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Chapter 2 · 6 hours

Electric Shocks

IOE past exam questions

Past questions and answers

16 questions set from this chapter, 3 of them more than once. Most asked first.

  • Asked 3 times
  • 2080 Chaitra · 4 marks
  • 2078 Chaitra · 8 marks
  • 2070 Bhadra · 8 marks

Explain Wenner's test method for measurement of earth resistivity with diagram and derive the relation for the resistivity.

Answer

Wenner's four-electrode method measures the average resistivity of soil by passing current between two outer electrodes and measuring the voltage between two inner electrodes, all equally spaced in a straight line. It removes the effect of electrode contact resistance.

Arrangement

Four electrodes C1, P1, P2, C2 are driven in a straight line at equal spacing aa, to a small depth bb (b≪ab \ll a, usually b<a/20b < a/20). A current II is passed through the outer (current) electrodes C1 and C2 from an earth tester, and the voltage VV between the inner (potential) electrodes P1 and P2 is measured.

        Earth tester (megger)
     +---I----------------------+
     |      +---V---+           |
     |      |       |           |
     C1     P1      P2          C2
  ---|------|-------|-----------|--- ground
     |<--a->|<--a-->|<----a---->|
     depth b (b << a)

Derivation

Treat each electrode as a point source of current at the surface. Current II from a point spreads into the earth over hemispheres. At distance rr the current density and field are

J=I2πr2,E=ρJ=ρI2πr2J = \frac{I}{2\pi r^2}, \qquad E = \rho J = \frac{\rho I}{2\pi r^2}

so the potential at distance rr is

V(r)=∫r∞E dr=ρI2πrV(r) = \int_r^\infty E\,dr = \frac{\rho I}{2\pi r}

Current +I+I enters at C1 and −I-I leaves at C2. Using superposition:

Potential at P1 (distance aa from C1, 2a2a from C2):

VP1=ρI2π(1a−12a)=ρI4πaV_{P1} = \frac{\rho I}{2\pi}\left(\frac{1}{a} - \frac{1}{2a}\right) = \frac{\rho I}{4\pi a}

Potential at P2 (distance 2a2a from C1, aa from C2):

VP2=ρI2π(12a−1a)=−ρI4πaV_{P2} = \frac{\rho I}{2\pi}\left(\frac{1}{2a} - \frac{1}{a}\right) = -\frac{\rho I}{4\pi a}

Voltage between P1 and P2:

V=VP1−VP2=ρI2πaV = V_{P1} - V_{P2} = \frac{\rho I}{2\pi a}

The tester reads R=V/IR = V/I, so

ρ=2πaR\rho = 2\pi a R

with ρ\rho in Ω-m\Omega\text{-m}, aa in m and RR in Ω\Omega.

If the electrode depth bb is not small compared with aa (image of each electrode taken), the general relation is

ρ=4πaR1+2aa2+4b2−aa2+b2\rho = \frac{4\pi a R}{1 + \dfrac{2a}{\sqrt{a^2+4b^2}} - \dfrac{a}{\sqrt{a^2+b^2}}}

which reduces to ρ=2πaR\rho = 2\pi aR when b≪ab \ll a.

Procedure and notes

  • Readings are taken at several spacings (aa = 1, 2, 5, 10 m …) and in different directions. The current spreads roughly to a depth equal to aa, so the change of ρ\rho with aa shows layered soil.
  • Measurements are made in dry season to get the worst (highest) value for earthing design.
  • An AC or reversing-DC tester is used to avoid polarisation and stray currents.

Example: with a=5a = 5 m and R=3.2 ΩR = 3.2\ \Omega, ρ=2π×5×3.2=100.5 Ω-m\rho = 2\pi \times 5 \times 3.2 = 100.5\ \Omega\text{-m}.

  • Asked 2 times
  • 2075 Bhadra
  • 2071 Bhadra · 6 marks

What are the physiological effects of electric shock? What are good safety practices to avoid shocks?

Answer

An electric shock is the physiological reaction of the body when electric current passes through it. The damage depends mainly on the current (not voltage), its path, duration and frequency.

Physiological effects (50/60 Hz, hand to feet, typical values)

Body currentEffect
about 1 mAThreshold of perception, slight tingling
1–9 mAPainful shock, but person can let go
9–25 mALet-go limit crossed; muscles contract, cannot release grip
25–60 mABreathing difficulty, severe pain, possible unconsciousness
60–100 mA and aboveVentricular fibrillation (heart beats irregularly), usually fatal
above about 1 ACardiac arrest, severe burns, nerve and tissue damage

Other effects: burns at entry and exit points, internal burns, damage to nervous system, fall from height due to reflex action, and later kidney damage.

Dalziel's relation for the fibrillation threshold (used in IEEE 80):

Ib=0.116t A (50 kg person),Ib=0.157t A (70 kg)I_b = \frac{0.116}{\sqrt{t}}\ \text{A (50 kg person)}, \qquad I_b = \frac{0.157}{\sqrt{t}}\ \text{A (70 kg)}

where tt is shock duration in seconds (0.03–3 s).

Good safety practices to avoid shock

  1. Proper earthing of all metal frames and enclosures, with low earth resistance, so a fault trips the protection quickly.
  2. Use of RCD/ELCB (30 mA) and fast protective relays/fuses.
  3. Isolation and lock-out/tag-out: switch off, lock, tag and test for "dead" before work; earth the conductors being worked on.
  4. Insulation: use double-insulated tools, insulated mats, rubber gloves and boots of proper voltage rating.
  5. Safe clearances and barriers around live parts; warning signs.
  6. Substation design: earth mat, surface layer of crushed gravel, and equipotential bonding to keep step and touch potentials within safe limits.
  7. Trained and authorised persons only; work permits for HV work.
  8. Avoid moisture: do not touch switches with wet hands; use weather-proof fittings.
  9. Regular inspection and testing of insulation, earthing and protective devices.
  10. First aid knowledge: cut the supply, separate the victim with a dry insulating object, and give CPR.
  • Asked 2 times
  • 2075 Bhadra · 8 marks
  • 2073 Bhadra · 8 marks

It is necessary to obtain a tower-footing resistance of 20 Ω in a soil of resistivity ρs = 100 Ω-m using the most common three types of electrodes. Take a = 1.25 cm for rods and counterpoises and a depth y = 0.5 m for counterpoise. Calculate the required dimensions.

Answer

The three common footing electrodes are a hemisphere, a single driven rod and a counterpoise (buried horizontal wire). Their resistances (Begamudre, EHV AC Transmission Engineering) are:

Hemisphere (radius r):R=ρ2πrDriven rod (length L,radius a):R=ρ2πLln⁡2LaCounterpoise (length L,depth y):R=ρπL[ln⁡2L2ay−1]\begin{aligned} \text{Hemisphere (radius } r):\quad & R = \frac{\rho}{2\pi r} \\ \text{Driven rod (length } L, \text{radius } a):\quad & R = \frac{\rho}{2\pi L}\ln\frac{2L}{a} \\ \text{Counterpoise (length } L, \text{depth } y):\quad & R = \frac{\rho}{\pi L}\left[\ln\frac{2L}{\sqrt{2ay}} - 1\right] \end{aligned}

Given: R=20 ΩR = 20\ \Omega, ρ=100 Ω-m\rho = 100\ \Omega\text{-m}, a=0.0125a = 0.0125 m, y=0.5y = 0.5 m.

(a) Hemisphere

r=ρ2πR=1002π×20=0.796 mr = \frac{\rho}{2\pi R} = \frac{100}{2\pi \times 20} = 0.796\ \text{m}

(b) Driven rod

20=1002πLln⁡2L0.012520 = \frac{100}{2\pi L}\ln\frac{2L}{0.0125}

This is solved by trial:

LL (m)RR (Ω\Omega)
4.025.71
5.021.28
5.3820.00
L≈5.38 m(ln⁡2×5.380.0125=6.757)L \approx 5.38\ \text{m} \quad \left(\ln\frac{2 \times 5.38}{0.0125} = 6.757\right)

(c) Counterpoise

2ay=2×0.0125×0.5=0.1118 m\sqrt{2ay} = \sqrt{2 \times 0.0125 \times 0.5} = 0.1118\ \text{m} 20=100πL[ln⁡2L0.1118−1]20 = \frac{100}{\pi L}\left[\ln\frac{2L}{0.1118} - 1\right]
LL (m)RR (Ω\Omega)
5.022.24
6.019.50
5.8020.00
L≈5.80 m(ln⁡2×5.800.1118=4.641)L \approx 5.80\ \text{m} \quad \left(\ln\frac{2 \times 5.80}{0.1118} = 4.641\right)

Answer: hemisphere radius = 0.796 m; driven rod length = 5.38 m; counterpoise length = 5.80 m (at 0.5 m depth).

In practice the rod or counterpoise is preferred, since a 0.8 m radius hemisphere is hard to make; in high-resistivity soil several rods or longer counterpoises are used in parallel.

  • 2082 Shrawan · 6 marks

Describe different types of earthing.

Answer

Earthing (grounding) is the connection of electrical systems and equipment to the general mass of earth through a low-resistance path. It is done for safety of people, protection of equipment and stable system operation.

A. By purpose

  1. System (neutral) earthing: the neutral of generators and transformers is connected to earth. It fixes the system voltage with respect to earth and allows earth-fault protection to work.
  2. Equipment (body) earthing: non-current-carrying metal parts (motor frames, panels, tower bodies) are earthed so that a fault does not leave a dangerous voltage on them.
  3. Lightning/surge earthing: lightning arresters, earth wires and towers are earthed to discharge surge currents.

B. Methods of neutral (system) earthing

MethodFeature
Ungrounded (isolated)No intentional earth; high overvoltage (arcing ground)
Solid earthingNeutral directly earthed; healthy phase voltage held low
Resistance earthingResistor limits earth-fault current
Reactance earthingReactor limits fault current
Resonant (Petersen coil)Reactor tuned to line capacitance; arc self-extinguishes
Earthing transformerZig-zag transformer gives a neutral where none exists

C. Earth electrode types (equipment earthing)

  1. Plate earthing: a copper (60 × 60 cm, 3 mm) or GI plate buried vertically about 3 m deep, surrounded by charcoal and salt to keep moisture.
  2. Pipe earthing: a GI pipe (about 38 mm diameter, 2.5 m long) with holes, driven vertically, with alternate layers of charcoal and salt; water poured through a funnel. Most common for buildings.
  3. Rod earthing: copper or copper-bonded steel rods driven into the soil; used in sandy soil.
  4. Strip or wire earthing: copper/GI strips buried horizontally at about 0.5 m depth; used in rocky soil.
  5. Counterpoise: buried wires along transmission lines to reduce tower footing resistance.
  6. Earth mat/grid: mesh of conductors under a substation, giving low resistance and equipotential surface.
   Pipe earthing
   funnel --o
            |  GI pipe with holes
   ground --+------------
            |   charcoal + salt
            |   layers
            |
           ~2.5 m deep

D. LV system earthing (IEC)

TN-S, TN-C, TN-C-S, TT and IT systems, depending on how source and exposed parts are earthed.

  • 2079 Chaitra · 2+6 marks

Describe the step and touch potentials. Why does a person have an electric shock and point out the safety precautions and regulations in electrical system?

Answer

Step and touch potentials

When fault current IgI_g flows into the earth through an electrode or substation grid, the soil around it rises in potential, with a gradient on the surface.

  • Step potential: the voltage between the two feet of a person standing on the ground, about 1 m apart, without touching any earthed object.
  • Touch potential: the voltage between a person's hand touching an earthed structure and their feet on the ground, at about 1 m horizontal distance.
  GPR  ___  structure
      |   |  hand
      |   |   \o    touch: hand-to-feet
  ____|___|___/|\___ ground
              / \      step: foot-to-foot
       potential falls with distance

Why a person gets an electric shock

A shock occurs when the body becomes part of a closed circuit and a current passes through it. This happens when:

  1. A person touches a live conductor and earth at the same time, or two conductors at different potential.
  2. Insulation fails and a metal frame becomes live, and the frame is not properly earthed or protection does not trip.
  3. During an earth fault, the ground near an electrode rises in potential and a person bridges a step or touch voltage.
  4. Induced voltage on de-energised lines near live lines, or stored charge in capacitors and cables.

The current is I=V/(Rb+Rcontact)I = V/(R_b + R_{contact}); human body resistance is about 1000 Ω1000\ \Omega. Above about 10 mA the person cannot let go, and around 50–100 mA ventricular fibrillation can occur.

Safety precautions

  1. Proper equipment and system earthing; low-resistance substation earth mat with equipotential bonding.
  2. Surface layer of crushed rock (high resistivity) in switchyards to cut step and touch current.
  3. Fast protective relays, fuses and RCDs (30 mA) to cut the fault quickly.
  4. Isolate, lock-out, tag-out, test for dead and earth the circuit before work.
  5. Use insulated tools, rubber gloves, boots, mats and helmets of proper rating.
  6. Keep safe clearance; use fences, barriers and danger signs.
  7. Only trained and authorised persons, with work permits for HV work.
  8. Regular inspection of insulation and earthing; avoid wet conditions.

Safety regulations

  • Safe step and touch voltage limits as per IEEE Std 80:
Estep=(1000+6Csρs)0.116ts,Etouch=(1000+1.5Csρs)0.116tsE_{step} = (1000 + 6C_s\rho_s)\frac{0.116}{\sqrt{t_s}},\qquad E_{touch} = (1000 + 1.5C_s\rho_s)\frac{0.116}{\sqrt{t_s}}
  • Wiring and earthing codes (IEC 60364, IS 3043, National Electrical Code).
  • In Nepal: the Electricity Act 2049 and Electricity Regulations 2050 set rules for safety, clearances and licensing, and NEA follows its own safety manual and earthing standards.
  • 2077 Chaitra · 8 marks

What are the physiological effects of electric shocks? Explain the effects of electrical current in human body.

Answer

An electric shock is the effect produced when current flows through the human body. The body reacts to the current, so its effects depend on the magnitude of current, the path, the duration and the frequency.

Physiological effects of electric shock

  1. Sensation and pain: small currents stimulate nerves and cause tingling and pain.
  2. Muscular contraction (tetanus): current stimulates muscles; the hand grips the conductor and cannot let go, increasing the duration of shock.
  3. Respiratory paralysis: current through the chest affects breathing muscles and the brain's breathing centre, causing suffocation.
  4. Ventricular fibrillation: current through the heart disturbs its rhythm; the ventricles quiver instead of pumping, and blood circulation stops. This is the main cause of death.
  5. Cardiac arrest: large currents stop the heart completely (it may restart after the current is removed).
  6. Burns: heating (I2RtI^2Rt) at the contact points and inside tissue; arc burns from high-voltage flashes.
  7. Nerve and brain damage, loss of consciousness, and later kidney failure due to muscle breakdown.
  8. Secondary injuries: falls from height and injuries from involuntary movement.

Effects of electric current in the human body (50/60 Hz, about 1 s, hand to feet)

CurrentEffect on body
0.5–1 mAThreshold of perception
1–5 mAMild shock, not painful; can let go
6–10 mA (women), 9–16 mA (men)Painful; let-go threshold
16–25 mACannot let go; strong muscle contraction
25–60 mABreathing difficulty, severe pain, possible suffocation
60–100 mAVentricular fibrillation likely
100 mA – 1 AFibrillation, cardiac arrest, burns
above 1 ASevere burns, heart stops, tissue destroyed

Dalziel's safe body current relation for fibrillation threshold:

Ib=kt,k=0.116 (50 kg), 0.157 (70 kg)I_b = \frac{k}{\sqrt{t}}, \quad k = 0.116 \text{ (50 kg)},\ 0.157 \text{ (70 kg)}

For example, for t=0.5t = 0.5 s and a 50 kg person, Ib=0.116/0.5=0.164I_b = 0.116/\sqrt{0.5} = 0.164 A.

Factors deciding the severity

  • Magnitude of current = voltage ÷ body resistance (dry skin up to 100 kΩ, wet skin about 1 kΩ).
  • Path: hand to hand or hand to foot (through the heart) is most dangerous.
  • Duration: longer time lowers the safe current.
  • Frequency: 50–60 Hz is the most dangerous; DC and high frequency are less likely to cause fibrillation.
  • Body condition: weight, health, heart condition, moisture and contact area.
  • 2074 Bhadra

Improper Earthing leads to high electric potential, justify. Also explain how the system grounding have been incorporated in common practice?

Answer

Why improper earthing leads to high electric potential

When an earth fault occurs, the fault current IfI_f flows through the earth electrode of resistance RgR_g. The electrode and all connected metal parts rise to the ground potential rise (GPR):

V=IfRgV = I_f R_g
  • If the earthing is poor (high RgR_g due to dry soil, corroded or loose connections, too few electrodes), VV becomes large. For example, If=1000I_f = 1000 A with Rg=10 ΩR_g = 10\ \Omega gives V=10V = 10 kV on the equipment frame, while with Rg=1 ΩR_g = 1\ \Omega it is only 1 kV.
  • High RgR_g also limits the fault current, so fuses or relays may not trip, and the frame stays live for a long time.
  • If the frame is not earthed at all, a body touching it becomes the only path to earth and gets the full phase voltage.
  • Poor earth-mat design gives steep surface voltage gradients, so step and touch potentials exceed safe values.
  • In ungrounded systems an earth fault raises healthy-phase voltage to line value, and arcing grounds can produce 5–6 pu overvoltages.
  • Lightning current through a high tower footing resistance causes high tower-top voltage and back-flashover.

So improper earthing creates dangerous potentials on equipment and ground, threatens people and damages insulation.

System grounding in common practice

  1. Generators: neutral earthed through a high resistance or a distribution transformer with a secondary resistor, to limit stator earth-fault current to a few amperes.
  2. Transmission (66 kV and above): solid (effective) earthing of transformer neutrals, so that healthy-phase voltage stays below 80% of line voltage (X0/X1≤3X_0/X_1 \le 3, R0/X1≤1R_0/X_1 \le 1). This allows reduced insulation and surge arrester ratings. NEA's 132/220/400 kV systems are effectively earthed.
  3. Distribution (11/33 kV): solid earthing, or resistance/reactance earthing to limit fault current; resonant (Petersen coil) earthing is used in some countries for overhead networks.
  4. LV (400/230 V): neutral solidly earthed at the transformer (TN or TT systems); consumer installations have equipment earthing and RCDs.
  5. Delta systems: a zig-zag earthing transformer provides an artificial neutral.
  6. Substations: an earth mat (grid) of copper or GI conductors buried 0.5–1 m deep, with earth rods, bonding all equipment frames, fences and structures, covered with crushed gravel; designed per IEEE 80 so that step and touch voltages are safe.
  7. Transmission lines: overhead earth wires bonded to each tower; tower footing resistance kept low (below about 10–20 Ω) with rods or counterpoises.
  • 2074 Magh · 4 marks

Explain the phenomenon of touch potential and step potential with necessary figures.

Answer

When fault current flows into the earth through an electrode, tower or substation grid, the soil near it rises in potential. The potential is highest at the electrode and falls with distance, so the ground surface has a voltage gradient. A person in this area can bridge part of this voltage.

Step potential

Step potential is the voltage difference between two points on the ground surface one step (about 1 m) apart, experienced between the feet of a person walking near the electrode. Current flows foot to foot (through legs), so it is less dangerous than touch, but a fall can turn it into a worse contact.

Touch potential

Touch potential is the voltage between an earthed metal structure that a person touches with the hand and the ground surface where the person stands (about 1 m away). Current flows hand to feet, through the heart, so it is more dangerous. The mesh potential (largest touch voltage in a grid mesh) and transferred potential are special cases.

 V (potential)
 |\
 | \   GPR at electrode
 |  \___
 |      \____
 |  Et       \______ Es
 +--|---|-----|-|------> distance
    hand feet  foot foot
   (touch)     (step, 1 m)
       structure at GPR

Equivalent circuits

 Touch:  structure--Rb--[feet in parallel: Rf/2]--earth
 Step:   foot--Rf--Rb--Rf--foot (feet in series)

With foot resistance Rf≈3ρsR_f \approx 3\rho_s and body resistance Rb=1000 ΩR_b = 1000\ \Omega, the tolerable limits (IEEE 80, 50 kg person) are:

Etouch=(Rb+1.5ρs)0.116ts,Estep=(Rb+6ρs)0.116tsE_{touch} = (R_b + 1.5\rho_s)\frac{0.116}{\sqrt{t_s}}, \qquad E_{step} = (R_b + 6\rho_s)\frac{0.116}{\sqrt{t_s}}

A surface layer of crushed rock (high ρs\rho_s) and a closely spaced earth grid keep these potentials safe.

  • 2074 Magh · 2+2 marks

What are the effects of electric shock? Why and how are the effects different for DC and AC different frequencies of AC?

Answer

Effects of electric shock

Current through the body causes tingling and pain (about 1 mA), muscular contraction and inability to let go (about 10–20 mA), breathing difficulty (about 25–60 mA), ventricular fibrillation (about 60–100 mA and above), cardiac arrest and burns at higher currents, plus nerve damage and secondary injuries from falls.

Why and how DC and AC of different frequencies differ

The body's nerves and heart work on electrical signals of low frequency. The effect depends on how strongly the current repeatedly stimulates nerves and heart muscle.

  1. DC: a steady current stimulates nerves mainly when it is switched on or off, so muscles do not stay in continuous spasm. The let-go and fibrillation thresholds are about 2–4 times higher than for 50 Hz AC (let-go roughly 50–90 mA). DC can still cause burns and electrolysis of body fluids.
  2. AC at 50–60 Hz: this is the most dangerous range. Each cycle restimulates the muscles, causing continuous tetanic contraction, so the victim grips the conductor and cannot let go, and the heart is easily thrown into fibrillation.
  3. Higher frequency (above about 1 kHz): nerves cannot respond to each cycle, and current tends to flow near the surface of the body (skin effect). Thresholds of perception, let-go and fibrillation rise with frequency. At very high frequency (above about 100 kHz) the main danger is heating and burns rather than fibrillation.
SupplyLet-go thresholdFibrillation risk
DCHigh (2–4× AC)Lower
50–60 Hz ACLowest (about 10–16 mA)Highest
kHz and aboveRises with frequencyLow; burns dominate
  • 2073 Bhadra · 8 marks

Explain different methods of earth resistivity measurement with diagrams.

Answer

Earth resistivity ρ\rho is measured to design earthing systems (substation grids, tower footings). The common methods use an earth tester that injects current II and measures voltage VV, giving R=V/IR = V/I.

1. Wenner four-electrode method

Four electrodes are placed in a straight line at equal spacing aa, driven to a depth b≪ab \ll a. Current is passed between the outer pair (C1, C2); voltage is measured between the inner pair (P1, P2).

   C1      P1      P2      C2
 --|-------|-------|-------|--
   |<- a ->|<- a ->|<- a ->|

Potential due to a point source at the surface is ρI2πr\frac{\rho I}{2\pi r}. By superposition:

V=ρI2π[(1a−12a)−(12a−1a)]=ρI2πaV = \frac{\rho I}{2\pi}\left[\left(\frac{1}{a}-\frac{1}{2a}\right) - \left(\frac{1}{2a}-\frac{1}{a}\right)\right] = \frac{\rho I}{2\pi a} ρ=2πaR\rho = 2\pi a R

Readings at increasing spacing give resistivity at increasing depth (about depth aa).

2. Schlumberger (Schlumberger–Palmer) method

The potential electrodes are kept close together (spacing dd), and the current electrodes are placed at distance cc outside each potential electrode. Only the current electrodes are moved for deeper readings, so it is faster and more sensitive for large spacings.

   C1          P1  P2          C2
 --|-----------|---|-----------|--
   |<--- c --->|<d>|<--- c --->|
ρ=πc(c+d)Rd\rho = \frac{\pi c(c+d)R}{d}

For c=d=ac = d = a this gives ρ=2πaR\rho = 2\pi aR, the Wenner case.

3. Driven-rod (three-point / variation-of-depth) method

A test rod of length LL and diameter dd is driven into the soil, and its resistance RR is measured by the fall-of-potential method using a current probe C (far away, at least 5–10 times rod length) and a potential probe P at about 62% of that distance.

  Earth tester
   E      P            C
 --|------|------------|--
   |<-0.62D->|
   |<-------- D ------->|
   test rod
ρ=2πLRln⁡(8Ld)−1\rho = \frac{2\pi L R}{\ln\left(\dfrac{8L}{d}\right) - 1}

Repeating with the rod driven deeper gives the change of resistivity with depth. It is simple and suited to small areas but measures only the soil near the rod.

Practical points

  • Take readings in several directions and at several spacings; use the dry-season (highest) value for design.
  • Use AC or reversing DC to avoid polarisation and errors from stray currents.
  • Keep test leads away from buried metal pipes and cables, which distort the readings.
  • 2073 Magh · 8 marks

List out the various types of earthing techniques commonly practiced in industry for personnel and equipment safety. Discuss how the properly designed earthing system in electric substation prevents the personnel from getting possible electric shock.

Answer

Earthing techniques commonly practised in industry

  1. Pipe earthing: perforated GI pipe (about 38 mm, 2–3 m long) in a pit with charcoal and salt; most common for buildings and small installations.
  2. Plate earthing: copper or GI plate (600 × 600 mm) buried about 3 m deep with charcoal and salt.
  3. Rod earthing: copper-bonded steel rods driven into the ground; easy where soil is soft.
  4. Strip/wire earthing: horizontal copper or GI strips buried 0.5 m deep; used in rocky soil.
  5. Chemical (maintenance-free) earthing: electrode with conductive backfill compound such as bentonite.
  6. Earth mat/grid: horizontal mesh with vertical rods for substations and power plants.
  7. Counterpoise and tower footing: buried wires/rods for transmission towers.
  8. Neutral earthing of the system: solid, resistance, reactance or resonant earthing of transformer and generator neutrals.
  9. Equipment earthing and bonding: frames, enclosures, cable sheaths and structures bonded to the earth bus; double earthing of major equipment.
  10. Lightning protection earthing: separate earth pits for arresters and lightning conductors, bonded to the main grid.

How a properly designed substation earthing protects people

During an earth fault, fault current IgI_g flows into the soil and the whole grid rises to GPR=IgRgGPR = I_g R_g. The danger to a person is not the GPR itself but the difference in potential across the body (step, touch, mesh and transferred potentials). A good design (IEEE Std 80) keeps these differences below safe limits:

   ____  ____  ____  earth mat (mesh)
  |    ||    ||    |  0.5-1 m deep
  |____||____||____|  + vertical rods
  |    ||    ||    |
  |____||____||____|
   gravel layer on top, fence bonded
  1. Low grid resistance: a large grid with many rods lowers RgR_g and thus the GPR.
  2. Equipotential area: the closely spaced mesh makes the ground surface nearly the same potential as the equipment, so touch and mesh voltages are small. Finer mesh is used near equipment operated by people.
  3. Bonding of all metal: frames, structures, fences, cable sheaths and gates are bonded to the grid, so a person touching them stands on ground at nearly the same potential.
  4. High-resistivity surface layer: 10–15 cm of crushed rock (2000–3000 Ω-m) greatly increases foot contact resistance and reduces body current.
  5. Grading rings and perimeter conductors outside the fence reduce steep gradients at the edge and step voltage outside.
  6. Fast fault clearing: low RgR_g gives enough fault current for relays to trip quickly; shorter tst_s allows larger safe voltages:
Etouch=(1000+1.5Csρs)0.116tsE_{touch} = (1000 + 1.5C_s\rho_s)\frac{0.116}{\sqrt{t_s}}
  1. Control of transferred potential: isolation of communication circuits and care with fences and rails leaving the yard.
  2. Low earth-conductor impedance so that lightning and switching surges are discharged safely.

Thus a well-designed earth mat makes the substation an equipotential zone where actual step and touch voltages remain below the tolerable values, so personnel do not receive a dangerous shock.

  • 2072 Magh · 4 marks

Mention basic electrical safety rules and regulations.

Answer

Electrical safety rules are practices and legal requirements that protect people and equipment from shock, burns, fire and explosion.

Basic safety rules

  1. Work on a circuit only after it is switched off, isolated, locked and tagged (lock-out/tag-out).
  2. Test for dead with a proper tester and earth/short the conductors before touching them; treat every circuit as live until proven dead.
  3. Use insulated tools, rubber gloves, boots, mats, helmets and face shields of correct voltage rating.
  4. Keep safe clearance from live parts; use barriers, fences and danger boards.
  5. All metal frames must be properly earthed; use RCD/ELCB on socket circuits.
  6. Use correctly rated fuses, MCBs and cables; never bypass protection.
  7. Do not work on electrical equipment with wet hands or in wet conditions.
  8. Only trained and authorised persons should operate or repair HV equipment; use work permits for HV work.
  9. Discharge capacitors and cables before handling.
  10. Keep fire extinguishers (CO2, dry powder) nearby; never use water on live equipment.
  11. Learn first aid and CPR; display shock-treatment charts.

Regulations

  • Nepal: the Electricity Act 2049 (1992) and Electricity Regulations 2050 (1993) cover licensing, safety, clearances of lines from buildings and ground, and responsibility for accidents; NEA safety rules and the Nepal Electricity Authority Act apply to utility work.
  • International/standards: IEC 60364 (LV installations), IEEE Std 80 (substation earthing), IS 3043 (earthing), IEC 61936 (HV installations), and occupational safety codes (e.g. NFPA 70E).
  • Regular inspection and testing of insulation resistance, earth resistance and protective devices is required by these codes.
  • 2072 Magh · 4 marks

Why is it necessary to measure earth resistivity?

Answer

Earth resistivity ρ\rho (Ω-m) is the resistance of a 1 m cube of soil. It must be measured because every earthing design depends on it, and it changes widely (from about 10 Ω-m in wet clay to over 1000 Ω-m in rock) with soil type, moisture, salts and temperature.

Reasons for measuring earth resistivity

  1. Design of earth electrodes and substation grids: the earth resistance is directly proportional to ρ\rho (e.g. rod R=ρ2πLln⁡2LaR = \frac{\rho}{2\pi L}\ln\frac{2L}{a}). Without ρ\rho the number, size and depth of electrodes cannot be chosen.
  2. Safety of personnel: step and touch potentials depend on ρ\rho of the soil and surface layer; IEEE 80 design needs ρ\rho to check safe limits.
  3. Tower footing resistance: lightning performance and back-flashover rate of lines depend on footing resistance, which depends on ρ\rho.
  4. Choice of site and best location for earth pits, and the best depth (layered soil), since resistivity changes with depth.
  5. Protection operation: a low earth resistance gives enough fault current for relays to operate quickly.
  6. Corrosion assessment: low-resistivity soil is more corrosive for buried metal and pipelines; also needed for cathodic protection design.
  7. Geophysical surveys: locating water, ore bodies and rock layers.

It is measured mainly by the Wenner four-electrode method (ρ=2πaR\rho = 2\pi aR), preferably in the dry season to get the worst-case value.

  • 2072 Magh · 8 marks

It is necessary to obtain a tower footing resistance of 16 Ω in a soil of resistivity ρs = 100 Ω-m using the three common types of electrodes. Calculate the required dimensions if the radius of rod and counterpoise is 1.25 cm and depth of counterpoise is 0.6 m.

Answer

The three common electrodes are a hemisphere, a driven rod and a counterpoise. Using the standard relations (Begamudre, EHV AC Transmission Engineering):

Hemisphere:R=ρ2πrDriven rod:R=ρ2πLln⁡2LaCounterpoise:R=ρπL[ln⁡2L2ay−1]\begin{aligned} \text{Hemisphere:}\quad & R = \frac{\rho}{2\pi r} \\ \text{Driven rod:}\quad & R = \frac{\rho}{2\pi L}\ln\frac{2L}{a} \\ \text{Counterpoise:}\quad & R = \frac{\rho}{\pi L}\left[\ln\frac{2L}{\sqrt{2ay}} - 1\right] \end{aligned}

Given: R=16 ΩR = 16\ \Omega, ρ=100 Ω-m\rho = 100\ \Omega\text{-m}, a=0.0125a = 0.0125 m, y=0.6y = 0.6 m.

(a) Hemisphere

r=ρ2πR=1002π×16=0.995 mr = \frac{\rho}{2\pi R} = \frac{100}{2\pi \times 16} = 0.995\ \text{m}

(b) Driven rod

16=1002πLln⁡2L0.012516 = \frac{100}{2\pi L}\ln\frac{2L}{0.0125}
LL (m)RR (Ω\Omega)
5.021.28
6.018.22
7.015.96
6.9816.00
L≈6.98 m(ln⁡2×6.980.0125=7.018)L \approx 6.98\ \text{m} \quad \left(\ln\frac{2 \times 6.98}{0.0125} = 7.018\right)

(c) Counterpoise

2ay=2×0.0125×0.6=0.1225 m\sqrt{2ay} = \sqrt{2 \times 0.0125 \times 0.6} = 0.1225\ \text{m} 16=100πL[ln⁡2L0.1225−1]16 = \frac{100}{\pi L}\left[\ln\frac{2L}{0.1225} - 1\right]
LL (m)RR (Ω\Omega)
7.017.00
8.015.41
7.6016.00
L≈7.60 m(ln⁡2×7.600.1225=4.821)L \approx 7.60\ \text{m} \quad \left(\ln\frac{2 \times 7.60}{0.1225} = 4.821\right)

Answer: hemisphere radius = 0.995 m; driven rod length = 6.98 m; counterpoise length = 7.60 m (buried 0.6 m deep).

  • 2071 Magh · 8 marks

Write about the physiological effects of electric shock and the factors affecting its magnitude of damage.

Answer

Electric shock is the body's reaction to current passing through it. The current disturbs the electrical signals of nerves, muscles and heart and heats the tissues.

Physiological effects

Current (50 Hz, about 1 s)Effect
0.5–1 mAPerception threshold, tingling
1–9 mAPainful shock; can let go
9–25 mACannot let go; muscle contraction (tetanus)
25–60 mABreathing difficulty, respiratory paralysis
60–100 mA+Ventricular fibrillation, usually fatal
above 1 ACardiac arrest, deep burns, tissue damage

Main effects explained:

  1. Nerve stimulation and pain.
  2. Tetanic muscle contraction: the hand grips the conductor, so the shock lasts longer.
  3. Respiratory arrest: chest muscles and the brain's breathing centre are paralysed.
  4. Ventricular fibrillation: the heart quivers and stops pumping blood; death follows in minutes without defibrillation.
  5. Burns: contact burns, arc burns and internal burns from I2RtI^2Rt heating.
  6. Other: unconsciousness, nerve and brain damage, kidney damage, falls and fractures.

Factors affecting the magnitude of damage

  1. Magnitude of current: the most important factor; I=V/RbodyI = V/R_{body}.
  2. Duration of current: the safe current falls with time. Dalziel's relation for fibrillation threshold:
Ib=0.116t A (50 kg),Ib=0.157t A (70 kg)I_b = \frac{0.116}{\sqrt{t}}\ \text{A (50 kg)}, \qquad I_b = \frac{0.157}{\sqrt{t}}\ \text{A (70 kg)}

e.g. for 1 s, 116 mA; for 0.1 s, 367 mA (50 kg). 3. Path of current: paths through the heart (hand–hand, left hand–feet) are most dangerous; foot–foot is less dangerous. 4. Body resistance: dry skin can be 100 kΩ, wet or broken skin about 1 kΩ (500 Ω internal). Moisture, sweat, contact area and pressure lower it. 5. Voltage: higher voltage breaks down skin resistance and drives more current; above about 500 V burns become severe. 6. Frequency: 50–60 Hz is most dangerous; DC and high frequencies need larger currents for the same effect. 7. Phase of heart cycle: shock during the T-wave (vulnerable period) easily causes fibrillation. 8. Body weight, age, sex and health: lighter persons, women, children and heart patients have lower thresholds. 9. Contact condition: footwear, floor material and surface layer resistivity (crushed rock) add series resistance. 10. Speed of rescue and first aid: early CPR and defibrillation greatly reduce fatality.

  • 2070 Magh · 8 marks

Explain how Improper Earthing may result in step potential and touch potential beyond safe value in power system with necessary diagram and formulae.

Answer

When an earth fault occurs, fault current IgI_g enters the soil through the earth electrode or substation grid. The grid rises to the ground potential rise

GPR=IgRgGPR = I_g R_g

and the ground surface potential falls with distance from the electrode. A person bridging part of this potential gets a shock. Improper earthing (high RgR_g, too few electrodes, wide mesh, no bonding, no surface layer, slow protection) makes these potentials exceed the safe values.

 V
 |\_ GPR = Ig Rg
 |  \       steep gradient
 |   \__    (poor earthing)
 |      \___
 |  Et      \____ Es
 +---|--|-----|-|------> distance
   structure  feet 1 m apart

Touch potential

Voltage between the hand on an earthed structure and the feet standing 1 m away. Equivalent circuit: body resistance RbR_b in series with two feet in parallel (Rf/2R_f/2), where Rf≈3ρsR_f \approx 3\rho_s:

Ib=EtouchRb+1.5ρsI_b = \frac{E_{touch}}{R_b + 1.5\rho_s}

Step potential

Voltage between two feet 1 m apart. Body RbR_b in series with two feet (2Rf2R_f):

Ib=EstepRb+6ρsI_b = \frac{E_{step}}{R_b + 6\rho_s}

Safe (tolerable) limits (IEEE 80, 50 kg person)

Using Dalziel's current Ib=0.116tsI_b = \frac{0.116}{\sqrt{t_s}} and Rb=1000 ΩR_b = 1000\ \Omega:

Etouch,50=(1000+1.5Csρs)0.116ts,Estep,50=(1000+6Csρs)0.116tsE_{touch,50} = (1000 + 1.5C_s\rho_s)\frac{0.116}{\sqrt{t_s}}, \qquad E_{step,50} = (1000 + 6C_s\rho_s)\frac{0.116}{\sqrt{t_s}}

CsC_s is the surface-layer derating factor and tst_s the fault duration.

How improper earthing makes them unsafe

  1. High grid resistance RgR_g: GPR =IgRg= I_gR_g is large, so actual touch voltage (a fraction of GPR) is large. For example, Ig=2I_g = 2 kA and Rg=5 ΩR_g = 5\ \Omega gives GPR = 10 kV; even 30% of this (3 kV) as touch voltage is far above the safe value of a few hundred volts.
  2. Wide mesh or no grid: surface potential is not uniform; mesh (touch) voltage
Em=ρKmKiIgLME_m = \frac{\rho K_m K_i I_g}{L_M}

rises when the total buried length LML_M is small. 3. Steep gradient at the edge (no grading ring): high step voltage

Es=ρKsKiIgLSE_s = \frac{\rho K_s K_i I_g}{L_S}
  1. No surface layer of crushed rock: low ρs\rho_s, so tolerable voltages drop.
  2. Unbonded equipment or fences: the full GPR may appear as touch voltage.
  3. High RgR_g reduces fault current so relays may be slow; longer tst_s lowers the tolerable limits.

Remedies

Lower RgR_g (more rods, larger grid, deep electrodes, chemical backfill), closer mesh spacing, bonding of all metal parts, a crushed-rock surface layer, grading rings, and fast fault clearing so that Eactual<EtolerableE_{actual} < E_{tolerable}.

Questions from Old Question Collection (EE 751) (IOE exam papers from 2066 Magh to 2082 Shrawan (2066–2069 papers from the older course)). Answers are written for this site; check them against your class notes.

Chapter titles and hours from the IOE syllabus ↗