Chapter 3 · 4 hours
Fuses
IOE past exam questions
Past questions and answers
17 questions set from this chapter. Most asked first.
- 2070 Bhadra · 2+2+2+2 marks
What is the main difference between Fuse and MCB? Define the following terms for HRC fuse: (i) cut off (ii) pre-arcing time (iii) arcing time.
Answer
Difference between fuse and MCB
| Point | Fuse | MCB (miniature circuit breaker) |
|---|---|---|
| Operating principle | Fuse element melts by I²R heating | Thermal (bimetal) trip for overload + magnetic trip for short circuit |
| Reuse | One-time; element must be replaced | Reset and reused after tripping |
| Speed on short circuit | Very fast; HRC fuse cuts off before first peak | Fast (magnetic, ~5–10 ms), but slower than HRC |
| Breaking capacity | Very high (HRC up to 80–120 kA) | Limited (typically 6–10 kA) |
| Switching | Cannot be used as a switch | Can be used as an ON/OFF switch |
| Indication | Not always visible | Handle position shows tripped state |
| Cost | Low initial cost | Higher initial, but no replacement cost |
| Single phasing | Possible when one fuse blows | Multi-pole MCB trips all poles |
HRC fuse terms
(i) Cut-off (cut-off current): the maximum instantaneous value of current actually reached before the fuse element melts, when the prospective fault current is large. An HRC fuse melts before the first peak of the fault current, so the cut-off current is much lower than the prospective peak; this limits thermal and electromagnetic stress on the circuit.
(ii) Pre-arcing time (melting time): the time from the start of the fault current until the fuse element melts and an arc is initiated. It depends on the magnitude of current; for heavy faults it is a few milliseconds.
(iii) Arcing time: the time from the instant the arc is initiated (element melts) until the arc is finally extinguished and the current becomes zero. Silica sand absorbs the arc energy and quenches the arc.
i prospective
│ ╱‾╲
│ cut-off ●╲ actual current
│ ╱│ ╲
│ ╱ │ ╲___
└──┴──┴──────┴────► t
pre-arc arcing
◄──────►◄────►
total operating time
Total operating (clearing) time = pre-arcing time + arcing time.
- 2074 Bhadra · 3+3 marks
Explain the construction and operation of HRC fuse. Also mention the difference between Fuse and MCB.
Answer
Construction of HRC fuse
An HRC (high rupturing capacity) cartridge fuse can safely interrupt very high fault currents (up to 80–120 kA) without external effects.
end cap end cap
┌──┐ ┌────────────────┐ ┌──┐
│ ├─┤ ·:·:·:·:·:·:· ├─┤ │
│ │ │ ══╪══╪══╪══╪══ │ │ │ silver
│ ├─┤ ·:·:·:·:·:·:· ├─┤ │ element
└──┘ └────────────────┘ └──┘
ceramic body filled
with quartz sand (·:)
- Ceramic (steatite) body: strong, heat-resistant tube that withstands arc pressure.
- Metal end caps with contact blades/tags, welded to the fuse element.
- Fuse element: silver (or copper) strips with reduced sections (notches) in parallel; sometimes a tin/low-melting-point bead (M-effect) for overload.
- Filling powder: pure quartz (silica) sand packed around the element.
- Indicator: a fine wire in parallel that melts and releases a striker/indicator.
Operation
- Under normal current, the heat produced is dissipated and the element stays below melting temperature.
- On fault, the reduced sections melt quickly (pre-arcing) and several arcs form in series.
- The heat of the arcs vaporises silver, which reacts with/fuses the quartz sand to form a high-resistance glass-like substance (fulgurite).
- The arc resistance rises rapidly, the current is forced to zero, and the arc is quenched (arcing time).
- For high faults the element melts before the first current peak, so the fuse limits current (cut-off).
Difference between fuse and MCB
| Point | Fuse | MCB |
|---|---|---|
| Principle | Element melts | Bimetal (thermal) + magnetic trip |
| After operation | Replace element | Reset |
| Breaking capacity | Very high | Limited (6–10 kA) |
| Current limitation | Strong (HRC) | Moderate |
| Switching duty | No | Yes, used as switch |
| Single phasing risk | Yes | No (multi-pole trips together) |
| Cost | Low | Higher |
- 2072 Magh · 6 marks
Explain the constructional details of HRC fuse. State advantages of HRC fuse.
Answer
An HRC (high rupturing capacity) fuse is a totally enclosed cartridge fuse that can safely interrupt very large fault currents (tens of kA) without any external flame or explosion.
Constructional details
contact contact
blade ┌──────────────┐ blade
═══╦═══╡ ::::::::::::: ╞═══╦═══
║ │ ═╦═╦═╦═╦═╦═ │ ║
║ │ ::::::::::::: │ ║
end cap└──────────────┘ end cap
ceramic body + silica sand
silver element with notches
● indicator / striker
- Body: a cylindrical tube of high-grade ceramic (steatite or porcelain), strong enough to withstand pressure and heat during arcing.
- End caps and contacts: metal (brass/copper) caps fixed to the ends, with knife-blade or bolted tags; the fuse element is welded to them.
- Fuse element: usually silver (sometimes copper) because silver does not oxidise, has low resistivity and a well-defined melting point. It is made as a strip with several reduced sections (notches); for large ratings, two or more strips in parallel.
- Overload "M-effect": a small tin or solder bead on the element; at moderate overloads it melts first and alloys with silver, lowering its melting point, giving controlled time delay.
- Filler: pure, dry quartz (silica) sand packed tightly around the element. It absorbs arc heat and, with the vaporised silver, forms a high-resistance fulgurite that quenches the arc.
- Indicator/striker: a thin resistance wire in parallel with the element; it melts after the main element and releases a spring-loaded indicator or striker that can also trip a switch.
Operation (brief)
On fault, the notches melt almost at once, several series arcs are formed, the sand cools them and the arc resistance increases, forcing the current to zero, often before the first peak.
Advantages of HRC fuse
- Very high breaking capacity at low cost.
- Current-limiting: cuts off before the prospective peak, reducing thermal and mechanical stress.
- Very fast operation on heavy faults (a few ms).
- Inverse time-current characteristic: time delay for small overloads, fast for short circuits.
- Consistent, non-deteriorating performance; calibration does not change with age.
- No maintenance; totally enclosed, so no external flame or noise.
- Good discrimination with other fuses and breakers.
- Simple, compact and cheaper than a circuit breaker of equal breaking capacity.
Disadvantages: must be replaced after operation; heating of contacts may cause discrimination problems.
- 2080 Chaitra · 6 marks
Explain the operating characteristics of HRC fuse. How do these characteristics influence their applications in electrical systems?
Answer
The operating characteristics of an HRC fuse describe how quickly it operates and how much current it lets through for different overcurrents. They are given mainly by the time–current characteristic and the cut-off (current-limiting) characteristic, together with the I²t values.
1. Time–current characteristic
time (s, log)
│╲
│ ╲ pre-arcing time
│ ╲
│ ╲ inverse region
│ ╲
│ ╲___
│ ‾‾──____ total operating
└──┬─────────────── current (log)
min. fusing current
- Inverse: the higher the current, the faster the fuse melts.
- Below the minimum fusing current (≈ fusing factor × rated current, e.g. 1.25–1.6 In for gG) the fuse never operates; overloads just above it melt in minutes to hours.
- For heavy faults the operating time is a few milliseconds.
- Characteristic categories: gG (general purpose, full-range), aM (motor circuit, short-circuit only, withstands starting current), gR/aR (semiconductor, very fast).
2. Cut-off characteristic
peak let-through (kA)
│ ╱ prospective peak (√2·2.5 × Irms)
│ ╱
│ ╱ ___ cut-off current
│ ╱_─‾
│ _─╱
└──────────────── prospective current (kA rms)
For large prospective currents the element melts before the first peak; the current is limited to the cut-off current, much less than the prospective peak.
3. I²t characteristic
Pre-arcing I²t (fixed for an element) and total I²t (depends on voltage and power factor) define the energy let-through.
Influence on applications
- Cable and equipment protection: low cut-off current and I²t mean the cables, busbars and switches need to withstand only small let-through energy, so smaller, cheaper equipment can be used.
- Motor circuits: aM fuses or time-delay fuses are chosen so the starting current (6–7 × In for 5–10 s) does not melt them; overload protection is left to a thermal relay.
- Transformer circuits: fuse must withstand magnetising inrush (10–12 × In for 0.1 s) yet clear secondary faults.
- Semiconductor protection: very fast fuses with total I²t below the device rating.
- Discrimination: fuses in series are selective if the total I²t of the minor fuse is less than the pre-arcing I²t of the major fuse (typically a 1.6:1 or 2:1 rating ratio for gG).
- Back-up for breakers/contactors: high breaking capacity allows HRC fuses to protect contactors and MCBs against currents beyond their breaking capacity.
- 2078 Chaitra · 6 marks
Explain the cut-off characteristics and the time-current characteristics of a fuse.
Answer
Time–current characteristic of a fuse
The time–current (T–I) characteristic shows the time taken by a fuse to operate for different values of current. It is plotted on log–log scales.
t (s, log)
10⁴│╲
│ ╲
10²│ ╲ pre-arcing curve
│ ╲ ╲ total clearing curve
1 │ ╲ ╲
│ ╲__╲__
0.01│ ‾‾‾──____
└──┬──────────────────► I (log)
I_min fusing
- It is an inverse characteristic: small overcurrent → long time; large current → very short time.
- Below the minimum fusing current the fuse does not operate at all.
- Two curves are usually given: pre-arcing (melting) time and total operating time (pre-arcing + arcing). The gap between them is the arcing time; it is important only for short times.
- The characteristic depends on element material, cross-section, shape (notches), the filler and the ambient temperature.
- It is used for coordination: the T–I curves of fuses, relays and breakers in series must not cross.
Cut-off characteristic of a fuse
When the prospective fault current is large, an HRC fuse melts before the current reaches its first peak. The maximum instantaneous current actually reached is the cut-off current.
i
│ ╱‾‾╲ prospective current
│ ╱ ╲
│ ●╲╱ cut-off
│ ╱ ╲
│ ╱ ╲___ actual (limited)
└┴──┴───┴──────► t
t_p t_a
The cut-off characteristic is a graph of cut-off current (peak) versus prospective current (r.m.s.):
I_cut-off (kA peak)
│ ╱ no cut-off line
│ ╱ (peak = 2.5 × Irms)
│ ╱ ___ 200 A fuse
│ ╱ __─‾ ___ 100 A fuse
│ ╱_─‾_──‾
└─────────────────► prospective (kA rms)
- Each fuse rating has a curve; above a certain prospective current, cut-off occurs and the curve falls below the "no cut-off" line.
- Smaller fuse ratings give lower cut-off currents.
Importance: it gives the maximum peak current (electromagnetic force) and let-through energy that the downstream cables, busbars and switches must withstand, and allows smaller equipment and cheaper switchgear to be used.
- 2073 Bhadra · 6 marks
Explain the following terms showing the sketch of the cut off characteristics of fuse: i) Minimum fusing current ii) Rated current iii) Prospective current iv) Cutoff current v) Fusing factor.
Answer
Sketch of cut-off characteristic of a fuse
i prospective current
│ ╱‾‾‾╲
│ ╱ ╲
│ cut-off ╲
│ ●──╲ ╲
│ ╱ ╲ ╲
│ ╱ ╲___ ╲___
└┴────┴────┴─────────────► t
0 melting arc
(pre-arc) extinction
Without fuse, current follows the
prospective curve; with fuse it is
cut off at a lower peak.
(i) Minimum fusing current
The minimum value of current that will cause the fuse element to melt (and thus operate) under specified conditions. Below this value the fuse carries current indefinitely. It depends on element material, length, cross-section, enclosure and ambient temperature.
(ii) Rated current (current rating)
The current that the fuse element can carry continuously without overheating or melting. It is always less than the minimum fusing current. Example: a 32 A fuse can carry 32 A for an indefinite time.
(iii) Prospective current
The r.m.s. value of the fault current (first loop, symmetrical) that would flow in the circuit if the fuse were replaced by a link of negligible impedance. It depends only on the source and circuit impedance, not on the fuse.
(iv) Cut-off current
The maximum instantaneous current actually reached before the fuse element melts, when the prospective current is large enough. Because an HRC fuse melts before the first peak of the fault current, the cut-off current is much smaller than the prospective peak, limiting electromagnetic and thermal stress.
(v) Fusing factor
Fusing factor = Minimum fusing current / Current rating
- It is always greater than 1.
- A value close to 1 gives closer protection, but too close causes nuisance blowing.
- Typical values: about 1.1–1.5 for HRC cartridge fuses, about 1.8–2 for semi-enclosed rewirable fuses.
- Example: a 100 A fuse with minimum fusing current 140 A has a fusing factor of 1.4.
- 2071 Bhadra · 6 marks
In relation to a fuse, explain what do you mean by (i) prospective current (ii) cut-off current (iii) time delay fuse.
Answer
(i) Prospective current
The prospective current is the r.m.s. value of the fault current that would flow in a circuit if the fuse were replaced by a link of negligible impedance. It is decided by the source voltage and the impedance up to the fault, not by the fuse.
I_prospective = V_phase / Z_fault-loop
e.g. 230 V / 0.023 Ω = 10 kA (rms)
With DC offset the first peak can be up to about 2.5 × I_rms (≈ 25 kA in this example).
(ii) Cut-off current
The cut-off current is the maximum instantaneous value of current actually reached before the fuse element melts, for a large prospective current. An HRC fuse melts in the first quarter cycle, so the current is "cut off" before the prospective peak.
i prospective peak
│ ╱‾‾╲
│ cut-off
│ ●╲╱ ╲
│ ╱ ╲ ╲
│ ╱ ╲__ ╲
└┴─────┴───────► t
pre-arc arc
- It reduces electromagnetic forces (∝ i²) and heating (I²t) in cables, busbars and switches.
- The cut-off characteristic gives cut-off current vs prospective current for each fuse rating; a lower fuse rating gives a lower cut-off.
(iii) Time-delay fuse
A time-delay (slow-blow, anti-surge) fuse is designed to carry short-time, harmless overcurrents (motor starting current, transformer inrush, capacitor charging) without melting, while still operating quickly on short circuits.
Construction:
- A dual-element design: a short-circuit element (silver strip with notches) in series with an overload element (a heat-absorbing mass, solder joint or spring-loaded link) that melts only after sustained overload.
- Or an element with an "M-effect" tin bead to delay melting at moderate overloads.
Characteristic: at 5–6 × rated current it may take 5–10 s to blow (allowing motor starting), but at 20 × rated current it operates in milliseconds.
Applications: motor circuits (aM fuses), transformer primaries, capacitor banks, and circuits with lamp or heater switch-on surges. It allows a fuse rating closer to full-load current, giving better overload protection.
- 2073 Magh · 6 marks
What factors govern the time-current characteristics of a fuse link? Explain what is meant by "fuse coordination".
Answer
Factors governing time–current characteristic of a fuse link
The time–current characteristic shows the operating time of a fuse for various currents (inverse curve). It depends on:
- Element material: melting point, resistivity, specific heat and thermal conductivity (silver, copper, tin, zinc, lead). Silver gives stable, predictable operation.
- Cross-section and shape of element: thinner sections melt faster; notches (reduced sections) concentrate heating and decide pre-arcing time.
- Length of element: a longer element has more resistance and less heat lost to end caps.
- Number of parallel strips: affects current sharing and speed.
- M-effect (tin/solder bead): lowers melting temperature at overload, giving time delay at small overloads.
- Filler and enclosure: silica sand, ceramic body and its thermal conductivity determine heat dissipation and arcing time.
- Ambient temperature and previous loading: a hot fuse operates faster.
- Contact resistance and mounting: poor contacts add heat.
- Circuit voltage and power factor: affect the arcing time (and thus total operating time) for high currents.
- Ageing / oxidation of the element in non-HRC fuses.
Fuse coordination (discrimination)
Fuse coordination means selecting fuses in series (e.g. main fuse upstream, branch fuse downstream) so that for any fault on a branch, only the branch fuse nearest the fault blows and the upstream (major) fuse remains intact, keeping other branches in service.
Supply ──[F1 major 200 A]──┬──[F2 minor 63 A]── load 1
└──[F3 minor 63 A]── load 2
Fault on load 1 → only F2 should blow
Conditions:
- The T–I curves of the two fuses must not intersect over the whole range of fault current.
- For short-time (high) faults, the total I²t of the minor fuse must be less than the pre-arcing I²t of the major fuse.
- A rating ratio of about 1.6:1 to 2:1 between successive gG fuses usually ensures discrimination (as per manufacturer tables).
Coordination is also needed between a fuse and a circuit breaker or overcurrent relay (fuse curve below the relay curve with a time margin).
- 2075 Baisakh · 6 marks
Describe the time/current I²t characteristics of a HRC fuse and explain how they are used to select the ratings of fuses in series.
Answer
Time–current and I²t characteristics of an HRC fuse
The time–current characteristic shows the operating time against current on log–log scales. It is inverse: long times for small overloads, a few milliseconds for heavy faults.
For short operating times (< about 10 ms) heat cannot escape from the element, so melting depends only on the energy I²t delivered:
Pre-arcing I²t = ∫₀^tp i² dt (energy to melt element)
Arcing I²t = ∫tp^tt i² dt (energy during arc)
Total I²t = pre-arcing I²t + arcing I²t
I²t (A²s, log)
│ total I²t ─────────────
│ ╱ (rises with voltage, pf)
│ pre-arcing I²t ────────
│ (constant at high current)
└─────────────────────────► prospective I
- Pre-arcing I²t is practically constant for a given fuse at high currents (fixed by element mass and material).
- Total (let-through) I²t depends on circuit voltage, power factor and prospective current.
- Manufacturers give both values for each rating.
Selection of fuses in series using I²t
For a minor (downstream) fuse F2 and a major (upstream) fuse F1:
Supply ──[F1]──┬──[F2]── load
fault after F2
- Both fuses carry the same fault current.
- For F1 not to be damaged or blow, the energy let through by F2 in clearing the fault must not melt F1:
Total I²t of F2 < Pre-arcing I²t of F1
- At lower currents (long times) the time–current curves must not cross; the major fuse curve must lie above the minor fuse curve with margin.
- Choose F2 from load current; then choose the smallest F1 satisfying the I²t rule. In practice a rating ratio of about 1.6–2 : 1 is sufficient for gG fuses.
Example (typical catalogue values): a 63 A fuse with total I²t ≈ 15 000 A²s and a 125 A fuse with pre-arcing I²t ≈ 30 000 A²s discriminate, since 15 000 < 30 000.
The total I²t of the fuse must also be less than the withstand I²t (k²S²) of the protected cable or semiconductor, so the I²t data also fix the maximum fuse rating for a given cable.
- 2077 Chaitra · 6 marks
Define fusing current and current carrying capacity of a fuse with an example.
Answer
Current carrying capacity (current rating)
The current carrying capacity of a fuse is the r.m.s. current that the fuse element can carry continuously without melting and without its temperature exceeding the permitted limit. It is marked on the fuse as the rated current.
- It depends on element material, cross-section, length, contacts, enclosure and ambient temperature.
- The fuse rating is chosen slightly above the normal full-load current of the circuit.
Fusing current
The fusing current (minimum fusing current) is the minimum current at which the fuse element melts and opens the circuit within a specified (long) time, e.g. 1 hour for small fuses or 4 hours for large ones.
- It is always greater than the current rating.
- It depends on the material, length, diameter, contact condition and surroundings of the element.
- Preece's formula for a round wire: I = k·d^(3/2), where d = diameter and k a material constant (e.g. k ≈ 80 for copper with d in mm, I in A).
Fusing factor
Fusing factor = Min. fusing current / Rating (> 1)
Typical: 1.1–1.5 for HRC fuses, about 2 for rewirable fuses.
Example
- A lighting circuit draws 12 A. A fuse with current rating 16 A is chosen; it carries 16 A indefinitely.
- If its fusing factor is 1.45, the minimum fusing current = 1.45 × 16 = 23.2 A.
- Thus currents up to 16 A are carried continuously; at about 23 A or more (overload) the fuse melts after a long time, and at a short-circuit current of, say, 1000 A it blows in milliseconds.
| Point | Current carrying capacity | Fusing current |
|---|---|---|
| Meaning | Carried continuously | Causes element to melt |
| Value | Rated value (lower) | Higher (rating × fusing factor) |
| Effect | No operation | Fuse operates |
| Example | 16 A | 23.2 A |
- 2075 Bhadra · 3 marks
Write a short note on low voltage HRC fuse.
Answer
A low-voltage HRC fuse is an enclosed cartridge fuse for circuits up to 1000 V AC (commonly 415 V) that can safely interrupt very high prospective fault currents (typically 50–80 kA, up to 120 kA) with current limitation.
Construction: a ceramic tube filled with pure quartz sand, containing one or more silver strip elements with notches, welded to brass end caps with knife-blade or bolted contacts; a tin bead (M-effect) gives time delay on overload; an indicator or striker shows operation.
Types:
- Knife-blade (NH) type: fitted into fuse bases with a handle puller; ratings up to about 1250 A.
- Bolted / tag type: bolted to the fuse carrier.
- Cylindrical (ferrule) type: small ratings for control circuits.
- Categories: gG (general purpose, cables), aM (motor back-up), gR (semiconductors).
Operation: on short circuit the notches melt within milliseconds, multiple arcs form, and the sand cools them and forms fulgurite, forcing current to zero before the first peak (cut-off).
Advantages: high breaking capacity, current limiting, fast and reliable, no maintenance, consistent characteristics, cheap compared with breakers.
Applications: LV distribution boards, transformer LV side, motor feeders (back-up to contactors), cable protection and semiconductor protection.
- 2075 Bhadra · 3 marks
Write a short note on drop-out fuse.
Answer
A drop-out fuse (expulsion fuse cut-out) is an outdoor fuse widely used on 11 kV and 33 kV distribution lines and pole-mounted distribution transformers in Nepal.
Construction:
- A porcelain or polymer insulator with upper and lower contacts, mounted on a cross-arm.
- A fuse tube (carrier) made of fibre-glass with an inner lining of horn fibre or boric acid, containing a tin or copper fuse link.
- The fuse link holds the tube in position against a spring at the top contact.
upper contact (latch)
┌┐
││ fuse tube with
││ fuse link inside
││
─┘└─ hinge at lower contact
tube drops (swings) down
when link melts
Operation:
- On overcurrent or fault, the fuse link melts and an arc forms inside the tube.
- The arc heat decomposes the fibre lining, producing de-ionising gases (expulsion action) that blow the arc out through the open bottom end at current zero.
- Melting of the link releases the latch; the tube drops out and hangs down by gravity, giving a visible isolation gap and clear indication from the ground.
Advantages: cheap, simple, visible indication of operation, provides isolation, easy re-fusing with an operating (hot) stick.
Limitations: noisy expulsion of gases and flame, limited breaking capacity, not current limiting, needs manual replacement of link.
Applications: protection of distribution transformers, spur lines and capacitor banks in rural/urban feeders.
- 2070 Magh · 6 marks
A transformer having capacity of 250 KVA, 11/0.415 KV, percentage impedance of 4.75% is required to be protected with fuses. Suggest guide line for selection of HV and LV side fuses. State any assumptions you make.
Answer
Given: 250 kVA, 11/0.415 kV, 3-phase, impedance 4.75 %.
Assumptions: delta–star (Dyn11) transformer; infinite source on 11 kV side (gives the highest fault current); HRC (gG/back-up) fuses used on both sides; magnetising inrush ≈ 12 × full-load for 0.1 s.
Step 1: Full-load currents
I_FL(HV) = 250 000 / (√3 × 11 000) = 13.12 A
I_FL(LV) = 250 000 / (√3 × 415) = 347.80 A
Step 2: Fault current for a fault at LV terminals
I_sc(LV) = I_FL(LV) / Z_pu
= 347.80 / 0.0475 = 7322 A ≈ 7.32 kA
Seen on HV side = 13.12 / 0.0475 = 276.2 A
Step 3: Inrush current
I_inrush ≈ 12 × 13.12 = 157.5 A for about 0.1 s (HV side)
Guidelines for HV fuse
- Rating must exceed full-load current with margin for permissible overload: about 1.5–2 × I_FL → 19.7–26.2 A. Choose standard 25 A HV HRC fuse.
- It must not melt on inrush: the 0.1 s point of its T–I curve must be above 157.5 A (for a 25 A fuse this is normally satisfied; otherwise use 31.5 A).
- It must clear a fault at the LV terminals (276 A on HV side, i.e. ≈ 11 × its rating) quickly, within the transformer short-circuit withstand time (2 s as per IEC 60076-5).
- Breaking capacity must exceed the 11 kV system fault level.
- It must discriminate with the LV fuse: for an LV feeder fault, the LV fuse must clear before the HV fuse melts (HV fuse curve, referred to LV via ratio 11/0.415 = 26.5, must lie above the LV fuse curve).
Guidelines for LV fuse
- Rating ≥ full-load current: 347.8 A → choose standard 400 A gG fuse (≈ 1.15 × I_FL), allowing normal overload.
- Breaking capacity ≥ 7.32 kA (a standard 80 kA LV HRC fuse is ample).
- It must clear an LV fault of 7.32 kA in a short time and protect the LV cables and busbars (its I²t below cable withstand).
- It must be selective with downstream feeder fuses (rating ratio ≈ 1.6:1 or more) and with the HV fuse.
Answer: HV fuse ≈ 25 A HRC (I_FL = 13.12 A); LV fuse ≈ 400 A HRC (I_FL = 347.8 A); LV terminal fault current = 7.32 kA (276 A on HV side). Final ratings are confirmed from manufacturer T–I curves.
- 2071 Magh · 6 marks
An induction motor is required to protect through a fuse and contactor. Suggest a suitable guide line for selection of ratings of fuse and contactor. State any assumptions you make.
Answer
Assumptions: 3-phase squirrel-cage induction motor, 15 kW, 415 V, 50 Hz, efficiency 90 %, power factor 0.85, direct-on-line (DOL) starting, starting current 6 × full-load for up to 10 s, prospective fault current at motor control centre 25 kA. The scheme is fuse (short-circuit protection) + contactor (switching) + thermal overload relay (overload protection).
415 V ─[HRC fuse]─[Contactor]─[O/L relay]─ M
short-ckt switching overload
Step 1: Full-load current
I_FL = P / (√3 · V · η · cosφ)
= 15 000 / (√3 × 415 × 0.90 × 0.85)
= 27.28 A
Starting current = 6 × 27.28 = 163.7 A for ~10 s
Step 2: Fuse selection guidelines
- Type: motor-rated HRC fuse (aM or gM) or time-delay gG fuse, because it must not melt on starting current.
- Rating: for DOL start, about 2–2.5 × I_FL for gG fuses → 54.6–68.2 A. Choose 63 A gG HRC (or a 32 A/40 A aM fuse).
- Check starting: the pre-arcing time of the chosen fuse at 163.7 A must be greater than the starting time (10 s) with margin, also allowing for repeated starts.
- Breaking capacity: must exceed the prospective fault current (25 kA); LV HRC fuses with 80 kA are suitable.
- Back-up to contactor: the fuse must clear all currents above the contactor's breaking capacity, and its cut-off current/I²t must be within the contactor's short-circuit withstand (type-2 coordination as per IEC 60947-4-1).
- Overload protection is not expected from the fuse.
Step 3: Contactor selection guidelines
- Utilisation category: AC-3 (starting squirrel-cage motors, switching off while running).
- Rated operational current Ie (AC-3) ≥ I_FL at 415 V → choose a 32 A AC-3 contactor (≥ 27.28 A).
- Making/breaking capacity: AC-3 contactor makes 10 × Ie (320 A) and breaks 8 × Ie (256 A), so it can make the 163.7 A starting current and break locked-rotor current.
- Coil voltage suited to control supply (230 V AC or 110 V AC/24 V DC) and number of auxiliary contacts for control/interlocks.
- Electrical and mechanical life suitable for number of operations per hour.
Step 4: Overload relay
Thermal (bimetal) relay with range including I_FL, set at 1.0–1.05 × I_FL ≈ 27.3–28.6 A, class 10 trip (operates within 10 s at 7.2 × setting), with single-phasing protection.
Coordination: overload relay curve protects for currents up to about 8–10 × I_FL; above the crossover point the fuse operates first, protecting the contactor.
Answer (for assumed motor): I_FL = 27.28 A; fuse ≈ 63 A gG HRC (or 40 A aM); contactor 32 A AC-3; overload relay set ≈ 28 A.
- 2079 Chaitra · 5 marks
Discuss the method of selecting the rating of HRC fuse in motor stator.
Answer
An HRC fuse in a motor (stator) circuit provides short-circuit protection only; overload protection is given by a thermal overload relay. The fuse must not blow during starting, but must clear faults and protect the contactor.
Method of selection
- Find full-load current:
I_FL = P_out / (√3 · V_L · η · cosφ)
- Find starting current and starting time: DOL starting current I_st ≈ 5–7 × I_FL for 5–10 s (longer for high-inertia loads); star–delta ≈ 2–2.5 × I_FL; soft starter ≈ 3 × I_FL.
- Choose fuse type: motor-circuit HRC fuse (aM, partial-range) or time-delay gG fuse. aM fuses withstand starting current and can be rated close to I_FL.
- Choose rating: the smallest standard rating whose pre-arcing time at I_st is greater than the starting time (with margin for frequent starts and hot condition). As rough guides for gG fuses:
- DOL start: fuse ≈ 2–3 × I_FL
- Star–delta start: fuse ≈ 1.5–2 × I_FL
- Check breaking capacity ≥ prospective fault current at the motor terminals.
- Coordinate with contactor and overload relay: the fuse must interrupt all currents above the contactor's breaking capacity (≈ 8 × Ie for AC-3) and limit the let-through I²t to the contactor and cable withstand (type-2 coordination).
- Check cable protection: fuse I²t below the cable's k²S².
Example
15 kW, 415 V motor, η = 0.9, pf = 0.85, DOL:
I_FL = 15 000 / (√3 × 415 × 0.9 × 0.85) = 27.28 A
I_st = 6 × 27.28 = 163.7 A for 10 s
Fuse (gG) ≈ 2.5 × 27.28 = 68.2 A → 63 A standard,
check its T–I curve: melting time at 163.7 A > 10 s ✓
An aM fuse of about 32–40 A could also be used.
Points to remember
- Too small a fuse blows during starting (nuisance tripping); too large a fuse may not protect the contactor and cables.
- Never rely on the fuse for overload; use the overload relay set at ≈ 1.0–1.05 × I_FL.
- 2076 Bhadra · 6 marks
What are the considerations in selecting a fuse for (i) Transformer protection (ii) capacitor protection (iii) lighting loads?
Answer
(i) Transformer protection
- Full-load current: fuse rating above full-load with allowance for permissible overload; HV side typically 1.5–2 × I_FL, LV side 1.1–1.25 × I_FL (next standard rating).
- Magnetising inrush: the fuse must not melt on inrush of about 10–12 × I_FL for 0.1 s (and 25 × I_FL for 0.01 s).
- Through-fault clearing: the HV fuse must clear a short circuit on the LV terminals (I_FL/Z_pu) before the transformer is damaged (within thermal withstand time, e.g. 2 s).
- Discrimination: HV fuse must coordinate with LV fuses or LV breaker (curves referred to the same voltage must not cross).
- Breaking capacity ≥ system fault level; HRC or expulsion (drop-out) fuses on 11 kV.
- Use striker-pin fuses with a switch to avoid single phasing.
(ii) Capacitor protection
- Overcurrent due to harmonics and over-voltage: capacitor current may be up to 1.3 × rated (IEC 60871/60831) plus tolerance, so fuse rating is about 1.43–1.65 × rated capacitor current (commonly 1.5–1.8 ×).
- Switching inrush: high-frequency inrush at energising (up to 100 × I_rated for back-to-back switching) must not melt the fuse.
- Discharge energy: the fuse must withstand discharge of parallel units into a faulted unit; individual-unit fuses must clear a failed unit before case rupture (fuse curve below tank-rupture curve).
- Use individual (internal or external) fuses for each unit in large banks and a group fuse/breaker for the whole bank.
- Voltage rating to suit system, considering neutral shift in ungrounded star banks.
(iii) Lighting loads
- Rating just above the normal full-load current of the circuit (e.g. 1.1–1.25 × I_FL), and not greater than the current-carrying capacity of the wiring.
- Switch-on surges: incandescent lamps (cold filament ≈ 10–15 × for a few ms), discharge lamps and LED drivers (capacitive inrush) – choose a fuse that withstands these surges.
- Lighting circuits have little overload, so a fuse with low fusing factor can be used; fast HRC or cartridge fuses/MCBs are suitable.
- Discrimination with the distribution-board main fuse (ratio ≥ 1.6:1).
- Breaking capacity ≥ prospective fault at the board; fuse must clear an earth fault quickly enough for shock protection (fault loop impedance check).
- Power factor of discharge lamps and harmonic currents in the neutral should be considered.
- 2072 Magh · 6 marks
Write the types of MCB used in resistive load and inductive loads. Why?
Answer
A miniature circuit breaker (MCB) is a small, re-settable automatic switch that trips on overload (thermal bimetal element) and on short circuit (magnetic coil element). MCBs are grouped into tripping curves B, C and D (IEC 60898-1), according to the current at which the instantaneous magnetic trip operates. The choice of curve depends on the inrush (starting) current of the load.
Types of MCB by trip curve
| Type | Magnetic trip range | Typical loads |
|---|---|---|
| B | 3 to 5 × rated current In | Resistive loads: heaters, geysers, incandescent lamps, domestic sockets |
| C | 5 to 10 × In | Inductive loads: small motors, fans, pumps, fluorescent/LED lighting, AC units |
| D | 10 to 20 × In | Highly inductive loads: large motors, transformers, welding sets, X-ray machines |
| K / Z | 8–12 × In / 2–3 × In | K: motors and transformers; Z: semiconductor and electronic circuits |
MCB for resistive loads: Type B
- A resistive load (heater, iron, filament lamp) draws almost its normal current at switch-on; there is no large inrush current.
- So the breaker can be made sensitive. A Type B MCB trips instantly at 3–5 In, which clears faults quickly even on long cables with high loop impedance (small fault current).
- This gives better protection of cables and people without nuisance tripping.
MCB for inductive loads: Type C (or D)
- An induction motor takes 5–7 times full-load current while starting; a transformer takes 8–12 times rated current as magnetising inrush when switched on. Fluorescent lamps with chokes also draw a surge.
- If a Type B MCB were used, this harmless inrush would cross 3–5 In and the MCB would trip falsely every time the load is switched on.
- A Type C MCB (5–10 In) rides through normal motor and lighting inrush but still trips on a true short circuit. For very large inrush (DOL motors, transformers, welders) a Type D MCB (10–20 In) is used.
- The thermal (overload) part works the same in all types, so sustained overloads are still cleared.
Why the difference matters
- Correct type gives selectivity between normal inrush and fault current.
- Too sensitive a type → nuisance tripping; too insensitive a type → slow clearing of low-level faults and possible cable damage.
Example: a 2 kW room heater circuit uses a 10 A Type B MCB, while a 2 hp water pump motor circuit uses a 16 A Type C MCB.
Questions from Old Question Collection (EE 651) (IOE EE 651 exam papers from 2070 Bhadra to 2080 Chaitra (16 papers)). Answers are written for this site; check them against your class notes.
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