Part 8 · Chapter 35

Switchgear and Circuit Breakers

A circuit breaker cannot simply pull its contacts apart and stop the current — the inductance of the network will not allow it — so every interruption is a short, violent negotiation with an arc, won or lost in the few microseconds after the current passes through zero.

Electric Power Systems Prof. Mithun Mondal Reading time ≈ 50 min
i What you'll learn
  • Why an arc is necessary rather than accidental, what sustains it, and why its voltage–current characteristic falls.
  • The two extinction strategies — high-resistance and current-zero interruption — and why every high-voltage breaker uses the second.
  • How the restriking voltage of a simple \(L\!-\!C\) circuit works out to \(E_m(1-\cos\omega_n t)\), giving a peak of \(2E_m\) and a maximum RRRV of \(E_m/\sqrt{LC}\).
  • Why current chopping converts \(\tfrac12 L i_{ch}^2\) into a surge of \(i_{ch}\sqrt{L/C}\), and how a shunt resistor of \(\tfrac12\sqrt{L/C}\) removes the overshoot entirely.
  • What each entry on a breaker nameplate means: rated symmetrical and asymmetrical breaking current, making current, short-time current, breaking capacity in MVA.
  • How air-break, air-blast, oil, vacuum and SF6 breakers each solve the same deionisation problem differently, and where each belongs.
  • What a fuse does that a breaker cannot, and the one rule that governs every isolator operation.
Section 35-1

What Switchgear Has to Do

Everything from Chapter 21 to Chapter 25 was arithmetic in service of a single decision. The sequence networks, the \(Z\)-bus, the subtransient reactances and the DC offset all existed to answer one question: how much current flows when this point of the network is faulted, and how much of it is still flowing when the contacts part? Chapter 25 ended with that number written on a nameplate. This chapter opens the nameplate and asks how the device behind it actually does the job.

Switchgear is the collective name for the apparatus that connects, disconnects and protects a circuit: circuit breakers, isolators, load-break switches, fuses, earthing switches, the instrument transformers of Chapter 36, the busbars they sit on and the enclosure around them. Within that family the circuit breaker is the only device that can make and break a circuit carrying fault current, on command, and repeatedly. Everything else either carries current, or breaks it once, or breaks it only when there is nothing worth breaking.

A breaker has three separate duties, and they conflict.

It must carry rated current continuously without its contacts or joints exceeding their temperature-rise limit — an argument about contact resistance and heat transfer that favours large, firmly pressed contacts. It must withstand fault current for the time the protection takes to decide, absorbing an \(I^2 t\) that may be a thousand times the rated \(I^2\) for a second — an argument about thermal mass and about the electromagnetic forces that try to blow the contacts apart at the first peak. And it must interrupt that current within a few cycles of being told to — an argument that favours light, fast-moving contacts with a long travel. No design satisfies all three comfortably; every breaker is a compromise among them.

The third duty is the hard one, and the reason is a single equation from Chapter 6. The network between the source and the fault is dominantly inductive, so

Why the current cannot simply be stopped
\[ v_L = L\,\frac{di}{dt}, \qquad W = \tfrac12 L i^2 \]

If a breaker could force \(i\) to zero in zero time, \(di/dt\) would be infinite and so would the voltage across the opening gap. Equivalently, the magnetic energy \(\tfrac12 Li^2\) stored in the network — for \(20\) kA in \(20\) mH, some \(4\) megajoules — has to be disposed of, and it cannot vanish. Something must either dissipate it or convert it. The arc that appears between the parting contacts is not a failure of the breaker; it is the controlled channel through which the network is persuaded to give up its energy at a survivable rate. Designing a breaker is therefore not the art of preventing an arc but the art of making one and then destroying it on schedule.

The relay decides, the breaker acts. Chapter 36 builds the intelligence that detects a fault and issues the trip signal; this chapter builds the muscle that executes it. The division matters in practice because the two fail differently — a relay failure is a wrong decision, a breaker failure is a decision that could not be carried out — and the backup schemes of Section 36-2 are designed around exactly that distinction.
Section 35-2

The Arc: What Makes It and What Feeds It

Follow the contacts as they separate. Long before any visible gap exists, the contact area has shrunk to a few microscopic bridges. The whole current crowds through them, the current density becomes enormous, and the bridges melt and then boil. What now fills the opening gap is not air but metal vapour at several thousand kelvin, and that vapour is already partly ionised.

Two mechanisms then keep the channel conducting. Thermal ionisation dominates in the body of the arc: at \(5000\)–\(20\,000\) K the gas molecules collide hard enough to strip electrons, and the free electrons accelerated by the field ionise still more. Field emission dominates at the cathode: the gap is short and the voltage across it, though small, gives a field of the order of \(10^6\) V/m at the contact surface, which pulls electrons directly out of the metal. Between them these supply the charge carriers the current needs.

Working against them are recombination of ions and electrons — favoured by high pressure, which brings the carriers together — and diffusion of charged particles out of the arc column into cooler surroundings. Everything a breaker does to extinguish an arc is a way of tipping this balance: cool the column, raise the pressure, blow the ionised gas away, replace it with an unionised medium, or lengthen the path so that the same voltage produces a weaker field.

The arc's terminal behaviour is peculiar and worth stating precisely, because it explains several later results. The voltage across an established arc consists of a cathode drop of roughly \(10\)–\(20\) V, an anode drop of a few volts, and a column drop proportional to length. Empirically the total obeys the Ayrton equation:

Arc voltage and arc resistance
\[ V_{arc} = A + B\,\ell + \frac{C + D\,\ell}{I}, \qquad R_{arc} = \frac{V_{arc}}{I} = \frac{A + B\ell}{I} + \frac{C+D\ell}{I^{2}} \]

Here \(\ell\) is the arc length and \(A,B,C,D\) are constants of the electrode material and the surrounding medium. The important feature is the sign: \(V_{arc}\) falls as \(I\) rises. An arc is a negative-resistance element. Push more current through it and it gets hotter, more fully ionised, and less resistive, so the voltage needed to sustain it drops. This is why an arc, once struck, is stable and will not extinguish itself, and why an arc in series with a purely resistive circuit cannot be controlled by voltage alone.

Reading the same equation the other way gives the recipe for killing it. Arc resistance rises when \(\ell\) rises (a longer arc), when the current is forced down, when the column is cooled so that thermal ionisation weakens, and when the cross-section is constricted so the same current must flow through less ionised gas. Those four levers are, in order, arc lengthening, series-resistance insertion, cooling, and splitting or constriction — and every breaker in Section 35-8 uses some combination of them.

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The arc is a negative resistance
\(V_{arc}\) decreases as \(I\) increases, so an arc cannot be extinguished by reducing the voltage across it; the current must be attacked instead, either by raising \(R_{arc}\) until the source cannot sustain it, or by exploiting the instant at which the current is already zero.

The arc also dissipates real power \(v_{arc}i\) throughout its life. Integrated over the arcing time this is the arc energy, and it is what erodes contacts, decomposes oil, and sets the number of full-current operations a breaker may perform before overhaul.

Section 35-3

Two Ways to Kill an Arc

Only two strategies exist, and the choice between them is settled by whether the current ever reaches zero on its own.

The high-resistance method attacks the arc directly. Lengthen it, cool it, constrict it and split it into a series string of short arcs, each with its own cathode and anode drop. The arc resistance climbs until the voltage the arc demands exceeds the voltage the source can supply, and the current falls smoothly to zero. Splitter plates in an air-break breaker do this: a stack of steel plates draws the arc in magnetically and cuts one long arc into a dozen short ones whose cathode drops add to perhaps \(200\) V, which at \(415\) V is enough to choke the arc. The method is simple and needs no timing, but all the energy of the arc is dissipated inside the breaker. That is tolerable at \(415\) V and impossible at \(400\) kV. It is also the only method available for direct current, which is precisely why DC circuit breaking is so much harder than AC and why the HVDC schemes of Chapter 38 avoid it wherever they can.

The low-resistance or current-zero method exploits a gift that alternating current makes twice per cycle. At \(50\) Hz the current is genuinely zero every \(10\) ms, and at that instant the arc is momentarily starved: no current, no ohmic input, no fresh ionisation. The energy input to the arc column has already been falling for the previous quarter cycle, and around the zero the column cools rapidly. If in the microseconds that follow the gap can regain insulating strength faster than the circuit rebuilds voltage across it, the arc never restarts and the interruption is complete. The breaker's job is reduced to keeping the arc voltage low and harmless for up to a cycle, and then winning a short race at the current zero.

That race has two rounds, and a breaker can lose either. In the first few microseconds the gap still contains hot residual plasma and a small post-arc current flows. If the power fed back into that plasma by the rising voltage exceeds the rate at which the gap can carry heat away, the column re-heats and the arc restarts: this is thermal reignition, and it is governed by the rate of rise of voltage. Survive that and the gap becomes a cold dielectric, but a dielectric whose strength is still building as the contacts continue to travel apart. If the voltage crest arriving tens or hundreds of microseconds later exceeds that strength, the gap breaks down again: this is dielectric restrike, and it is governed by the peak of the voltage rather than its slope.

t (µs) v 2Eₘ Eₘ restriking voltage gap strength — fast recovery ⇒ cleared slow recovery reignition 0 current zero + t
Interruption is a race between the recovering gap strength and the restriking voltage
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The interruption criterion
The arc is extinguished at a current zero if and only if the dielectric strength of the gap exceeds the restriking voltage at every instant after that zero.

Two quantities therefore describe the duty the circuit imposes: the rate of rise of the restriking voltage, which decides the thermal round, and its peak, which decides the dielectric round. Section 35-4 computes both.

If the breaker loses the race it does not usually explode. The arc simply restarts, carries current for another half cycle, and gets a second chance at the next zero — by which time the contacts are further apart and the gap is stronger. A well-designed medium-voltage breaker may take two or three current zeros to clear. What it must not do is fail repeatedly until the contacts are fully open with the arc still burning, because then there is no more travel left to buy and the fault energy has nowhere to go but into the breaker itself.

Section 35-4

Restriking Voltage, Recovery Voltage and RRRV

Three terms are used loosely in conversation and must be kept apart in analysis.

The restriking voltage is the transient voltage that appears across the breaker contacts in the instants immediately after the arc current reaches zero. It is a high-frequency oscillation, it lasts a few hundred microseconds at most, and it is what the gap must withstand. Modern standards call the same thing the transient recovery voltage or TRV. The recovery voltage is the power-frequency voltage that remains across the open contacts once that transient has died away — for a terminal fault it is essentially the system voltage. And the rate of rise of restriking voltage, universally written RRRV, is the slope \(dv/dt\) of the transient, quoted in kV/µs.

To see where the transient comes from, reduce the network to the simplest circuit that contains the essential physics. Between the source and the breaker there is series inductance \(L\) — the source, transformer and line reactance the fault current flows through. Across the breaker terminal to earth there is capacitance \(C\) — busbar capacitance, bushing capacitance, the winding capacitance of transformers, the stray capacitance of the switchgear itself. Before interruption the arc short-circuits that capacitance and holds its voltage near zero. At the current zero the arc vanishes, and the capacitance must now charge up to whatever the source is offering. It does so through \(L\), and the result is an oscillation.

the circuit at current zero e = Eₘ sin ωt L breaker C v(t) fault at the current zero the source voltage is at its peak Eₘ, and C starts from v = 0, dv/dt = 0 v(t) = Eₘ(1 − cos ωₙt) 2Eₘ Eₘ slope = Eₘ/√(LC) = max RRRV π√(LC)
The lumped L–C model of a terminal fault and the restriking voltage it produces

Write Kirchhoff's voltage law around the loop after interruption. The current now flowing is the charging current of the capacitance, \(i = C\,dv/dt\), so

Derivation of the restriking voltage
\[ e(t) = L\frac{di}{dt} + v = LC\,\frac{d^{2}v}{dt^{2}} + v \]
\[ \text{with}\quad v(0)=0 \quad\text{(the arc held it there)}, \qquad \left.\frac{dv}{dt}\right|_{0} = \frac{i(0)}{C} = 0 \quad\text{(current zero)} \]

Two observations make the solution immediate. First, because the fault current is limited by \(L\) it lags the source voltage by very nearly \(90^\circ\); at the instant the current passes through zero the source voltage is therefore at its peak \(E_m\). Second, the transient we are about to compute lasts tens of microseconds while the power-frequency source takes \(10\) ms to move appreciably, so over the whole transient \(e(t)\) may be treated as the constant \(E_m\). The equation becomes a step-driven undamped oscillator:

Solution
\[ LC\,\frac{d^{2}v}{dt^{2}} + v = E_m \qquad\Longrightarrow\qquad v(t) = E_m\big(1 - \cos\omega_n t\big), \qquad \omega_n = \frac{1}{\sqrt{LC}} \]
\[ \frac{dv}{dt} = E_m\,\omega_n \sin\omega_n t \]

Everything the switchgear engineer needs is contained in those two lines. The capacitance overshoots the source voltage by exactly as much as it started below it, so the peak restriking voltage is \(2E_m\) — twice the crest of the phase voltage, reached after half a period of the natural oscillation. The slope is greatest where the cosine crosses its own mean, at \(\omega_n t = \pi/2\), and there the transient is climbing at \(E_m\omega_n\). An "average RRRV" is often quoted as well: the peak divided by the time taken to reach it.

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The four numbers of a terminal-fault TRV
\[ f_n = \frac{1}{2\pi\sqrt{LC}}, \qquad v_{peak} = 2E_m, \qquad t_{peak} = \pi\sqrt{LC} \]
\[ \left(\frac{dv}{dt}\right)_{\max} = \frac{E_m}{\sqrt{LC}} = E_m\,\omega_n, \qquad \left(\frac{dv}{dt}\right)_{avg} = \frac{2E_m}{\pi\sqrt{LC}} = \frac{2}{\pi}\left(\frac{dv}{dt}\right)_{\max} \]

The average is \(2/\pi = 0.637\) of the maximum, which is why a specification written in terms of one must never be compared directly with a test result quoted in terms of the other.

Three refinements convert this idealisation into something a standard would recognise.

Damping. Real circuits have resistance, so the overshoot never quite reaches \(2E_m\). The peak is written \(v_{peak} = k\,E_m\) with an amplitude factor \(k\) between about \(1.3\) and \(1.7\), the smaller values belonging to circuits with cable or several lines connected at the bus.

The first pole to clear. A three-phase breaker does not clear all three phases at the same instant; the poles reach a current zero at different times. The pole that clears first has to hold off more than the phase voltage, because the other two phases are still connected through the fault and drag the neutral. The multiplier is the first-pole-to-clear factor \(k_{pp}\), equal to \(1.5\) for a system with an isolated or high-impedance-earthed neutral and \(1.3\) for an effectively earthed system, the difference being exactly the neutral-shift argument of Chapter 23.

Short-line fault. If the fault is a kilometre or two down the line rather than at the breaker terminals, the line behaves as the distributed circuit of Chapter 12, not as a lumped capacitance. Travelling waves reflect from the fault and return, so the line-side voltage is a sawtooth whose initial slope is \(2 Z_0\,di/dt\) with \(Z_0\) the surge impedance. That slope is far steeper than the terminal-fault value even though the current is smaller, which is why the short-line fault is a separate and often decisive test duty.

The lumped model overstates the slope, and usefully so. A single \(L\!-\!C\) pair gives one natural frequency; a real substation with several lines and transformers on the bus has many, and the extra capacitance slows the rise. A calculation like Example 1 will typically return an RRRV two or three times the value a standard specifies for the same voltage class. Treat it as an upper bound that shows which parameters matter — halve the inductance and the slope halves; quadruple the capacitance and it halves again — rather than as a number to put in a purchase specification.
Section 35-5

Current Chopping

The whole argument of Section 35-4 rested on the breaker waiting politely for a natural current zero. A breaker with a very powerful arc-quenching mechanism does not always wait. When the current to be interrupted is small — the magnetising current of an unloaded transformer, a few amperes; the current of a shunt reactor; a lightly loaded motor — the cooling applied to the arc is out of all proportion to the energy the arc is receiving. The arc becomes unstable, and the current is forced abruptly to zero some microseconds before its natural zero. This is current chopping, and the value \(i_{ch}\) at which it happens is a property of the interrupter, not of the circuit.

Chopping is a problem because it violates the premise that made interruption gentle. At the moment of the chop the inductance is carrying \(i_{ch}\) and holds magnetic energy \(\tfrac12 L i_{ch}^{2}\). The current is gone in an instant, so that energy has nowhere to go except into the capacitance across the interrupter. Equating the two stores gives the overvoltage directly.

Energy balance at a chop
\[ \tfrac12 L\,i_{ch}^{2} \;=\; \tfrac12 C\,V^{2} \qquad\Longrightarrow\qquad V = i_{ch}\sqrt{\frac{L}{C}} \]
\[ \text{and if } C \text{ already stands at } v_0 \text{ when the chop occurs,}\qquad V_{\max} = \sqrt{\,v_0^{2} + i_{ch}^{2}\,\frac{L}{C}\,} \]

The dangerous factor is \(\sqrt{L/C}\), the surge impedance of the trapped circuit. For a loaded feeder it is a few hundred ohms and the overvoltage is negligible. For an unloaded transformer or a reactor, where \(L\) is the magnetising inductance — tens of henries — and \(C\) is only the winding and bushing capacitance of a few nanofarads, \(\sqrt{L/C}\) can exceed \(50\) kΩ. A chop of four or five amperes then produces hundreds of kilovolts across equipment rated for a fraction of that, as Example 3 works out. The failure that follows is usually not in the breaker but in the transformer winding it was switching.

Three defences are used, and they are used together. Contact materials are chosen to chop as gently as possible: the chromium–copper alloys of a modern vacuum interrupter chop at \(3\)–\(5\) A where the pure-copper contacts of early designs chopped at \(10\)–\(15\) A. Surge arresters are installed at the terminals of the switched equipment, clipping the transient at a level the insulation can accept. And the resistance switching of the next section damps the oscillation before it can develop a crest.

The gentlest duty is the most dangerous one. A breaker asked to clear \(25\) kA is working within a duty it was type-tested for. The same breaker asked to switch out an unloaded \(20\) MVA transformer at four amperes is doing something the fault-level calculation of Chapter 25 never considered, and it is the switching operation most likely to destroy a winding. Chopping, capacitor-bank restrike and reactor switching are all small-current, high-consequence duties, and they are specified separately from the breaking capacity for exactly that reason.
Section 35-6

Resistance Switching and the Critical Resistor

Both problems so far — a restriking voltage that overshoots to \(2E_m\), and a chopping surge of \(i_{ch}\sqrt{L/C}\) — have the same root cause: an undamped \(L\!-\!C\) oscillation. The cure is therefore the same in both cases, and it is the oldest trick in switchgear. Connect a resistor \(R\) permanently across the breaker's main contacts, and let it stay in circuit for the few milliseconds between the main contacts parting and a small auxiliary interrupter opening in series with the resistor. This is resistance switching.

What the resistor does can be read straight out of the differential equation. After the main arc is extinguished the source current divides between the capacitance and the resistor, \(i = C\,dv/dt + v/R\), and the loop equation becomes

The damped restriking-voltage equation
\[ E_m = L\frac{d}{dt}\!\left(C\frac{dv}{dt} + \frac{v}{R}\right) + v \qquad\Longrightarrow\qquad \frac{d^{2}v}{dt^{2}} + \frac{1}{RC}\,\frac{dv}{dt} + \frac{v}{LC} = \frac{E_m}{LC} \]

This is the standard second-order form of Chapter 27's swing equation and of every damped oscillator: the natural frequency is unchanged at \(\omega_n = 1/\sqrt{LC}\), and the damping term is \(1/RC\). The response is oscillatory only while the damping term is smaller than \(2\omega_n\). Setting the two equal locates the boundary.

The critical resistance
\[ \left(\frac{1}{RC}\right)^{2} = \frac{4}{LC} \qquad\Longrightarrow\qquad R^{2} = \frac{L}{4C} \qquad\Longrightarrow\qquad \boxed{\;R_{c} = \frac12\sqrt{\frac{L}{C}}\;} \]

A shunt resistance equal to half the surge impedance of the interrupting circuit makes the response critically damped. Below that value the response is overdamped and slower still; above it the oscillation reappears. Note the direction of the inequality carefully: smaller resistance gives more damping, because the resistor is in parallel with the capacitance, not in series with the loop.

At exactly \(R = R_c\) the two roots of the characteristic equation coincide at \(-\alpha\) with \(\alpha = 1/(2RC)\), and substituting \(R_c\) gives \(2R_cC = \sqrt{L/C}\cdot C = \sqrt{LC}\), so \(\alpha = \omega_n\) exactly. The response and its slope are then

Critically damped response
\[ v(t) = E_m\Big[1 - (1+\omega_n t)\,e^{-\omega_n t}\Big], \qquad \frac{dv}{dt} = E_m\,\omega_n^{2}\,t\,e^{-\omega_n t} \]
\[ \frac{d^{2}v}{dt^{2}} = 0 \;\text{ at }\; t = \frac{1}{\omega_n} \qquad\Longrightarrow\qquad \left(\frac{dv}{dt}\right)_{\max} = \frac{E_m\,\omega_n}{e} = \frac{1}{e}\left(\frac{dv}{dt}\right)_{\max,\;R=\infty} \]
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What the critical resistor buys
With \(R = \tfrac12\sqrt{L/C}\) the restriking voltage rises monotonically to \(E_m\) instead of overshooting to \(2E_m\), and the maximum RRRV falls by a factor \(e = 2.718\) — both rounds of the race in Section 35-3 are won at once.

The price is a residual current \(v/R\) which the auxiliary contacts must then interrupt. Because \(R\) is enormous compared with the fault-limiting reactance \(\omega L\), that residual current is a fraction of a per cent of the fault current, and the second interruption is trivial by comparison.

ωₙt v/Eₘ 1 2 2468 R = ∞ (no resistor): peak 2Eₘ R = ½√(L/C): critical R = ¼√(L/C): overdamped
A shunt resistor at half the surge impedance removes the overshoot and cuts the peak RRRV by a factor e

Resistance switching is not free and is not universal. The resistor must absorb real energy for the few milliseconds it is in circuit, so on an EHV breaker it is a substantial stack of ceramic elements with its own thermal rating and its own auxiliary interrupter to maintain. Where the duty permits, designers prefer the cheaper relative: a grading capacitor connected across each interrupting chamber. Its first job is to share the recovery voltage equally among the two or four breaks in series, since without it the stray capacitance to earth would load them unequally and the outermost break would take most of the voltage. Its second job is to add to \(C\) in the formulae above, lowering \(\omega_n\) and therefore lowering the RRRV. Example 2 quantifies the second effect.

Section 35-7

What the Ratings Mean

Chapter 25 produced the fault currents; the nameplate is where they are compared with what the breaker can do. Every entry answers a specific physical question, and a breaker is suitable only if every entry clears its corresponding duty.

RatingWhat it limitsChecked against
Rated voltage \(U_r\) (kV rms, line)Insulation and recovery-voltage withstandHighest system voltage, not nominal
Rated insulation level (kV BIL / power-frequency)Withstand of lightning and switching surgesInsulation coordination study (Chapter 37)
Rated normal current (A rms)Continuous temperature rise of contacts and jointsMaximum load current including future growth
Rated short-circuit breaking current \(I_{sc}\) (kA rms)Arc energy the interrupter can extinguishSymmetrical fault current at contact separation
Rated DC component (%)Asymmetry the interrupter tolerates at that instantNetwork \(X/R\) and breaker opening time
Rated short-circuit making current \(i_p\) (kA peak)Contact welding and electromagnetic repulsion on closurePeak of the fully offset first loop
Rated short-time withstand current \(I_k / t_k\)Thermal \(I^2t\) while the fault persistsFault current × total clearing time
Rated TRV envelope \((u_c, t_3)\)Peak and rate of rise the gap can outrunRRRV of the network (Section 35-4)
Rated operating sequenceMechanical and thermal recovery between operationsAutoreclose policy of the feeder

Four of these deserve their arithmetic written out.

Breaking capacity in MVA. Historically a breaker was rated not in kiloamperes but in megavolt-amperes, the product of the rated line voltage and the rated breaking current in three-phase form. The number is fictitious as a power — no real \(3\phi\) apparatus delivers it — but it is the quantity a fault-level study naturally produces, and it survives in specifications and in examination questions.

Breaking capacity
\[ S_{br}\;[\text{MVA}] = \sqrt{3}\;V_{r}\,[\text{kV}]\;\times\;I_{sc}\,[\text{kA}] \]

Symmetrical and asymmetrical breaking current. Chapter 21 showed that the fault current in the first cycles is an AC component plus a decaying DC offset. The rms of that combination, evaluated at the instant of contact separation, is the asymmetrical breaking current. Writing the DC component as a fraction \(\delta\) of the AC peak, so \(I_{dc} = \delta\sqrt2 I_{ac}\),

Asymmetrical from symmetrical
\[ I_{asym} = \sqrt{I_{ac}^{2} + I_{dc}^{2}} = I_{ac}\sqrt{1 + 2\delta^{2}} \]
\[ \delta = 0.5 \;\Rightarrow\; I_{asym} = 1.225\,I_{ac}; \qquad \delta = 1 \;\Rightarrow\; I_{asym} = \sqrt3\,I_{ac} = 1.732\,I_{ac} \]

Making current. Closing onto an existing fault is a purely mechanical and magnetic duty, decided by the crest of the very first loop of a fully offset wave. The theoretical crest is \(2\sqrt2 I_{ac} = 2.83 I_{ac}\), but the DC offset decays measurably during the first half cycle, so a doubling factor of \(1.8\) rather than \(2\) is used, giving \(1.8\sqrt2 = 2.55\). IEC 62271-100 rounds this to \(2.5\) at \(50\) Hz and \(2.6\) at \(60\) Hz, the shorter \(60\) Hz half cycle allowing less decay.

Making current
\[ i_{p} = 2.5\,I_{sc}\;\;(50\ \text{Hz}), \qquad i_{p} = 2.6\,I_{sc}\;\;(60\ \text{Hz}), \qquad i_p = 2.55\,I_{sc}\;\;\text{(classical } 1.8\sqrt2) \]

Short-time rating. While the fault persists the breaker is simply a conductor being heated. For durations under a second or so the heating is adiabatic, so the temperature rise depends on \(\int i^2 dt\) alone and a rating at one duration converts to another by holding \(I^2t\) constant.

Short-time equivalence
\[ I_1^{2}\,t_1 = I_2^{2}\,t_2 \qquad\Longrightarrow\qquad I_2 = I_1\sqrt{\frac{t_1}{t_2}} \]

Finally, the rated operating sequence states how quickly the duty may be repeated. The standard non-autoreclosing duty is \(\text{O}-3\,\text{min}-\text{CO}-3\,\text{min}-\text{CO}\): open, wait three minutes, close-and-immediately-open, wait, repeat. For a line with rapid autoreclosing the interval shrinks to \(\text{O}-0.3\,\text{s}-\text{CO}-3\,\text{min}-\text{CO}\), and the \(0.3\) second dead time is what allows the arc path of a transient flashover to deionise before the line is re-energised — the same deionisation physics as Section 35-2, applied to the air around an insulator string instead of to the inside of an interrupter.

The two encounters, restated. Chapter 25 introduced the distinction between making duty and breaking duty from the network side; here it appears from the breaker side. Making is decided in the first half cycle by a peak current and is a question of mechanics; breaking is decided a few cycles later at a current zero and is a question of dielectrics. The same fault produces both numbers, and a breaker adequate for one may be inadequate for the other.
Section 35-8

The Five Interrupting Media

Every breaker in service solves the same problem — remove charge carriers from the gap faster than the restriking voltage can exploit them — and the classification of breakers is simply a classification of the substance used to do it.

Air-break breakers interrupt in air at atmospheric pressure and rely on the high-resistance method. The arc is drawn into an arc chute by the magnetic field of its own current, sometimes helped by a blow-out coil, and there it is lengthened, cooled against insulating side walls and split by steel splitter plates into a series of short arcs. Adding up the cathode and anode drops of a dozen short arcs produces an arc voltage exceeding the supply, and the current is choked. This is the mechanism of every low-voltage moulded-case and air circuit breaker and of medium-voltage designs up to about \(15\) kV. Above that the arc voltage needed becomes impractical.

Air-blast breakers switch to the current-zero method and attack the arc with a jet of compressed air at \(20\)–\(30\) bar, blown either along the arc axis or across it. The blast removes the ionised gas bodily and replaces it with cold air, so the dielectric recovery is very fast and the arcing time very short. Air-blast breakers dominated the \(220\)–\(400\) kV range for a generation. They have no fire risk and no carbonising medium, but they require a compressed-air plant with its own reliability problems, they are extremely noisy, and their fast recovery is accompanied by a strong tendency to chop small currents. They have been almost entirely displaced by SF6.

Oil breakers strike the arc under insulating mineral oil. The arc decomposes the oil into a gas that is roughly \(70\%\) hydrogen — which has an exceptionally high thermal conductivity and is therefore an excellent arc cooler — and the gas bubble raises the local pressure, which drives recombination. In a bulk-oil breaker the tank holds a large volume of oil that serves both as the interrupting medium and as the insulation to earth. In a minimum-oil breaker the arc is confined to a small insulating chamber and the oil volume falls by an order of magnitude, with cross-jet or axial explosion pots directing the gas flow across the arc at the current zero. Oil breakers are robust and cheap but they carbonise their own medium, so the oil must be tested and replaced, the contacts eroded by each operation must be inspected, and the fire risk of hot oil under pressure is real. New installations are rare.

Vacuum breakers remove the medium altogether. At \(10^{-7}\) torr there are too few molecules to ionise, so the only conducting material in the gap is metal vapour boiled off the contacts themselves. That vapour is generated only while current flows; at the current zero the supply stops and the vapour condenses on the shields in a few microseconds. Dielectric recovery is therefore extraordinarily fast, of the order of tens of kilovolts per microsecond, and a gap of \(8\)–\(20\) mm suffices at \(36\) kV. A vacuum interrupter is sealed for life, needs no maintenance, has no fire risk, produces no by-products, and tolerates tens of thousands of load-current operations, which makes it the natural choice for the medium-voltage distribution switchgear and motor-starting duty of Chapter 2's distribution level. Its weaknesses are the chopping of Section 35-5 and, historically, a voltage ceiling — although vacuum interrupters at \(72.5\) kV and above are now in service and the ceiling is rising.

SF6 breakers use sulphur hexafluoride, a heavy, inert, non-toxic gas that is strongly electronegative: a free electron colliding with an SF6 molecule is captured to form a heavy, slow negative ion. Removing free electrons is exactly what deionisation means, so SF6 both quenches the arc quickly and, once quenched, insulates far better than air — about two and a half times the dielectric strength at the same pressure, and more at the \(5\)–\(7\) bar used in practice. In a puffer design the moving contact drives a piston that compresses gas and blows it through the arc; self-blast and rotating-arc designs let the arc's own energy generate the blast pressure, cutting the mechanical operating energy sharply. SF6 now covers the whole range from \(12\) kV to \(800\) kV and makes gas-insulated substations possible. Its drawback is environmental: SF6 has a global warming potential some \(23\,500\) times that of carbon dioxide and an atmospheric lifetime of millennia, so leakage is tightly regulated and the industry is actively developing fluoronitrile mixtures, CO2-based gases and clean air as replacements.

TypeExtinction principleUsual rangeStrengthsLimitations
Air-breakHigh resistance: lengthen, cool, split the arc0.4–15 kVSimple, visible, no special mediumLarge arc energy inside the breaker; not viable at HV
Air-blastCurrent zero: compressed air removes ionised gas33–400 kVVery short arcing time, no fire riskCompressor plant, noise, marked current chopping
Bulk / minimum oilCurrent zero: hydrogen from decomposed oil cools the arc3.3–220 kVRugged, inexpensive, long-establishedFire risk, carbonisation, frequent maintenance
VacuumCurrent zero: metal vapour condenses when current stops3.3–72.5 kV (rising)Sealed for life, maintenance-free, very high operation countChopping and reignition on small inductive currents
SF6Current zero: electronegative gas captures free electrons12–800 kVHighest interrupting capability, compact, enables GISPotent greenhouse gas; liquefies in very cold climates
Why the two survivors are the two extremes. Vacuum and SF6 have displaced everything else because they attack the deionisation problem at opposite ends. Vacuum removes the gas so there is nothing left to ionise; SF6 keeps the gas but chooses one that swallows free electrons. Both need very little mechanical energy and neither consumes its own medium, so both can be sealed and left alone for decades — and in switchgear, a device that needs no maintenance is a device that will still work on the day the fault arrives.
Section 35-9

Fuses and Isolators

A breaker is not the only way to open a circuit, and on most of the network it is not the cheapest.

A fuse is a deliberately weak conductor that melts, vaporises and then extinguishes the resulting arc. It is detector and interrupter in one component, it needs no relay, no CT, no trip supply and no moving parts, and its operating time falls extremely steeply with current — which is exactly the characteristic wanted, and hard to obtain from a relay. Its vocabulary is small and precise:

TermMeaning
Rated (carrying) currentCurrent the element carries indefinitely without deterioration
Minimum fusing currentLeast current that will eventually melt the element
Fusing factorMinimum fusing current ÷ rated current; typically 1.25–2.0
Pre-arcing (melting) timeFrom fault inception to the instant the element parts
Arcing timeFrom element parting to final current zero
Cut-off currentPeak actually let through, well below the prospective peak
Let-through \(I^2t\)\(\int i^2\,dt\) over the whole operation; the thermal stress passed downstream
Breaking capacityHighest prospective current the fuse can interrupt safely

The HRC (high rupturing capacity) fuse is the engineered version: a silver or silver-plated copper element, punched with restricted sections so that melting starts simultaneously at several points, embedded in graded quartz sand inside a ceramic body. The sand absorbs the arc energy and quenches the several short arcs almost instantly, and a small blob of tin alloyed to the element — the M-effect — lowers the melting point locally so that a sustained mild overload also clears without needing the element to reach the melting point of silver.

The property that a breaker cannot match is current limiting. At high prospective currents the element melts within the first two or three milliseconds, long before the first peak of the fault wave is reached, so the current is cut off at a value far below the prospective peak. Both the electromagnetic force, which goes as \(i^2\), and the let-through \(I^2t\) are then reduced by an order of magnitude or more. That is why a fuse is used to back up a contactor or a moulded-case breaker whose own breaking capacity is modest, and why cable protection at a switchboard is so often a fuse rather than a relay. The trade is that a fuse is single-shot, cannot be tripped by anything but its own current, and can leave a three-phase load single-phased if only one phase clears.

🔑
The cable-protection criterion
\[ \left(I^{2}t\right)_{\text{let-through}} \;\le\; k^{2}S^{2} \]

with \(S\) the conductor cross-section in mm² and \(k\) a material-and-insulation constant (\(115\) for PVC-insulated copper, \(143\) for XLPE copper, \(76\) for PVC-insulated aluminium). The right-hand side is the adiabatic energy the conductor can absorb before its insulation reaches the limiting temperature. Example 6 shows a case where a fuse passes this test and a breaker of the same nominal rating does not.

An isolator, or disconnector, is at the opposite extreme of sophistication: a switch with no arc-quenching capability whatever. Its purpose is not to interrupt current but to provide a visible, verifiable, air-insulated break so that maintenance staff can work on a circuit knowing it is dead. Because it cannot quench an arc it must be operated only when the current through it is already essentially zero, which means only after the associated circuit breaker has opened. Opening an isolator on load is one of the classic switchgear accidents, and it is prevented by mechanical and electrical interlocking rather than by instruction.

The standard sequence for taking a feeder out of service runs: open the circuit breaker; confirm it is open; open the line-side isolator; open the bus-side isolator; close the earthing switch; apply the safety document. Restoration reverses it exactly. The interlock scheme enforces two rules — an isolator may move only when its breaker is open, and an earthing switch may close only when both isolators are open — and those two rules eliminate the entire class of accidents. Between the isolator and the breaker sits the load-break switch, which has enough arc control to interrupt load current but not fault current, and which combined with a fuse makes the cheap and very common switch-fuse unit of a distribution substation.

Isolator before earth, breaker before isolator. The interlocking sequence is the one piece of switchgear practice that is memorised in the field rather than derived, and it exists because the consequence of getting it wrong is not equipment damage but a fatality. Chapter 37 takes it further into the substation layout and the earthing grid that makes the earthing switch safe to close.
Section 35-10

Worked Examples

1 The restriking voltage of a terminal fault

Problem. A \(132\) kV, \(50\) Hz system has an effective inductance of \(20\) mH per phase between the source and the breaker, and the total capacitance to earth at the breaker terminal is \(0.02\) µF. A three-phase terminal fault is cleared. Find the natural frequency of the restriking voltage, its peak value, the time to that peak, and the maximum and average RRRV. Check that the fault level implied by the inductance is sensible.

Solution. The peak of the phase voltage is what the transient is built on:

Driving voltage and time constant
\[ E_m = \sqrt2\;\frac{132}{\sqrt3} = \sqrt2 \times 76.21 = 107.78\ \text{kV} \]
\[ \sqrt{LC} = \sqrt{(20\times10^{-3})(0.02\times10^{-6})} = \sqrt{4\times10^{-10}} = 2.0\times10^{-5}\ \text{s} = 20\ \mu\text{s} \]

Everything else follows from those two numbers:

The four results
\[ f_n = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi(20\ \mu\text{s})} = 7958\ \text{Hz} \approx 7.96\ \text{kHz} \]
\[ v_{peak} = 2E_m = 215.6\ \text{kV}, \qquad t_{peak} = \pi\sqrt{LC} = \pi(20) = 62.8\ \mu\text{s} \]
\[ \left(\frac{dv}{dt}\right)_{\max} = \frac{E_m}{\sqrt{LC}} = \frac{107.78}{20} = 5.39\ \text{kV}/\mu\text{s} \]
\[ \left(\frac{dv}{dt}\right)_{avg} = \frac{2E_m}{\pi\sqrt{LC}} = \frac{215.6}{62.8} = 3.43\ \text{kV}/\mu\text{s} = \frac{2}{\pi}\times 5.39 \;\;\text{(as expected)} \]

The sanity check: the same inductance fixes the fault current, since \(\omega L = 314.2 \times 0.02 = 6.283\) Ω gives \(\hat I = 107.78\times10^{3}/6.283 = 17.16\) kA peak, or \(12.13\) kA rms. The fault level is \(\sqrt3 \times 132 \times 12.13 = 2773\) MVA, entirely plausible for a \(132\) kV bus — so the \(20\) mH is a realistic figure and the RRRV computed from it is a realistic upper bound.

2 First pole to clear, and what a grading capacitor is worth

Problem. For the system of Example 1, (a) correct the results for a first-pole-to-clear factor of \(1.3\) and an amplitude factor of \(1.4\); (b) find the new natural frequency and maximum RRRV if a grading capacitor raises the terminal capacitance from \(0.02\) µF to \(0.08\) µF.

Solution (a). The first pole to clear sees \(k_{pp}\) times the phase voltage, and damping reduces the overshoot from the ideal factor \(2\) to the amplitude factor \(k\):

Corrected peak and slope
\[ u_c = k_{pp}\,k\,E_m = 1.3 \times 1.4 \times 107.78 = 196.2\ \text{kV} \]
\[ \left(\frac{dv}{dt}\right)_{\max} = k_{pp}\,\frac{E_m}{\sqrt{LC}} = 1.3 \times 5.39 = 7.01\ \text{kV}/\mu\text{s} \]

The peak has come down because the damping factor \(1.4\) beats the pole factor \(1.3\), while the initial slope has gone up, damping having little effect on a slope measured before the oscillation has developed. A published TRV envelope for this voltage class specifies a rate near \(2\) kV/µs, so our lumped estimate is roughly three times conservative — the real bus carries several lines whose capacitance is not in the model.

Solution (b). Quadrupling \(C\) doubles \(\sqrt{LC}\):

With the grading capacitor
\[ \sqrt{LC} = \sqrt{(0.02)(0.08\times10^{-6})} = \sqrt{1.6\times10^{-9}} = 4.0\times10^{-5}\ \text{s} = 40\ \mu\text{s} \]
\[ f_n = 3.98\ \text{kHz}, \qquad \left(\frac{dv}{dt}\right)_{\max} = \frac{107.78}{40} = 2.69\ \text{kV}/\mu\text{s} \]

The rate of rise has been halved and the time to peak doubled to \(125.7\) µs, but the peak is untouched at \(2E_m\). Capacitance therefore buys victory in the thermal round of Section 35-3 and nothing at all in the dielectric round — which is exactly why grading capacitors and opening resistors are used together rather than as alternatives.

3 Current chopping on an unloaded transformer

Problem. A vacuum breaker switches out an unloaded transformer on a \(33\) kV system. The magnetising inductance referred to the \(33\) kV side is \(20\) H and the terminal capacitance is \(0.005\) µF. The interrupter chops at \(4\) A. Find the prospective overvoltage, express it in per unit of the peak phase voltage, and size a surge capacitor that would hold the transient below \(80\) kV.

Solution. The surge impedance of the trapped circuit is the whole story:

Surge impedance and overvoltage
\[ \sqrt{\frac{L}{C}} = \sqrt{\frac{20}{5\times10^{-9}}} = \sqrt{4\times10^{9}} = 63\,246\ \Omega = 63.2\ \text{k}\Omega \]
\[ V = i_{ch}\sqrt{\frac{L}{C}} = 4 \times 63\,246 = 253.0\ \text{kV} \]

Including the voltage already standing on the capacitance — the magnetising current lags by nearly \(90^\circ\), so the chop occurs close to the voltage peak \(E_m = \sqrt2\times 33/\sqrt3 = 26.94\) kV — barely changes it:

Total crest
\[ V_{\max} = \sqrt{(26.94)^2 + (253.0)^2} = \sqrt{725.8 + 64\,009} = 254.4\ \text{kV} \]
\[ \frac{V_{\max}}{E_m} = \frac{254.4}{26.94} = 9.4\ \text{per unit} \]

The lightning impulse withstand level of \(33\) kV equipment is \(170\) kV, so this transient destroys the transformer it was switching. Two remedies, applied together. A surge arrester rated \(30\) kV clips the crest to a residual of roughly \(85\) kV and has to absorb only \(\tfrac12 L i_{ch}^2 = \tfrac12(20)(4)^2 = 160\) J, a trivial duty. Alternatively, add capacitance at the transformer terminals:

Surge capacitor
\[ 4\sqrt{\frac{20}{C}} = 80\times10^{3} \;\Longrightarrow\; \sqrt{\frac{20}{C}} = 2\times10^{4} \;\Longrightarrow\; C = \frac{20}{4\times10^{8}} = 0.05\ \mu\text{F} \]

A \(0.05\) µF surge capacitor — a small, cheap component — divides the overvoltage by \(3.2\), because the surge impedance falls as \(1/\sqrt{C}\). This is the standard protection for vacuum switching of motors and unloaded transformers.

4 Resistance switching

Problem. For the circuit of Example 1 (\(L=20\) mH, \(C=0.02\) µF, \(E_m = 107.78\) kV, fault current \(12.13\) kA rms), find the resistance that must be connected across the contacts to give critical damping, the resulting peak restriking voltage and maximum RRRV, and the current the auxiliary contacts must then interrupt.

Solution. The critical value is half the surge impedance:

Critical resistance
\[ R_c = \frac12\sqrt{\frac{L}{C}} = \frac12\sqrt{\frac{0.02}{2\times10^{-8}}} = \frac12\sqrt{10^{6}} = \frac{1000}{2} = 500\ \Omega \]

Check the damping directly: \(\alpha = 1/(2R_cC) = 1/(2\times500\times2\times10^{-8}) = 5\times10^{4}\ \text{s}^{-1}\), and \(\omega_n = 1/\sqrt{LC} = 1/(2\times10^{-5}) = 5\times10^{4}\ \text{s}^{-1}\). They are equal, as the algebra of Section 35-6 promised.

The damped transient
\[ v(t) = E_m\Big[1-(1+\omega_n t)e^{-\omega_n t}\Big] \;\longrightarrow\; v_{peak} = E_m = 107.8\ \text{kV}\;\;(\text{was } 215.6) \]
\[ \left(\frac{dv}{dt}\right)_{\max} = \frac{E_m\omega_n}{e} = \frac{5.389}{2.718} = 1.98\ \text{kV}/\mu\text{s}\;\;(\text{was } 5.39) \]

Now the residual duty. With the main contacts open, the source drives current through \(R_c\) in series with the fault-limiting reactance \(\omega L = 6.28\ \Omega\), which is negligible beside \(500\) Ω:

Duty on the auxiliary interrupter
\[ \hat I_R = \frac{E_m}{R_c} = \frac{107.78\times10^{3}}{500} = 215.6\ \text{A} \;\Rightarrow\; I_R = 152.4\ \text{A rms} \]
\[ \frac{I_R}{I_{fault}} = \frac{152.4}{12\,130} = 1.26\%, \qquad P_R = I_R^{2}R_c = (152.4)^2(500) = 11.6\ \text{MW} \]

The auxiliary contacts interrupt \(152\) A instead of \(12.1\) kA — a duty smaller by a factor of eighty. The resistor stack absorbs \(11.6\) MW while it is in circuit, so over a \(10\) ms insertion it takes \(116\) kJ, which fixes its thermal mass. Both halves of the design are now determined by the single number \(\sqrt{L/C} = 1000\ \Omega\).

5 Reading a nameplate

Problem. A breaker is offered as \(36\) kV, \(1250\) A, \(25\) kA, \(3\) s. Express its breaking capacity in MVA on a \(33\) kV system, find its peak making current, its asymmetrical breaking current for DC components of \(30\%\) and \(50\%\), and its equivalent \(1\) s short-time withstand. Then decide whether it may be used on a bus whose fault level is \(1200\) MVA and whose protection clears in \(0.5\) s.

Solution.

Nameplate arithmetic
\[ S_{br} = \sqrt3\,(33)(25) = 1429\ \text{MVA} \]
\[ i_p = 2.5 \times 25 = 62.5\ \text{kA peak}\quad(50\ \text{Hz}); \qquad 2.55\times25 = 63.8\ \text{kA by the classical } 1.8\sqrt2 \]
\[ I_{asym}\big|_{\delta=0.3} = 25\sqrt{1+2(0.3)^2} = 25\sqrt{1.18} = 27.2\ \text{kA} \]
\[ I_{asym}\big|_{\delta=0.5} = 25\sqrt{1+2(0.5)^2} = 25\sqrt{1.5} = 30.6\ \text{kA} \]
\[ I^{2}t = (25)^2(3) = 1875\ \text{kA}^2\text{s} \;\Longrightarrow\; I_{1\text{s}} = \sqrt{1875} = 43.3\ \text{kA} \]

Now the application check, one row at a time. The bus fault current is

Duty at the bus
\[ I_f = \frac{1200}{\sqrt3 \times 33} = \frac{1200}{57.16} = 21.0\ \text{kA rms symmetrical} \]

Breaking: \(21.0\) kA against \(25\) kA leaves a margin of \(19\%\) — acceptable, though a rise in generation would erode it. Making: the peak duty is \(2.5\times21.0 = 52.5\) kA against \(62.5\) kA rated. Short time: the thermal duty is \((21.0)^2(0.5) = 220\ \text{kA}^2\text{s}\) against \(1875\) — a factor of eight in hand. Voltage: \(36\) kV rated covers a \(33\) kV nominal system whose highest voltage is \(36\) kV. Every row clears, so the breaker is suitable.

One trap is worth naming. A duty of \(30\) kA for \(1\) s gives \(I^2t = 900\ \text{kA}^2\text{s}\), comfortably inside the \(1875\) the breaker can absorb — yet the breaker is not suitable, because \(30\) kA exceeds its \(25\) kA breaking rating. The thermal check and the interrupting check are independent, and passing one says nothing about the other.

6 Why a fuse sometimes beats a breaker

Problem. A \(415\) V board has a prospective symmetrical fault current of \(25\) kA rms; the peak factor for that current is \(2.2\). A \(50\) mm² PVC-insulated copper cable (\(k = 115\)) leaves the board. Compare protection by a \(100\) A HRC fuse, whose cut-off is \(9\) kA peak and whose let-through is \(45\,000\ \text{A}^2\text{s}\), with protection by a non-current-limiting breaker clearing in three cycles.

Solution. First the cable's own capability:

What the cable can absorb
\[ k^{2}S^{2} = (115)^2(50)^2 = 13\,225 \times 2500 = 33.06\times10^{6}\ \text{A}^2\text{s} \]

The fuse lets through \(45\,000\ \text{A}^2\text{s}\), smaller by a factor of \(735\); the cable is never at risk. The breaker allows the full fault current to flow for its whole clearing time:

What a three-cycle breaker lets through
\[ I^{2}t = (25\times10^{3})^{2}\times 0.06 = 6.25\times10^{8}\times0.06 = 37.5\times10^{6}\ \text{A}^2\text{s} \]
\[ 37.5\times10^{6} \;>\; 33.06\times10^{6} \quad\Longrightarrow\quad \text{the cable insulation is exceeded} \]

The breaker fails by \(13\%\); it would need to clear in \(33.06\times10^{6}/6.25\times10^{8} = 52.9\) ms, under two and a half cycles, to be acceptable. The mechanical comparison is even more lopsided. The prospective peak is \(2.2\times25 = 55\) kA, while the fuse cuts off at \(9\) kA, so the electromagnetic force on the busbars and cable cleats — which goes as \(i^2\) — is reduced by \((55/9)^2 = 37\) times.

This is the entire case for the fuse. It is not that a fuse is a better protective device than a relay and breaker; it plainly is not, since it cannot be graded finely, cannot be reset and cannot be told to trip by anything else. It is that a fuse acts before the first peak arrives, and no mechanical device can. Where the fault level is high and the equipment downstream is modest, that single property outweighs everything a breaker offers — which is why fuse-backed contactors and switch-fuse units remain standard on distribution boards a century after the circuit breaker was perfected.

Review

Chapter Summary

The arc is necessary

\(\tfrac12 Li^2\) cannot vanish; the arc is the controlled channel through which it leaves.

Negative resistance

Ayrton: \(V_{arc}=A+B\ell+(C+D\ell)/I\) — arc voltage falls as current rises.

Two strategies

High resistance for DC and LV; current-zero interruption for everything at HV.

The race

Gap strength must exceed restriking voltage at every instant, in slope then in peak.

Restriking voltage

\(v=E_m(1-\cos\omega_n t)\): peak \(2E_m\) at \(\pi\sqrt{LC}\), max slope \(E_m/\sqrt{LC}\).

Chopping

\(V=i_{ch}\sqrt{L/C}\) — small inductive currents are the dangerous duty, not big faults.

Critical resistor

\(R_c=\tfrac12\sqrt{L/C}\) removes the overshoot and divides the peak RRRV by \(e\).

Ratings

\(S_{br}=\sqrt3 VI\), \(i_p=2.5I_{sc}\), \(I_{asym}=I_{ac}\sqrt{1+2\delta^2}\), \(I^2t\) constant.

Media

Vacuum removes the gas; SF6 captures the electrons. Both need no maintenance.

Fuse and isolator

The fuse cuts off before the first peak; the isolator never opens on load.

Practice

Practice Problems

Take \(50\) Hz throughout and treat the source voltage as constant at its peak over the duration of any restriking transient.

  1. A \(220\) kV breaker clears a terminal fault. The inductance up to the fault is \(12\) mH per phase and the capacitance to earth at the terminal is \(0.015\) µF. Find the natural frequency, the peak restriking voltage, the time to that peak, and the maximum and average RRRV.
  2. A breaker on a \(132\) kV system shows a peak restriking voltage of \(216\) kV and a maximum RRRV of \(4\) kV/µs, and the fault current is \(10\) kA rms. Deduce \(\sqrt{LC}\), then \(L\) and \(C\) separately, and state the natural frequency.
  3. A vacuum breaker chops at \(5\) A while switching out an unloaded \(66\) kV transformer whose magnetising inductance is \(30\) H and whose terminal capacitance is \(4\) nF. Find the prospective overvoltage in kV and in per unit of the peak phase voltage, and the surge capacitance that would hold the crest below \(200\) kV.
  4. For a circuit with \(L = 5\) mH and \(C = 0.01\) µF, find the resistance required for critical damping. State the peak restriking voltage and maximum RRRV with and without it, given \(E_m = 107.8\) kV, and find the current the auxiliary interrupter must clear.
  5. A breaker is rated \(145\) kV, \(31.5\) kA, \(3\) s. Find its breaking capacity in MVA, its peak making current, its asymmetrical breaking current for a \(40\%\) DC component, and its equivalent \(1\) s short-time withstand.
  6. 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 for it to become adequate.
  7. A \(415\) V board with a prospective fault current of \(31.5\) kA rms feeds a \(35\) mm² XLPE copper cable (\(k = 143\)). Find the cable's \(I^2t\) withstand, and the longest clearing time a non-current-limiting breaker could have. Would a fuse with a let-through of \(60\,000\ \text{A}^2\text{s}\) be acceptable?
  8. Show from the critically damped solution \(v = E_m[1-(1+\omega_n t)e^{-\omega_n t}]\) that the maximum rate of rise occurs at \(t = 1/\omega_n\) and equals \(E_m\omega_n/e\). Explain physically why adding capacitance reduces the RRRV but not the peak, while adding the critical resistance reduces both.
Tip: almost every numerical question in this chapter reduces to two quantities, \(\sqrt{LC}\) and \(\sqrt{L/C}\). The first is a time and controls everything about the restriking transient — frequency, time to peak, rate of rise. The second is an impedance and controls everything about energy transfer — chopping overvoltage, critical resistance. Compute both before touching the specific question, and check that the fault current the inductance implies is a number a real system could produce.