4 kV/µs. The fault current was 10 kA rms. Deduce \(\sqrt{LC}\), then \(L\) and \(C\) separately, and state the natural frequency.
The peak gives \(E_m\) and the slope gives the time. Since \(v_{peak}=2E_m\) in the undamped model,
Two measurements have fixed the pair \((E_m,\sqrt{LC})\). But \(\sqrt{LC}\) alone cannot separate \(L\) from \(C\) — a third fact is needed, and the fault current supplies it.
The fault current fixes \(L\). The current is limited by \(\omega L\) alone in this model:
And then \(C\) follows from the product:
Thirty nanofarads — bushings, current transformers, a few metres of busbar and the capacitance of the transformer winding behind the breaker. A plausible substation figure.
Cross-check with the surge impedance, which the two separated values now allow:
The two quantities multiply to \(L\) and divide to \(C\). Any pair of them determines the circuit, which is the practical reason for computing both at the start of every problem in this set.
The fault level implied is \(\sqrt3(132)(10) = 2287\) MVA, an ordinary 132 kV bus. The test therefore reproduced a realistic duty rather than an artificially severe one — which matters, because a test at lower current with the same \(C\) would need a larger \(L\) and would give a gentler slope. The least severe RRRV occurs at the highest fault current, an inversion worth remembering.
The rated TRV envelope of a 145 kV breaker is quoted as a peak \(u_c\) reached in \(t_3 = 124\) µs at an average rate of 2 kV/µs. Find \(u_c\), the maximum RRRV the equivalent undamped circuit would show, the natural frequency and \(\sqrt{LC}\). Taking the first-pole-to-clear factor as 1.3, deduce the amplitude factor the envelope implies.
The envelope is defined as peak over time, so the quoted rate is the average one by construction:
Convert to the maximum slope using the exact \(2/\pi\) of the cosine solution:
A breaker whose gap can outrun 3.14 kV/µs satisfies an envelope written as 2 kV/µs. Reading the specification as if 2 were the instantaneous maximum would over-specify the breaker by 57%.
The equivalent circuit constants. Since \(t_3\) plays the role of the time to peak,
Note the shortcut: the time to peak is half a natural period, so \(f_n = 1/(2t_3)\) without any square roots at all.
Now back out the two multipliers. The crest of the phase voltage at rated voltage is
An amplitude factor of 1.61 sits in the upper half of the 1.3–1.7 band, which is what a bus with few connected lines gives — little extra capacitance, little damping.
The consistency test. An undamped ideal circuit would have \(k = 2\) and give \(u_c = 1.3\times2\times118.39 = 307.8\) kV. The standard's 248 kV is 81% of that, and the missing 19% is exactly the damping the amplitude factor represents. Nothing in the envelope is arbitrary: it is the lumped model with two named corrections.
Explain why the first pole of a three-phase breaker to reach a current zero has to withstand more than the phase voltage, derive the factors 1.5 and 1.3, and evaluate the corrected TRV peak and slope for the 132 kV circuit of Problem 2 with an amplitude factor of 1.4.
The three poles do not clear together. The three phase currents pass through zero at 120° intervals, so one pole interrupts first while the other two are still conducting through the fault. That surviving pair, joined in series through the fault point, drags the potential of the fault point away from earth — and the first pole stands between the system voltage and that shifted point.
Isolated neutral, three-phase ungrounded fault. After pole \(a\) clears, phases \(b\) and \(c\) remain joined at the fault, so their common point sits at \((V_b+V_c)/2\). With a balanced source \(V_a+V_b+V_c=0\):
Exactly the neutral-shift argument of Set 22, read in the phase domain. In sequence terms the same result is \(k_{pp} = 3X_0/(X_1+2X_0)\), which tends to 1.5 as \(X_0\to\infty\).
Effectively earthed system. Now the zero-sequence path is not open. Substituting the effective-earthing limit \(X_0 \le 3X_1\) into the same expression:
The standard rounds it to 1.3. The two numbers are therefore not conventions but the two limits of one formula, and the boundary between them is precisely the effective-earthing criterion.
Apply to the 132 kV circuit. With \(E_m = 107.78\) kV and \(k = 1.4\):
And the slope is multiplied too, because the whole transient is scaled by \(k_{pp}\) while \(\sqrt{LC}\) is unchanged. From Problem 2's circuit, \((dv/dt)_{\max} = E_m/\sqrt{LC} = 108/27 = 4.00\) kV/µs, so
Note that damping reduces the peak but not the initial slope, while \(k_{pp}\) raises both. The two corrections act differently and must each be applied to the right quantity.
The design consequence. A breaker type-tested on an effectively earthed system at \(k_{pp}=1.3\) is not automatically suitable for an isolated or resonant-earthed network of the same voltage and current. It is the commonest way a correctly rated breaker is misapplied, and the check costs nothing: read the earthing arrangement before reading the fault level.
A 132 kV breaker has two interrupting chambers in series. The stray capacitance across each open break is 40 pF and the capacitance from the mid-point of the pair to earth is 200 pF. The circuit behind the breaker has \(L = 30\) mH and a terminal capacitance to earth of 3200 pF.
- Find how the recovery voltage divides between the two breaks with no grading capacitors.
- Repeat with a 1600 pF grading capacitor across each break.
- Find the change in maximum RRRV that the two grading capacitors produce, and state what happens to the peak.
Set up the voltage division. Let the source-side terminal be at zero, the line terminal at \(V\) and the mid-point at \(v_m\). Each break carries a capacitance \(C_g\), and the mid-point has \(C_e\) to earth. Charge conservation at the mid-point node gives
The line-side break \(A\) always takes more, because the current leaving it splits between break \(B\) and the earth capacitance. The ratio depends on nothing but \(C_e/C_g\).
(i) Without grading capacitors, \(C_g = 40\) pF against \(C_e = 200\) pF:
The two chambers are identical and share the duty six to one. The outer break is doing almost all the work and will fail first, and the second chamber contributes almost nothing — the breaker behaves as a single-break unit that happens to cost twice as much.
(ii) With 1600 pF across each break:
Six to one becomes nine to eight. This is the grading capacitor's first and principal duty, and it is why every multi-break EHV breaker has one across every chamber.
(iii) The effect on the RRRV. Seen from the terminal, the two 1600 pF capacitors are in series, adding 800 pF to the 3200 pF already there:
A reduction of 10.6%, the factor being \(1-\sqrt{3200/4000}\). The natural frequency falls from 16.24 to 14.53 kHz in the same proportion.
And the peak does not move at all:
Because \(v = E_m(1-\cos\omega_n t)\) has an amplitude set only by \(E_m\). Capacitance stretches the time axis and leaves the voltage axis alone.
Reading the three results together. The grading capacitor is bought for the voltage sharing, which it transforms; the RRRV reduction is a bonus of 10%, and the dielectric round of the interruption race is not helped at all. That is why a breaker fitted with grading capacitors may still need opening resistors — the two components address different rounds, as Problems 8 and 9 show.
A vacuum breaker switches out an unloaded transformer on a 66 kV system. The magnetising inductance referred to the 66 kV winding is 30 H and the terminal capacitance is 4 nF. The interrupter chops at 5 A. Find the prospective overvoltage in kilovolts and in per unit of the peak phase voltage, the energy an arrester would have to absorb, and the surge capacitance that would hold the crest below 200 kV.
Everything follows from the surge impedance of the trapped circuit. At the instant of the chop the inductance carries \(i_{ch}\) and the current vanishes, so its magnetic energy must transfer to the only other store present:
Eighty-six kilohms. For a loaded feeder the same quantity is a few hundred ohms, and that single difference is the whole of the chopping problem.
The overvoltage:
The capacitance is not at zero volts when the chop occurs. A magnetising current lags by nearly 90°, so the chop happens close to the voltage crest:
The standing voltage adds 0.8% and can usually be neglected — a quadrature sum is dominated by whichever term is larger, and here they differ by a factor of eight.
In per unit:
The lightning impulse withstand level of 66 kV equipment is 325 kV. The transient exceeds it by 34%, so this operation destroys the transformer it was switching — the failure being in the equipment, not in the breaker.
The arrester's energy duty, by contrast, is trivial:
Three hundred and seventy-five joules against a surge arrester's several hundred kilojoules of capability. The chopping surge is a voltage problem and never an energy problem — which is why a small distribution-class arrester at the transformer terminals is a complete remedy.
Sizing the surge capacitor. Adding capacitance lowers \(\sqrt{L/C}\) as \(1/\sqrt{C}\). Requiring \(V_{\max}\le200\) kV and allowing for the standing 53.89 kV:
Specify the next standard value, 0.022 µF. Checking it: \(\sqrt{30/(0.022\ \mu\text{F})} = 36\,928\ \Omega\), \(V = 184.6\) kV, \(V_{\max} = 192.3\) kV. Comfortably inside the target.
What the capacitor achieves. Raising \(C\) from 4 nF to 22 nF is a factor of 5.5, so the crest falls by \(\sqrt{5.5} = 2.35\). It also lowers the ringing frequency from \(1/(2\pi\sqrt{30\times4\times10^{-9}}) = 459\) Hz to 196 Hz, a further benefit because the steepness of the front — not only its height — determines how unevenly the surge distributes itself across the first turns of the winding.
A 33 kV vacuum interrupter switches an unloaded transformer for which \(L = 8\) H and \(C = 5\) nF. An early design with pure copper contacts chops at 12 A; a modern chromium–copper interrupter chops at 4 A. Compare the two overvoltages against the 170 kV impulse level of 33 kV equipment, and find the surge capacitance the modern interrupter would need to bring the crest below 120 kV.
The circuit is the same for both; only \(i_{ch}\) differs:
The two overvoltages:
The overvoltage is strictly proportional to the chopping level, so a three-fold reduction in \(i_{ch}\) is a three-fold reduction in volts. Nothing else in the circuit had to change.
Against the impulse level of 170 kV:
The contact alloy alone moves the duty from catastrophic to marginal. That is why the choice of contact material is a switchgear specification item and not a manufacturing detail.
Five per cent is not a margin. The chopping level is statistical — a quoted 4 A means a distribution with a tail — and the BIL is a withstand under a standard 1.2/50 µs wave, not under a kilohertz oscillation that stresses the first turns unevenly. Bring the crest to 120 kV, about 70% of BIL:
Check: \(\sqrt{8/10^{-8}} = 28\,284\ \Omega\), \(V = 113.1\) kV, \(V_{\max} = 116.3\) kV. A ten-nanofarad component — a capacitance a hundred thousand times smaller than a power-factor bank — settles it.
And the energies, for completeness:
Both negligible. The energy scales as \(i_{ch}^2\) and the voltage as \(i_{ch}\), so reducing the chopping level helps the voltage problem linearly and the (already absent) energy problem quadratically.
A 132 kV breaker clears a terminal fault in a circuit with \(L = 25\) mH and \(C = 0.02\) µF. Find the resistance that must be connected across the main contacts for critical damping, the peak restriking voltage and maximum RRRV with and without it, the current the auxiliary interrupter must then clear, and the energy the resistor absorbs in a 10 ms insertion.
Both governing quantities, at once:
The critical resistance. With \(R\) in shunt the source current divides, \(i = C\,dv/dt + v/R\), and the loop equation becomes
Verify the damping directly: \(\alpha = 1/(2R_cC) = 1/(2\times559\times2\times10^{-8}) = 4.472\times10^{4}\ \text{s}^{-1} = \omega_n\). The two roots coincide, as critical damping requires.
The transient, before and after:
The peak is halved and the slope divided by \(e = 2.718\). Both rounds of the interruption race are won at once, which is what distinguishes resistance switching from the grading capacitor of Problem 5.
The price: a residual current. With the main contacts open, the source drives current through \(R_c\) in series with \(\omega L = 7.85\ \Omega\), which is negligible beside 559 Ω:
The fault current the main contacts interrupted was
The auxiliary interrupter clears 136 A instead of 9.7 kA — a duty smaller by a factor of seventy-one, and one that a small series break handles without any arc control at all.
The resistor's thermal duty fixes its physical size:
Ten megawatts for ten milliseconds. A resistor stack rated for a hundred kilojoules per operation, with its own thermal time constant and its own operating-sequence limit — which is why resistance switching is confined to EHV breakers where the duty justifies the hardware.
The implied fault level is \(\sqrt3(132)(9.703) = 2218\) MVA, an ordinary 132 kV bus, so the whole design is self-consistent. Note that \(R_c\) depends on \(\sqrt{L/C}\) only, while the residual current depends on \(R_c\) and \(E_m\) — one number, 1118 Ω, settles the resistor's ohms, its current, its power and its energy.
Starting from the critically damped response \(v = E_m\left[1-(1+\omega_n t)e^{-\omega_n t}\right]\), show that the maximum rate of rise occurs at \(t = 1/\omega_n\) and equals \(E_m\omega_n/e\). Then explain, in terms of the two governing quantities, why adding capacitance reduces the RRRV but not the peak while adding the critical resistance reduces both — and identify the one circumstance in which capacitance is nevertheless the better remedy.
Differentiate once. Writing \(u = \omega_n t\) for brevity, and using the product rule on \((1+u)e^{-u}\):
The two exponentials cancel except for the term in \(u\). Notice that \(dv/dt = 0\) at \(t=0\) — the response starts flat, exactly as the undamped one does, because the initial condition \(\dot v(0)=i(0)/C=0\) is the same.
Differentiate again and set to zero:
The factor is exactly \(1/e = 0.3679\), independent of every circuit value. It is the peak of the function \(ue^{-u}\), which occurs at \(u=1\) whatever the units.
Why capacitance moves only the time axis. The undamped solution is \(v = E_m(1-\cos\omega_n t)\). The amplitude of that expression contains \(E_m\) and nothing else:
Adding \(C\) stretches the horizontal axis and leaves the vertical one untouched. Quadruple \(C\) and the slope halves, the frequency halves, the time to peak doubles — and the peak is precisely where it was. The physical statement is that a capacitance is an energy store and not an energy sink; it delays the arrival of the source's energy without absorbing any of it.
Why resistance moves both. A resistor dissipates. The overshoot to \(2E_m\) exists because an undamped \(L\!-\!C\) pair conserves the energy it exchanges, so the capacitance must overshoot by as much as it undershot. Remove the energy on the way and the overshoot cannot occur:
And the slope falls too, because part of the source current that would have charged \(C\) now flows in \(R\) instead — the same \(i = C\,dv/dt + v/R\) that produced the damping term.
The comparison, side by side, using the Problem 8 circuit:
Voltages in kV, slopes in kV/µs. To match the resistor's slope with capacitance alone would need \(C\) multiplied by \((4.82/1.77)^2 = 7.4\), and even then the peak would remain at 215.6 kV.
The circumstance in which capacitance wins. Where the breaker is losing the thermal round and not the dielectric one — an interrupter with plenty of gap but a slow post-arc recovery, or a short-line fault whose peak is modest and whose slope is brutal — capacitance addresses the entire problem. It is passive, it needs no auxiliary interrupter, it does not have to be re-cooled between operations, and it doubles as the grading capacitor Problem 5 required in any case. That is why grading capacitors are near-universal on EHV breakers and opening resistors are not.
A breaker is offered as 145 kV, 31.5 kA, 3 s. Find its breaking capacity in MVA, its rated peak making current at 50 Hz, its asymmetrical breaking current for a DC component of 40%, and the current it could carry for 1 s and for 0.5 s. State which of these numbers is fictitious and why it is still quoted.
Breaking capacity in MVA is the product of the rated line voltage and the rated breaking current in three-phase form:
This is the fictitious one. No three-phase apparatus delivers 7911 MVA — at the instant of maximum current the voltage across the breaker is near zero, and when the voltage recovers the current is zero. The product is a bookkeeping quantity, but it is exactly what a fault-level study produces, which is why it survives on specifications and in examinations.
Making current is decided by the crest of the first loop of a fully offset wave. The theoretical crest is \(2\sqrt2 I_{sc}\), but the DC offset decays measurably within the first half cycle:
At 60 Hz the factor is 2.6, the shorter half cycle allowing less decay. The rated figure is 78.75 kA peak, and it is a mechanical and magnetic rating — contact welding and the electromagnetic repulsion that tries to blow the contacts apart at closure.
Asymmetrical breaking current. Writing the DC component as a fraction \(\delta\) of the AC peak, so that \(I_{dc} = \delta\sqrt2 I_{ac}\):
A 40% DC component raises the rms by 15%. The limits are worth carrying: \(\delta=0\) gives \(I_{ac}\) and \(\delta=1\) gives \(\sqrt3 I_{ac} = 1.732 I_{ac}\), so the whole range is a factor of 1.73.
Short-time equivalents. For durations under a second or so the heating is adiabatic, so \(\int i^2dt\) alone matters and the rating converts by holding \(I^2t\) constant:
These are thermal currents, not interrupting currents. The breaker can carry 54.6 kA for a second; it cannot break it.
The five numbers describe five independent tests:
Passing one says nothing about another, and the selection procedure of Set 33 checks every row against its own duty.
A fault study gives 18 kA symmetrical at the instant of contact separation, with a DC component of 40% at that instant. A breaker is offered with a symmetrical rating of 20 kA and a declared DC capability of 30%. Determine whether it is adequate, and say what would have to change in the study — or in the breaker — for it to become so.
The symmetrical comparison passes easily:
Which is why this case is a trap. Every purchaser who checks only this row buys the breaker.
The asymmetrical comparison is the one that matters, because the interrupter has to deal with the total current at the current zero it is attacking:
On total rms the breaker just holds — but on the declared DC percentage, 40% against 30%, it does not. The two tests disagree, and the disagreement is not academic.
Why the percentage test is the binding one. The DC component does not merely raise the rms; it displaces the current zeros. With \(\delta = 0.4\) the major loop is longer and the minor loop shorter than a half cycle, so the arcing time before a usable zero appears is extended and the \(di/dt\) at that zero is reduced — and it is \(di/dt\) at the zero, not the rms, that determines how much residual plasma the gap must clear. A breaker type-tested at 30% has never demonstrated it can handle the delayed zero that 40% produces.
What would have to change — route one, a slower breaker. The DC component decays as \(\delta = e^{-t/T}\) with \(T = L/R = (X/R)/\omega\). Deduce the network's \(X/R\) from the study, taking contact separation at 60 ms:
Delaying contact separation by 19 ms — roughly one cycle — brings the DC component within the breaker's declaration. That means a slower opening mechanism, or one extra relay time step, and it costs a cycle of fault energy everywhere else on the system.
Route two, a lower \(X/R\). A value of 20.6 is high — typical of a bus fed directly from generator transformers with little line resistance in the path. Sixty milliseconds at the standard \(X/R = 15\) would give
— inside the declaration. So the offered breaker is adequate for an ordinary network and inadequate for this one, and the difference is entirely in the resistance of the path, not in the fault current.
Route three, and the honest one: buy the right breaker. A 25 kA unit with a 50% DC capability costs a fraction more than a 20 kA unit and removes both objections at once. The engineering point is that the reason for rejecting the offer must be recorded correctly. Rejecting it as "18 kA is too close to 20 kA" invites the supplier to argue about margins; rejecting it as "declared DC capability 30% against a duty of 40% at contact separation" is a statement about a type test, and cannot be argued with.
A 33 kV substation bus has a fault level of 750 MVA. Find the symmetrical fault current, the peak making current the breaker must survive if it is closed onto the fault, and select a rating from the standard series 12.5 / 16 / 20 / 25 kA. State what physically limits the making duty, and why it is not the same thing the breaking duty limits.
The symmetrical current follows from the fault level directly — the step Set 33 takes from the sequence networks:
The making current. Closing onto an existing fault the wave is fully offset in the worst pole, and the first crest arrives half a cycle later:
The 2.5 rather than 2.83 is the DC offset decaying during the first half cycle. Using 2.83 is not wrong, merely conservative by 13%.
Select from the series. Both duties must be checked against the same candidate:
The making check and the breaking check give the same verdict here, because both scale linearly with \(I_{sc}\) and both are compared against ratings derived from the same 2.5. That coincidence is what makes the making rating look redundant — and it is not.
What limits the making duty, and it is nothing to do with arcs. Two mechanical effects:
Both go as the square of the instantaneous peak, which is why the rating is a peak in kA and not an rms. The repulsion force at 32.8 kA peak is \((32.8/13.12)^2 = 6.25\) times what the rms value would suggest.
Why the breaking duty is a different question entirely. Making is settled in the first half cycle by a mechanical peak; breaking is settled a few cycles later at a current zero by a dielectric race. The same fault produces both numbers, and a breaker adequate for one may be inadequate for the other — most obviously in a design with a fast, light mechanism (good breaking, poor making) or a heavy latched one (the reverse).
One further check the fault level hides. The 750 MVA is the source contribution. Suppose the 33 kV bus also carries 20 MVA of induction motors with \(x'' = 0.20\) pu. Their contribution is \(20/0.20 = 100\) MVA, that is \(100/57.16 = 1.75\) kA:
Still inside the 16 kA / 40 kA unit, but the margin has fallen from 22% to 7.6%. The motor contribution decays over the first few cycles, so it appears at full value in the making duty and reduced in the breaking duty — which is exactly why the making rating is quoted separately rather than inferred from the breaking one.
A 12 kV indoor breaker is rated 25 kA for 1 s. It is to be installed on a bus whose symmetrical fault current is 18 kA. Find the equivalent 3 s rating, check the breaker against a backup protection time of 1.2 s and again against 2.0 s, and find the longest fault duration it can withstand at that bus. Then show that a duty of 30 kA for 0.5 s passes the thermal check and is nevertheless unacceptable.
The rating as an energy. Under a second the heating is adiabatic, so the whole rating is one number:
The equivalent 3 s rating:
Three times the duration at \(1/\sqrt3\) of the current. The same breaker is honestly described as "25 kA, 1 s" or "14.43 kA, 3 s"; the two statements carry identical information.
The duty at 1.2 s backup:
Utilisation 62%. Comfortable.
At 2.0 s backup:
It fails by 3.7%, and the remedy is a protection setting rather than a bigger breaker — the backup grading must be brought inside 1.93 s. Note how sharply the answer depends on a number the switchgear engineer does not own.
Now the trap. A duty of 30 kA for 0.5 s:
The breaker can absorb the heat and cannot interrupt the current. It would carry the fault faithfully for half a second, receive the trip signal, draw an arc it has no capacity to extinguish, and fail with the contacts fully open — the one failure mode Chapter 35 identifies as unrecoverable, because there is no travel left to buy and the fault energy has nowhere to go but into the breaker.
The general rule, therefore:
Two independent inequalities. Passing the first says nothing about the second, and the first can be satisfied at any current whatever by shortening the time.
A breaker recovered from an 11 kV switchboard is described on its plate as "500 MVA". It is proposed for a 6.6 kV bus whose fault level is 350 MVA. The proposal is defended on the ground that 500 exceeds 350. Show that the defence is wrong, find the breaker's true capability at 6.6 kV, and state the MVA rating an 11 kV breaker would need in order to be usable there.
Convert the plate to what the breaker actually does, which is interrupt a current:
The interrupter's real property. The MVA figure was that current multiplied by the voltage of the switchboard it was designed for, and the second factor travels with the installation, not with the breaker.
The same breaker on a 6.6 kV bus still interrupts 26.24 kA, so its breaking capacity expressed in MVA falls with the voltage:
The identical device is a "500 MVA breaker" on one bus and a "300 MVA breaker" on another. Quoting MVA without the voltage it refers to is not a rating at all.
The duty at the new bus:
Although 500 comfortably exceeds 350. The comparison was made between two numbers referred to different voltages, which is the same error as comparing per-unit impedances on different bases.
What rating would be needed. Scale the required 350 MVA to the breaker's own voltage base:
So an 11 kV breaker plated at 583 MVA or more — equivalently 30.6 kA or more — is what the 6.6 kV duty demands. The ratio 11/6.6 = 1.667 is the whole of the correction, and it is the ratio of voltages, not of currents.
The inverse case, for symmetry. Take the same 350 MVA fault level but at 11 kV:
Here the breaker is comfortable with 43% in hand. The same fault level, the same breaker, opposite verdicts — because the current, not the apparent power, is what an interrupter faces.
Two further rows that must also be re-checked before any breaker is moved between voltage classes:
Insulation improves when a breaker moves down in voltage; every current-based rating does not change at all. The failure mode of the proposal is entirely on the current rows.
A 415 V board has a prospective symmetrical fault current of 31.5 kA rms, for which the peak factor is 2.2. A 35 mm² XLPE-insulated copper cable (\(k = 143\)) leaves the board. Find the cable's \(I^2t\) withstand and the longest clearing time a non-current-limiting breaker could have. Then assess a 100 A HRC fuse whose cut-off current is 12 kA peak and whose let-through energy is 60 000 A²s, including the reduction in electromagnetic force it achieves.
What the cable can absorb. The adiabatic limit before the insulation reaches its limiting temperature is
The constant \(k\) carries the conductor material and the insulation together: 143 for XLPE copper, 115 for PVC copper, 76 for PVC aluminium. XLPE tolerates 250 °C against PVC's 160 °C, and the whole of the difference between 143 and 115 is that.
The breaker's allowance. A non-current-limiting device lets the full prospective current flow for its whole clearing time:
A breaker must clear in under one and a quarter cycles. No mechanical device does — the fastest moulded-case breakers take two to three cycles, and a relay-and-breaker combination four or more. The cable is unprotected.
The fuse, by contrast:
Four hundred and eighteen times in hand. The cable is never remotely at risk, and the reason is not that the fuse is faster at deciding — it is that the fuse acts before the first peak arrives, so the current never reaches the value the breaker has to live with.
The mechanical comparison is more lopsided still. The prospective peak is
while the fuse cuts off at 12 kA. Since the force on busbars, cleats and terminations goes as \(i^2\):
A factor of thirty-three in mechanical stress, which decides the cleat spacing, the busbar thickness and whether the switchboard needs internal-arc classification at all.
Where the current-limiting comes from. At 31.5 kA prospective the fuse element reaches melting within the first two or three milliseconds, long before the first crest at 5 ms. The element parts, several short arcs strike in the graded quartz sand, and the arc voltage they develop opposes the source before the wave has risen:
Total operating time under 5 ms. The sand absorbs the arc energy; the restricted sections of the silver element ensure melting starts at several points at once so the arc is split from the first instant.
What the fuse gives up. It is single-shot, it cannot be graded as finely as a relay, it cannot be tripped by anything but its own current, and if only one phase clears a three-phase motor is left single-phased and will burn out. The standard answer is the switch-fuse unit or the fuse-backed contactor: the fuse handles the high-current end where its cut-off is decisive, and a relay-driven device handles the overload end where discrimination matters.
A circuit is wired in a cable of continuous current-carrying capacity \(I_z = 120\) A and carries a design current of 100 A. Protection must satisfy \(I_b \le I_n \le I_z\) and \(I_2 \le 1.45\,I_z\), where \(I_2\) is the current that operates the device in the conventional time. Compare a 100 A HRC fuse of fusing factor 1.25 with a 100 A rewirable fuse of fusing factor 2.0, and state the rating the rewirable fuse would have to take.
Definitions first, since the whole problem is a matter of reading them correctly:
The rated current is what the element carries indefinitely without deterioration; the minimum fusing current is the least current that will eventually melt it. Between the two lies a band in which the fuse neither carries safely nor operates — and the fusing factor is the width of that band.
The two candidates against the two rules:
The first rule passes for both devices; the second fails for the rewirable one. It would allow 200 A — 167% of the cable's rating — to flow indefinitely without operating, and the cable would run far above its design temperature until the insulation failed.
The rating the rewirable fuse must be reduced to:
But the design current is 100 A, so an 87 A fuse would blow in normal service. The circuit cannot be protected by a rewirable fuse at all unless the cable is enlarged.
How much cable would be needed. Inverting the same inequality with \(I_n = 100\) A fixed by the load:
Against 120 A actually installed — the next cable size up. This is the historical origin of the derating factor of 0.725 (that is, \(1.45/2.0\)) applied to circuits protected by semi-enclosed rewirable fuses, a rule that predates the modern wording and says exactly the same thing.
Why the two fusing factors differ so much. A rewirable element is a bare tinned-copper wire in open air, whose melting point is high, whose cooling is by natural convection and whose actual diameter depends on who rewired it. An HRC element is a punched silver strip with restricted sections and an alloy blob whose local melting point is depressed — the M-effect — so a sustained mild overload clears without the whole element ever reaching the melting point of silver:
And note what the fusing factor does not tell you. It is entirely an overload property. The behaviour that made the fuse worth having in Problem 15 — the cut-off current and the let-through \(I^2t\) — belongs to the short-circuit end of the characteristic and is unrelated to \(FF\). A device can have an excellent fusing factor and no current-limiting capability whatever, and a miniature circuit breaker is exactly that.
A 245 kV breaker clears a fault 2 km down a line whose surge impedance is 350 Ω, the breaking current being 20 kA rms. Wave propagation is at \(3\times10^{8}\) m/s. Find \(di/dt\) at the current zero, the initial rate of rise of the line-side voltage and the peak of the sawtooth it produces, and confirm the peak independently from the line's power-frequency reactance. Compare the result with the 2 kV/µs terminal-fault envelope of the class and say why the short-line fault is a separate test duty.
The rate of change of current at its own zero is the slope of a sinusoid there, and it is the quantity the travelling wave responds to:
Nearly nine amperes per nanosecond. Note that this is proportional to the current, so a smaller short-line fault current gives a gentler slope — the severity and the current move together, unlike the terminal fault, where they move oppositely.
Why the line cannot be a lumped capacitance. The transit time to the fault is
Comparable with the whole restriking transient. A lumped model assumes information crosses the element instantly; here it takes a third of the time to peak, so the line must be treated as the distributed circuit of Chapter 12.
The line-side voltage. Interrupting the current launches a voltage wave into the line whose magnitude is the surge impedance times the rate at which the current is being removed:
It rises linearly until the wave reflects from the short-circuit at the fault and returns, at \(t = 2\tau\):
After which the reflected wave reverses the slope and the voltage falls, giving the characteristic sawtooth rather than the cosine of a terminal fault.
The independent check, from the power-frequency drop along the faulted section. The line's series inductance follows from its surge impedance:
The sawtooth reaches exactly twice the voltage that stood across the line section before interruption — the standard doubling at an open end. The two routes agree, which is the check worth doing whenever a travelling-wave result is quoted.
The total across the breaker. The source side is also rebuilding voltage over the same microseconds, so the gap sees the sum of the two. Chapter 35 quotes the combined initial slope in the conservative form
Why it is a separate duty. Set the two side by side:
A duty with a quarter of the peak and three times the slope. It therefore tests the thermal round of the interruption race — the microseconds in which residual plasma decides everything — while the terminal fault tests the dielectric round. Passing one is no evidence at all of passing the other, and a breaker's short-line fault rating is stated separately for that reason.
And the remedy is the one Problem 9 identified. Because the failure is on slope and not on peak, capacitance is the appropriate cure: a few nanofarads across the line-side terminal slows the initial rise without needing to absorb any energy. This is why line-side capacitors and grading capacitors are the standard fix for a marginal short-line fault performance, and opening resistors are not.
The breaker of Problem 1 has a contact speed of 5 m/s and reaches its current zero after 10 ms of arcing. Its SF6 gap withstands 25 kV/mm at the working pressure once cold. Taking \(k_{pp} = 1.3\) and an amplitude factor of 1.4, decide whether the gap length wins the dielectric round; then find the rate at which contact travel alone increases the gap's strength, compare it with the maximum RRRV, and state what actually wins the first microseconds.
The duty, corrected. From Problem 1, \(E_m = 179.63\) kV and \((dv/dt)_{\max} = 13.39\) kV/µs:
The gap at the current zero. The contacts have been travelling throughout the arcing period:
A margin of 3.8 times. The dielectric round — decided by the peak, tens or hundreds of microseconds after the zero — is won comfortably, and it is won by geometry that was already established before the current reached zero.
Now the rate at which that strength is still growing:
The restriking voltage climbs one hundred and seven times faster than contact travel can build strength. Over the first microsecond the gap grows by five micrometres — worth 0.125 kV — while the transient rises by 13 kV.
What speed would be needed if travel alone had to win:
A hundred times any real mechanism, and faster than sound in SF6. No breaker has ever been built that wins the thermal round mechanically, and none ever will.
So what does win it. In the first microseconds the gap is not a cold dielectric at all — it is filled with residual plasma left by the arc, and its conductance is finite. The race is between the power the recovering voltage feeds back into that plasma and the rate at which the interrupter removes heat and charge carriers from it:
Measured dielectric recovery rates for these mechanisms are of the order of tens of kilovolts per microsecond — two to three orders above the 0.125 kV/µs that travel provides, and comfortably above the 13.39 kV/µs of the duty.
The two rounds, and which number decides each:
Which is exactly why a TRV specification states two numbers and not one, and why the short-line fault of Problem 17 — all slope, little peak — attacks a different weakness from the terminal fault.
And what happens if the race is lost. The arc restarts, carries current for another half cycle and gets a second attempt at the next zero — by which time the contacts are 10 mm further apart and the gap is 250 kV stronger. A medium-voltage breaker may legitimately take two or three current zeros. The unrecoverable case is a repeated failure until the contacts are fully open with the arc still burning, because then no travel remains to be bought.
An SF6 breaker clears 25 kA rms with an arcing time of 12 ms, the measured arc voltage being 1.2 kV. The network inductance is 20 mH per phase. Find the arc energy, compare it with the magnetic energy stored in that inductance at the current crest, and hence explain quantitatively why direct-current interruption is the harder problem.
The arc energy is the arc voltage — nearly constant, because Ayrton's \(V_{arc} = A+B\ell+(C+D\ell)/I\) is dominated by the length terms at kiloampere currents — multiplied by the charge that passes:
Three hundred and twenty-four kilojoules deposited in the contacts and the gas in twelve milliseconds — a mean power of 27 MW. This is what erodes contacts and fixes the number of full-current operations before overhaul.
The magnetic energy at the current crest:
Thirty-nine times larger. If the breaker had to dissipate the network's stored energy, its arc energy would have to be forty times what it is.
And it does not have to, because of when it acts. The stored energy is \(\tfrac12Li^2\), and the interruption is attempted at \(i=0\):
The whole 12.5 MJ has already been returned to the source during the preceding quarter cycle. The alternating supply does the energy disposal for free, twice per cycle, and the breaker merely has to survive the voltage that follows. That is the entire content of the current-zero method.
Direct current has no such instant. To interrupt 25 kA DC in the same 20 mH:
Every joule of it has to pass through the arc, or through an absorbing element, because there is no current zero at which it can be handed back. A DC breaker must therefore either dissipate 6.25 MJ internally — which is the high-resistance method of Section 35-3, and impractical above a few hundred volts — or manufacture an artificial current zero.
How an artificial zero is made, and it is the basis of every HVDC breaker of Chapter 38: a pre-charged capacitor is discharged through the arc in opposition to the load current, forcing a momentary zero, and the energy is then diverted into a metal-oxide arrester stack sized for the full \(\tfrac12Li^2\):
A twenty-fold difference in the energy the device must own, which is why an HVDC breaker is a cabinet-sized assembly and an AC breaker of the same rating is a single interrupter.
A final consequence for maintenance. Contact erosion is proportional to \(W_{arc}\), so a breaker rated for, say, 20 full-current operations has a lifetime arc-energy budget of about 6.5 MJ. Load-current switching at 1250 A produces \(1200\times0.9\times1250\times0.010 = 13.5\) kJ per operation, twenty-four times less, which is why the same interrupter is permitted ten thousand load operations and twenty fault operations.
Write the switchgear specification for a 132 kV, 50 Hz, effectively earthed bus whose three-phase fault level is 3500 MVA and whose \(X/R\) is 15. Relay time is 20 ms, breaker opening time 40 ms, and backup protection clears in 1 s. Terminal capacitance is 0.02 µF. Determine every rating the breaker must meet, including the TRV, and check the offered unit 145 kV / 20 kA / 3 s row by row.
The symmetrical fault current, from which everything else descends:
The DC component at contact separation. The contacts part at relay time plus opening time:
A standard 30% DC declaration covers the 28.5% duty. Had the opening time been 30 ms rather than 40, \(\delta\) would have been 35% and a special declaration required — the sensitivity is steep.
Making, short-time and MVA:
The TRV duty. The fault level fixes the inductance, and the inductance with the stated capacitance fixes the transient:
The 1.3 rather than 1.5 because the bus is effectively earthed — the single cheapest reduction in TRV duty available, and it is a decision made in the earthing design, not in the switchgear order.
The specification, assembled:
The offered 145 kV / 20 kA / 3 s unit, row by row:
The short-time rating is the least binding, at 20% utilisation — which is typical, because 3 s ratings are set by the manufacturer's standard range rather than by any duty.
The one row to interrogate. The TRV rate of 7.87 kV/µs from the lumped model is well above the roughly 2 kV/µs a 145 kV standard envelope specifies, for the reason Problem 1 gave: the real bus carries several lines whose capacitance is not in the 0.02 µF. Ask the manufacturer for the type-test TRV envelope rather than accepting the lumped figure — and if the bus really is a single radial feed with no other circuits, the lumped figure is the honest one and a grading capacitor should be specified.
What has not been specified, and should be. The operating sequence (\(\text{O}-0.3\ \text{s}-\text{CO}-3\ \text{min}-\text{CO}\) if the feeder autorecloses), the insulation level for the 145 kV class, the rated normal current including load growth, and the small-current switching duties — unloaded transformer and line charging — which Problems 6 and 7 showed are specified separately and are the ones most likely to be omitted.
Practice Problems
Take 50 Hz throughout and treat the source voltage as constant at its peak over the duration of any restriking transient. Work each on paper before opening the answer; the answer is given so you can check yourself, and the method is deliberately not.
P1. A 132 kV breaker clears a terminal fault with \(L = 20\) mH and \(C = 0.025\) µF. Find \(\sqrt{LC}\), \(f_n\), the peak restriking voltage, the time to peak and the maximum RRRV.
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\(\sqrt{LC} = \mathbf{22.36}\) µs, \(f_n = \mathbf{7118}\) Hz, \(v_{peak} = \mathbf{215.6}\) kV at \(t = \mathbf{70.25}\) µs, \((dv/dt)_{\max} = \mathbf{4.82}\) kV/µs. Problem 1.P2. An oscillogram shows a peak restriking voltage of 300 kV reached 60 µs after the current zero. State the average and the maximum RRRV.
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Average \(300/60 = \mathbf{5.0}\) kV/µs; maximum \((\pi/2)(5.0) = \mathbf{7.85}\) kV/µs. Problem 3.P3. A circuit has \(L = 10\) mH and \(C = 0.04\) µF. Find its surge impedance and the resistance needed for critical damping.
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\(\sqrt{L/C} = \mathbf{500}\ \Omega\), \(R_c = \mathbf{250}\ \Omega\). Problem 8.P4. An interrupter chops at 6 A in a circuit whose surge impedance is 30 kΩ. What is the prospective overvoltage, ignoring the standing voltage?
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\(V = 6\times30\,000 = \mathbf{180}\) kV. Problem 6.P5. A grading capacitor quadruples the terminal capacitance. What happens to \(f_n\), the maximum RRRV, the time to peak and the peak itself?
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\(f_n\) halves, RRRV halves, \(t_{peak}\) doubles, and the peak is unchanged at \(2E_m\). Problems 5 and 9.P6. A 33 kV breaker is rated 20 kA. Give its breaking capacity in MVA and its rated making current at 50 Hz.
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\(S_{br} = \sqrt3(33)(20) = \mathbf{1143}\) MVA; \(i_p = 2.5(20) = \mathbf{50}\) kA peak. Problems 10 and 12.P7. Find the asymmetrical breaking current corresponding to 25 kA symmetrical with a 60% DC component.
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\(25\sqrt{1+2(0.36)} = 25\sqrt{1.72} = \mathbf{32.79}\) kA. Problem 10.P8. A breaker is rated 40 kA for 1 s. What current may it carry for 3 s?
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\(40/\sqrt3 = \mathbf{23.09}\) kA. Problem 13.P9. A 63 A fuse has a fusing factor of 1.6. What is its minimum fusing current, and what does the gap between the two figures mean physically?
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\(63\times1.6 = \mathbf{100.8}\) A. Between 63 and 100.8 A the fuse neither carries indefinitely without deterioration nor operates. Problem 16.P10. Find the \(I^2t\) withstand of a 25 mm² PVC-insulated copper cable, \(k = 115\).
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\((115\times25)^{2} = (2875)^{2} = \mathbf{8.27\times10^{6}}\) A²s. Problem 15.P11. By what factor does a critically damped shunt resistor reduce the maximum RRRV, and by what factor the peak?
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Slope by \(e = \mathbf{2.718}\); peak by 2, from \(2E_m\) to \(E_m\). Problems 8 and 9.P12. A 33 kV system with an isolated neutral suffers a three-phase fault. Give the ideal undamped TRV peak on the first pole to clear.
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\(E_m = 26.94\) kV, so \(1.5\times2\times26.94 = \mathbf{80.83}\) kV. Problem 4.
Challenge Problems
Three problems where the formula is not the difficulty. Each requires deciding which of the two governing quantities is in charge before any arithmetic can begin.
C1 — Reactor switching, and why it is not the transformer case. A 400 kV, 50 MVAr shunt reactor is switched out by an SF6 breaker that chops at 3 A. The terminal capacitance is 3 nF. Compute the chopping overvoltage in per unit, compare it with the 8.1 pu of the unloaded transformer in Problem 6, and identify what the real hazard of reactor switching is.
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The reactor's inductance. Per phase, \(Q = 50/3 = 16.67\) MVAr at \(V_{ph} = 400/\sqrt3 = 230.9\) kV:
\[ I = \frac{16.67\times10^{6}}{230.9\times10^{3}} = 72.17\ \text{A}, \qquad L = \frac{V_{ph}}{\omega I} = \frac{230\,940}{314.16\times72.17} = 10.19\ \text{H} \]The surge impedance and the crest:
\[ \sqrt{\frac{L}{C}} = \sqrt{\frac{10.19}{3\times10^{-9}}} = 58\,270\ \Omega, \qquad V = 3\times58\,270 = 174.8\ \text{kV} \]\[ E_m = \sqrt2(230.9) = 326.6\ \text{kV}, \qquad V_{\max} = \sqrt{326.6^{2}+174.8^{2}} = 370.4\ \text{kV} = \mathbf{1.13\ pu} \]Why it is so much milder than the transformer. The two cases differ in one factor only. A 400 kV reactor draws 72 A and so has \(L = 10.2\) H; an unloaded transformer's magnetising branch draws a fraction of an ampere and so has \(L = 30\) H at a fifth of the voltage. The reactor's \(E_m\) is six times larger while its surge impedance is only two-thirds as large, so the chopping term is small compared with the standing voltage rather than eight times it. The quadrature sum then barely moves.
\[ \frac{i_{ch}\sqrt{L/C}}{E_m} = \frac{174.8}{326.6} = 0.54 \quad\text{against}\quad \frac{433.0}{53.89} = 8.03\ \text{for the transformer} \]So what is the real hazard. Not the first chop, which 1.13 pu makes harmless. It is reignition. After the chop the reactor and its capacitance ring at
\[ f = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{10.19\times3\times10^{-9}}} = 910\ \text{Hz} \]and the voltage across the breaker becomes the difference between this ringing reactor voltage and the 50 Hz source — a quantity that reaches nearly \(2E_m\) within half a millisecond of the chop, while the contacts have barely moved. If the gap reignites, a high-frequency current flows, is chopped again at the next high-frequency zero, and the trapped charge is higher than before. Successive reignitions escalate the reactor voltage in steps, and a stack of three or four such steps reaches insulation-threatening levels from a first chop that was entirely benign.
The remedies follow from that diagnosis and not from the energy balance: open the contacts fast enough that the gap strength outruns the recovering voltage (a minimum arcing time is specified for reactor duty), use controlled or point-on-wave switching so the contacts part at a chosen instant, and fit surge arresters at the reactor terminals. Note that a surge capacitor, the correct answer for the transformer of Problem 6, does very little here — it lowers a chopping overvoltage that was never the problem.
The lesson. Two duties described by the same formula, and the dominant term is different in each. Compute \(i_{ch}\sqrt{L/C}\) and \(E_m\) and see which is larger before deciding what to install.
C2 — Designing the resistor stack. A 400 kV breaker clears a terminal fault in a circuit with \(L = 18\) mH and \(C = 0.05\) µF. Specify the opening resistor completely: its resistance, the current and power it carries, the energy it absorbs in a 10 ms insertion, and the mass of ceramic needed if the specific heat is 800 J/kg·K and a 400 K rise is allowed. State what the resistor achieves and what it costs.
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The two governing quantities:
\[ \sqrt{LC} = \sqrt{(0.018)(5\times10^{-8})} = 30.0\ \mu\text{s}, \qquad \sqrt{\frac{L}{C}} = \sqrt{\frac{0.018}{5\times10^{-8}}} = 600\ \Omega \]\[ E_m = \sqrt2\,\frac{400}{\sqrt3} = 326.6\ \text{kV}, \qquad R_c = \tfrac12(600) = \mathbf{300\ \Omega} \]What it achieves:
\[ \begin{array}{lcc} & R=\infty & R=R_c \\ \hline v_{peak} & 653.2\ \text{kV} & 326.6\ \text{kV} \\ (dv/dt)_{\max} & 10.89\ \text{kV}/\mu\text{s} & 4.00\ \text{kV}/\mu\text{s} \end{array} \]What it costs. The residual current and the power in the stack:
\[ \hat I_R = \frac{326.6\times10^{3}}{300} = 1088.7\ \text{A} \;\Rightarrow\; I_R = 769.8\ \text{A rms}, \qquad P_R = (769.8)^{2}(300) = 177.8\ \text{MW} \]\[ W = 177.8\times10^{6}\times0.010 = 1.78\ \text{MJ per operation, per phase} \]\[ m = \frac{1.78\times10^{6}}{800\times400} = 5.56\ \text{kg of ceramic per phase} \]The auxiliary interrupter's duty. The fault current the main contacts cleared was
\[ I_f = \frac{E_m}{\sqrt2\,\omega L} = \frac{326.6\times10^{3}}{\sqrt2(5.655)} = 40.84\ \text{kA rms} \quad\Rightarrow\quad \frac{769.8}{40\,840} = 1.88\% \]so the auxiliary break interrupts 770 A against 40.8 kA — a duty smaller by a factor of 53, and one a simple series contact handles. The implied fault level, \(\sqrt3(400)(40.84) = 28\,300\) MVA, is a strong 400 kV bus, so the design is self-consistent.
The engineering judgements hidden in the numbers. The 10 ms insertion is not free to choose: it must exceed the duration of the transient (a few hundred microseconds) by a wide margin and must be shorter than the interval before the auxiliary contacts can part cleanly. Lengthen it and the mass grows proportionally; shorten it and the resistor may still be carrying current when the transient has gone, achieving nothing. And 5.6 kg per phase is the thermal mass only — the mechanical assembly, the insulation and the auxiliary interrupter multiply the installed volume several times, which is the real reason resistance switching is confined to EHV.
What it does not do. A grading capacitor would still be required for voltage sharing between the four breaks a 400 kV interrupter uses, and the resistor does nothing for that. The two components coexist; they are not alternatives.
C3 — The breaker that passed its type test and destroyed a motor. A 6.6 kV vacuum contactor, type-tested at 25 kA, repeatedly causes inter-turn failures in a 1 MW motor (power factor 0.85, efficiency 0.94) when it is switched out while stalled. The interrupter chops at 5 A and the motor plus cable capacitance is 5 nF. Diagnose the failure, quantify it, and specify the remedy.
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The currents:
\[ I_{FL} = \frac{10^{6}}{0.85\times0.94\times\sqrt3\times6600} = 109.5\ \text{A}, \qquad I_{LR} = 6I_{FL} = 657\ \text{A} \]The inductance the stalled motor presents, which is the locked-rotor value and is far smaller than a magnetising inductance:
\[ L = \frac{V_{ph}}{\omega I_{LR}} = \frac{3810}{314.16\times657} = 18.46\ \text{mH} \]The first chop:
\[ \sqrt{\frac{L}{C}} = \sqrt{\frac{0.01846}{5\times10^{-9}}} = 1922\ \Omega, \qquad V = 5\times1922 = 9.61\ \text{kV} \]\[ E_m = \sqrt2(3810) = 5.39\ \text{kV}, \qquad V_{\max} = \sqrt{5.39^{2}+9.61^{2}} = 11.02\ \text{kV} = 2.04\ \text{pu} \]And 2.04 pu does not destroy a motor. The impulse level of 6.6 kV rotating plant is around 20 kV, so the first chop is inside it with margin. The diagnosis must therefore be something else, and it is: multiple reignition.
The escalation mechanism. A vacuum gap that has been open for only tens of microseconds is short, and the recovery voltage after the chop is the difference between the ringing motor voltage and the 50 Hz source. That difference reaches 2 pu within a quarter of the ringing period:
\[ f_{\text{ring}} = \frac{1}{2\pi\sqrt{(0.01846)(5\times10^{-9})}} = 16.6\ \text{kHz} \quad\Rightarrow\quad \text{crest in } 15\ \mu\text{s} \]The gap reignites, a high-frequency current of tens of amperes flows through the cable's surge impedance, the interrupter — which quenches high-frequency currents readily — chops it at its next zero, and the trapped voltage is higher than before. Three or four such steps take the terminal voltage to 4–6 pu, that is 22–33 kV, at the impulse level of the winding.
Why the damage is to the first turns. Each reignition front rises in a microsecond or less, so the surge does not distribute itself uniformly along the winding; the first few turns of the line-end coil take a disproportionate share, and it is there that the inter-turn failures are found. That distribution of damage is the diagnostic signature, and it distinguishes reignition escalation from a simple overvoltage, which stresses the whole winding to earth.
The remedy: an RC surge suppressor at the motor terminals, not merely a capacitor. Take \(C_s = 0.25\) µF:
\[ \sqrt{\frac{L}{C_s}} = \sqrt{\frac{0.01846}{2.5\times10^{-7}}} = 271.7\ \Omega \quad\Rightarrow\quad V = 5(271.7) = 1.36\ \text{kV},\quad V_{\max} = 5.56\ \text{kV} = 1.03\ \text{pu} \]\[ R_s \approx \tfrac12\sqrt{L/C_s} = 136\ \Omega \quad\text{— take a standard } 100\text{–}150\ \Omega \text{ element} \]The capacitor removes the chopping overvoltage and, more importantly, slows every reignition front so that the surge distributes evenly along the winding; the resistor damps the ringing so that the escalation ladder has no rungs. Both parts are necessary — a capacitor alone lowers the crest but leaves an undamped 16 kHz oscillation for the gap to reignite into.
The moral. The 25 kA type test certified the interrupter against the duty a fault study computes. The duty that destroyed the motor was five amperes, and no fault study ever mentions it. Small-current switching is a separate specification, and it is the one that is left off the order.
Multiple-Choice Questions
MCQ 1. The peak restriking voltage of an ideal undamped terminal fault is:
(a) \(E_m\) (b) \(\sqrt2 E_m\) (c) \(2E_m\) (d) \(3E_m\)Show answer
(c). The capacitance starts at zero and the source stands at \(E_m\), so it overshoots by as much as it started below. Note that (b) is the trap for anyone who confuses this transient with a peak-to-rms conversion — \(E_m\) is already a peak. Problem 1.MCQ 2. Quadrupling the capacitance across the breaker:
(a) halves the peak (b) halves the RRRV (c) halves both (d) changes neitherShow answer
(b). \(\sqrt{LC}\) doubles, so the slope halves and the time to peak doubles; the amplitude \(2E_m\) contains no circuit value at all. Problems 5 and 9.MCQ 3. The ratio of average to maximum RRRV in the lumped model is:
(a) \(1/2\) (b) \(2/\pi\) (c) \(1/\sqrt2\) (d) \(1/e\)Show answer
(b) \(= 0.637\), the mean of a half sine over its own half period. Distractor (d) is the factor by which critical damping reduces the maximum slope — a different quantity that appears two problems later. Problems 3 and 9.MCQ 4. The first-pole-to-clear factor for a system with an isolated neutral is:
(a) 1.0 (b) 1.3 (c) 1.5 (d) \(\sqrt3\)Show answer
(c). From \(k_{pp} = 3X_0/(X_1+2X_0)\) as \(X_0\to\infty\). Answer (b) is the effectively earthed value, \(9/7\); answer (d) is the healthy-phase rise on an earth fault, a different quantity from a different chapter. Problem 4.MCQ 5. The chopping overvoltage is proportional to:
(a) \(i_{ch}\sqrt{LC}\) (b) \(i_{ch}\sqrt{L/C}\) (c) \(i_{ch}^{2}L\) (d) \(i_{ch}/\sqrt{LC}\)Show answer
(b), from \(\tfrac12Li_{ch}^2 = \tfrac12CV^2\). Answer (c) is the energy, which is trivially small — that a duty is harmless in energy and lethal in voltage is the whole point of Problem 6.MCQ 6. The critical resistance for resistance switching is:
(a) \(\sqrt{L/C}\) (b) \(2\sqrt{L/C}\) (c) \(\tfrac12\sqrt{L/C}\) (d) \(\sqrt{LC}\)Show answer
(c). Setting \((1/RC)^2 = 4/LC\). Answer (d) is not even an impedance — it has the dimensions of time, which is the fastest way to eliminate it. Problem 8.MCQ 7. Increasing the shunt resistance above \(R_c\):
(a) increases the damping (b) reduces the damping (c) leaves the damping unchanged (d) makes the response aperiodicShow answer
(b). The damping term is \(1/RC\), so damping falls as \(R\) rises — the resistor is in parallel with the capacitance, not in series with the loop. The reversed intuition catches almost everyone once. Problem 8.MCQ 8. A breaker rated 25 kA has an asymmetrical breaking current, at a 50% DC component, of:
(a) 25 kA (b) 27.5 kA (c) 30.6 kA (d) 43.3 kAShow answer
(c) \(= 25\sqrt{1.5}\). Answer (d) is \(\sqrt3\times25\), the value at \(\delta=1\) — the theoretical maximum, reached only for a fully offset wave. Problem 10.MCQ 9. The rated making current of a 50 Hz breaker rated 20 kA is:
(a) 20 kA rms (b) 28.3 kA peak (c) 50 kA peak (d) 56.6 kA peakShow answer
(c) \(= 2.5\times20\). Answer (d) is \(2\sqrt2\times20\), the undecayed theoretical crest; answer (b) is a single symmetrical peak, which describes nothing on the nameplate. Problem 12.MCQ 10. A 25 kA, 3 s breaker faces a 30 kA fault cleared in 0.5 s. It is:
(a) suitable, since \(I^2t\) is within rating (b) unsuitable, since 30 exceeds 25 kA (c) suitable, since 0.5 s is short (d) suitable if the voltage is lowerShow answer
(b). The thermal duty is 450 kA²s against 1875 available, so (a) and (c) are both arithmetically true and both irrelevant — the breaker cannot interrupt 30 kA at any duration. Two independent checks. Problem 13.MCQ 11. A "500 MVA" 11 kV breaker moved to a 6.6 kV bus has a breaking capacity of:
(a) 500 MVA (b) 833 MVA (c) 300 MVA (d) unchanged in kA but undefined in MVAShow answer
(c) \(= \sqrt3(6.6)(26.24)\). Answer (d) is the more honest description of reality but the question asks for a number, and the number is 300. Answer (b) scales the wrong way. Problem 14.MCQ 12. A fuse limits current because:
(a) it operates faster than a relay (b) it melts before the first peak of the fault wave (c) its arc voltage exceeds the supply (d) the sand cools the arcShow answer
(b). Options (c) and (d) are true statements about how the fuse then extinguishes the arc, but the current limiting itself is a matter of timing — the element parts within two or three milliseconds, before the crest at five. Problem 15.
Key Formulas
| Statement | Relation | Notes |
|---|---|---|
| Why an arc is unavoidable | \(v_L = L\,di/dt\), \(W = \tfrac12Li^2\) | Zero fall time implies infinite voltage |
| Arc voltage (Ayrton) | \(V_{arc} = A + B\ell + (C+D\ell)/I\) | Falls as \(I\) rises — a negative resistance |
| Restriking voltage | \(v = E_m(1-\cos\omega_n t)\), \(\omega_n = 1/\sqrt{LC}\) | Lumped terminal fault, undamped |
| Peak and time to peak | \(v_{peak} = 2E_m\) at \(t = \pi\sqrt{LC}\) | Peak independent of \(L\) and \(C\) |
| Natural frequency | \(f_n = 1/(2\pi\sqrt{LC}) = 1/(2t_{peak})\) | Typically 3–20 kHz |
| Maximum and average RRRV | \((dv/dt)_{\max} = E_m/\sqrt{LC}\); average \(= (2/\pi)\) of it | \(2/\pi = 0.637\), exact |
| Standard TRV envelope | \(u_c = k_{pp}k\,E_m\), reached in \(t_3\) | \(k = 1.3\text{–}1.7\) amplitude factor |
| First pole to clear | \(k_{pp} = 3X_0/(X_1+2X_0)\) | 1.5 isolated, \(9/7 \approx 1.3\) effectively earthed |
| Two breaks in series | \(V_A/V_B = 1 + C_e/C_g\) | Why grading capacitors exist |
| Chopping overvoltage | \(V = i_{ch}\sqrt{L/C}\); \(V_{\max} = \sqrt{v_0^2 + i_{ch}^2L/C}\) | From \(\tfrac12Li_{ch}^2 = \tfrac12CV^2\) |
| Critical resistance | \(R_c = \tfrac12\sqrt{L/C}\) | Shunt, so smaller \(R\) damps more |
| Critically damped response | \(v = E_m[1-(1+\omega_n t)e^{-\omega_n t}]\) | Peak \(E_m\); slope max \(E_m\omega_n/e\) at \(t=1/\omega_n\) |
| Short-line fault | \((du/dt)_{\text{line}} = Z_0\,di/dt\), \(di/dt = \sqrt2\omega I\) | Sawtooth peaks at \(2\ell/v\); Chapter 35 quotes \(2Z_0\,di/dt\) across the breaker |
| Breaking capacity | \(S_{br} = \sqrt3\,V_r I_{sc}\) MVA | Fictitious as a power; scales with \(V\) |
| Asymmetrical breaking current | \(I_{asym} = I_{ac}\sqrt{1+2\delta^2}\) | \(\delta=1 \Rightarrow \sqrt3 I_{ac}\); \(\delta = e^{-t/T}\), \(T = (X/R)/\omega\) |
| Making current | \(i_p = 2.5I_{sc}\) (50 Hz), \(2.6I_{sc}\) (60 Hz) | \(1.8\sqrt2 = 2.55\) classical; a peak, not an rms |
| Short-time equivalence | \(I_1^2t_1 = I_2^2t_2\) | Adiabatic below about 1 s |
| Fusing factor | \(FF = I_2/I_n\) | 1.25–1.45 HRC, 1.8–2.0 rewirable |
| Cable protection criterion | \((I^2t)_{\text{let-through}} \le k^2S^2\) | \(k = 115\) PVC-Cu, 143 XLPE-Cu, 76 PVC-Al |
| Overload coordination | \(I_b \le I_n \le I_z\) and \(I_2 \le 1.45I_z\) | The second is what a large \(FF\) violates |
Common Mistakes
Comparing an average RRRV with a maximum RRRV. They differ by \(\pi/2 = 1.571\), and a specification written one way against a test result quoted the other over-states or under-states the duty by 57% — Problem 3.
Expecting a grading capacitor to lower the peak. Capacitance stretches the time axis only; the peak is \(2E_m\) whatever the circuit. Only a resistor, which dissipates, moves the peak — Problems 5 and 9.
Reversing the resistance-switching inequality. Smaller \(R\) gives more damping, because the resistor is in parallel with \(C\). Writing \(R_c = 2\sqrt{L/C}\) or treating \(R > R_c\) as safer inverts the design — Problem 8.
Using \(\sqrt{LC}\) where \(\sqrt{L/C}\) belongs. The first is a time and governs the restriking transient; the second is an impedance and governs chopping and the critical resistor. Check the dimensions before substituting — Problems 6 and 8.
Judging a chopping duty by its energy. Three hundred and seventy-five joules destroys a 66 kV transformer, because the danger is 436 kV of voltage and not the energy behind it. The arrester is sized for volts, and its energy duty is trivial — Problem 6.
Assuming the largest fault current is the severest duty. A short-line fault at 90% of the current has three times the slope, and a five-ampere magnetising current produces eight per unit. Both are outside the fault study altogether — Problems 6 and 17.
Reading a symmetrical rating and stopping. 18 kA against a 20 kA breaker passes; 40% DC against a 30% declaration does not, and it is the second row that decides — Problem 11.
Treating the short-time rating and the breaking rating as one check. A 30 kA, 0.5 s duty is thermally comfortable inside a 25 kA, 3 s breaker and is nonetheless fatal to it — Problem 13.
Comparing MVA ratings across voltage classes. "500 MVA exceeds 350 MVA" is true and useless: at 6.6 kV the breaker is a 300 MVA device facing a 30.6 kA duty. Divide by \(\sqrt3V\) first, always — Problem 14.
Assuming a breaker protects a cable that a fuse protects. The cable of Problem 15 needs clearance in 1.26 cycles, which no mechanical device achieves; the fuse passes 418 times less \(I^2t\) because it acts before the first peak — Problem 15.
Believing contact speed wins the interruption. Travel builds gap strength at 0.125 kV/µs against an RRRV of 13 kV/µs. The first microseconds are won by deionisation; travel wins only the late dielectric round — Problem 18.
Applying \(k_{pp} = 1.3\) to an isolated-neutral system. The correct 1.5 raises the TRV peak by 15% and the slope with it, and a breaker type-tested at 1.3 has no evidence for the harder duty — Problem 4.
The breaker is now specified as a device: it will interrupt 15 kA symmetrical, survive 38 kA peak on closure, absorb 234 kA²s while the protection thinks, and outrun a transient that reaches 196 kV in fifty microseconds. What it will not do is decide when to operate. Every number in this set was conditional on a trip signal arriving at a stated instant — the 60 ms of Problem 20, the 1 s backup of Problem 13, the 25 ms the cable of Problem 15 could not be given.
Set 37 supplies that decision. Overcurrent and earth-fault relays with their inverse-time characteristics and grading margins, distance relays whose reach is set from the line impedance of Part 3, differential schemes and the current transformers that either serve them or saturate and mislead them. The division is clean and it is worth carrying: a relay failure is a wrong decision, a breaker failure is a decision that could not be executed, and the backup schemes of Chapter 36 are built around exactly that distinction.