Part 103-108: Resistors and Resistance Worksheet 18B - Answers with Commentary

Introduction

Conductor resistance is an important part of any electrical installation - excessive conductor resistance increases power losses and voltage drop. Apart from being wasteful of energy, excessive heating can cause deterioration of the insulation around wires.

Insulation is the other essential component of an installation. Insulation protects against short circuit, leakage current, and electric shock. All installations must be tested to make sure that the separation between live parts and earth is sufficient to protect from short circuit, leakage current, and electric shock.

Insulation acts as a large number of resistances in parallel with the power supply. The path these currents take is sometimes referred to as leakage. This also means that the more length of insulation there is, the lower the insulation resistance, since there are more leakage paths in a longer cable.

Insulation Resistance Testing

NOTE: The information in this section is a simplified summary. The full requirements for insulation resistance testing are laid out in § 8.3.6 of AS/ANZ3000:2007 (The Wiring Rules). The Wiring Rules should always be consulted in the first instance.

Insulation resistance testing is carried out at a high voltage. The normal voltage for testing insulation resistance in a single phase installation is 500 V dc. The figure of 500 V is used because the peak value of the 230 V ac waveform is abour 325 V. A 500 V potential stresses the insulation beyond it's rating, so that if it passes at 500 V, it will also be ok with 230 V ac.

The device that carries out insulation testing is called an insulation resistance tester. An insulation resistance tester may also be referred to as a Megger, but Megger is a trademark name. That said, "to Megger an installation" will be universally understood as performing insulation testing.

The requirements for insulation resistance testing are laid out in § 8.3.6 of AS/ANZ3000:2007 (The Wiring Rules).

A table showing simplified insulation resistance requirements is below. All insulation resistance tests are between live parts (live and/or neutral) and earth. Unless you know specifically otherwise, always use a minimum insulation resistance value of 1 MΩ. Most of the time, insulation resistance values are given and processed in megohm (MΩ) values.

A table showing simplified insulation resistance testing voltage requirements is below. All insulation resistance tests are between live parts (live and/or neutral) and earth. Unless you know specifically otherwise, always use a test voltage of 500 V.

Insulation Resistance of Cables

The insulation resistance of a cable is proportional to how much current can leak through the cable insulation. The shorter the cable, the fewer paths there are.

• The shorter the cable, the higher the insulation resistance.
  
  
• The longer the cable, the lower the insulation resistance.
  
  

The insulation resistance is also affected by the following features of the cable itself.

• Cable length (already discussed above).
  
  
• Insulation material. Different insulation materials have different resistivity.
  
  
• Insulation condition. Insulation that is in good condition has higher insulation resistance.
  
  
• Insulation age. Older insulation tends to have lower insulation resistance.
  
  
• Insulation thickness. Thicker insulation has higher resistance.
  
  
• Moisture. Moisture in or around cables and terminations can decrease insulation resistance.
  
  
• Contamination. Contamination (e.g. salt, animal residues) in or around cables and terminations can decrease insulation resistance.
  
  
• Physical damage. Physical damage (e.g. physical impact, animal activity) in or around cables and terminations can decrease insulation resistance.
  
  
• Electrical damage. Electrical damage (e.g. overload, overheating, overvoltage) can cause damage to the insulation that can decrease insulation resistance.
  
  

All cables have a "cable constant" \( \rho \), that governs how the insulation resistance varies with length. \( \rho \) has the unit Ωm (or any suitable prefix), and is equal to the product of the insulation resistance and the length of the cable. The value is equivalent to resistivity.

The basic equation for \( \rho \) is given below.

\(\rho = R_1 L_1 \)

where \( \rho \) is the cable constant, \( R_1 \) is the insulation resistance of the cable sample, and \( L_1 \) is the length of the cable sample.

When \( \rho \) is known, the insulation resistance, or length of cable can be calculated for a different sample of cable.

• \( R_2 = \frac{\rho}{L_2} \), where \( R_2 \) is the insulation resistance of the different cable sample, and \( L_2 \) is the length of the different cable sample.
  
  
• \( L_2 = \frac{\rho}{R_2} \), where \( R_2 \) is the insulation resistance of the different cable sample, and \( L_2 \) is the length of the different cable sample.

Insulation Resistance of Appliances

NOTE: The information in this section is a simplified summary. The full requirements for insulation resistance testing of appliances are laid out in AS/ANZ3760:2010 (In-Service Safety Inspection and Testing of Electrical Equipment). The standard should always be consulted in the first instance.

Different types of appliances have their own standards, but the fundamentals above still apply. In general, the minimum insulation resistance between live parts and earth is 1 MΩ.

The normal test voltage is 500 V dc.

Some appliances have surge protectors (e.g. MOV), that will "clamp" the input voltage to about 400 V. In this case, the insulation test will appear to fail, as the insulation resistance tester is designed to output a maximum current of about 1.5 mA, which will not be able to drive the MOV to 500 V. If this is suspected, the test voltage can be reduced to 250 V dc.

It always pays to have an understanding of what the appliance is when judging insulation test results. A computer power supply with MOV protection may need testing at 250 V, but a heater may be tested at 500 V.

Some electrical equipment is Class II, or double insulated. Class II appliances are generally easier and cheaper to produce because of the lack of requirement for an earth connection. They are also safer in countries where enforcement of electrical standards may be variable. Some appliances may have no external conductive parts at all (e.g. enclosure made entirely of non-conductive material such as plastics), rendering an earth connection moot for safety.

A double insulated appliance will always say it is double insulated on the nameplate, and/or will have the symbol "⧈" (one square nested inside another) on the name plate.

The exact details will not be discussed here, but fundamentally Class II appliances do not have an earth connection between accessible metal parts (e.g. the chuck on a drill) and mains earth.

NOTE: Never assume an appliance is class II, even if it lacks an earth connection. An explicit indication of appliance class will be given on the nameplate.

Class II appliances are tested between live parts, and accessible metal parts. On some appliances such as power supplies, the "accessible metal parts" include the power supply output.

Appliances are tested as if they are "live" and connected to the mains. This means that all switches are in the "on" position wherever possible. On many appliances, this is difficult, because of electronic control modules and such. This is the reason that live and neutral are connected together during testing. Doing so gives the test voltage the chance to "back-propagate" up the neutral connection, and hopefully "touch" all internal parts of the appliance.

Question 1

Fill in the blanks. The answers are in bold.

When testing an installation or an appliance, the IR tester is connected between phase/neutral (clamped together) and earth. An important testing consideration is that all switches must be in the on position and all circuitry must be tested as being live.

Question 2

Fill in the blanks. The answers are in bold.

The minimum IR test result for an installation or an appliance is 1 MΩ. When this minimum IR test result is recorded, what is the current that will flow in the earthing conductor of an appliance or the protective earthing conductors for an installation, when 230V ac is connected? 230 µA.

Comments

If you don't know any exception to the 1 MΩ rule (e.g. you know a device with a functional earth is present), always assume the minimum insulation resistance is 1 MΩ.

The leakage current is calculated using Ohm's Law. The worst case is a 1 MΩ leakage path from 230 V to earth i.e. \( I = \frac{230}{1000000} = \) 230 µA. Note that the actual leakage current will depend on the location of the leakage path. A leakage path from the neutral side to earth will cause negligible current flow, due to New Zealand's use of a connection between neutral and earth at the main switchboard. However, it is still a leakage path from live parts to earth nonetheless.

Question 3

Fill in the blanks. The answers are in bold.

A normal test result for a new appliance/installation could be expected to be well over 1 MΩ e.g. 100 MΩ +.

A test result less than the minimum would indicate:

• Insulation failure.
• Contamination of connections.
• Cross-connections between live parts and earth.
• Incorrect wiring e.g. transposed neutral and earth.

Comments

1 MΩ is quite a low reading for insulation resistance. Most new appliances should be at least 1000 times that high. There is little economy on trying to skimp as much on insulation as possible. In order to be durable enough to be useful, most insulation will automatically have enough resistance.

I have only listed "actual" faults above. Errors due to equipment failure or incorrect use of equipment are not shown above.

I personally don't like the use of "would indicate". Saying "could indicate" is better, as the modes of failure are not a closed set.

An appliance/installation could fail for a variety of reasons, not all of which are actual "faults". Some of these are listed below.

• Incorrect use of test equipment. Don't assume everything your test device says is correct for your situation!
• Incorrect setup of the test. Make sure you're testing what you think you are testing!
• Forgetting to remove an earth-neutral link on a sub-main, especially if the sub-main is under RCD (electric shock) protection. Old sub-mains used to link neutral and earth together at the sub-main switchboard. This is not done today.
  
  
• Appliance or installation has a functional earth connection. This may cause a reading of as low as 0.05 MΩ per device. Multiple devices will make the insulation resistance even lower.
  
  
• Forgetting to disconnect the neutral when testing a sub-main installation. You will get a short through the "MEN" (earth-neutral) link in the main switchboard.
  
  
• A stove or hot water cylinder with mineral-insulated heating elements. This may cause a reading of as low as 0.01 MΩ per device. Multiple devices will make the insulation resistance even lower.

Question 4

Why should the main earthing conductor for an installation be connected to the protective earthing conductor being tested when an IR test is being performed?

When testing to earth, you want to do the hardest possible test to pass. If the test is harder to pass, and the installation passes it, the results will be more reliable. In this case, connecting the protective earthing conductor makes the test harder to pass because the insulation is being tested against the entire general mass of the earth, and everything connected to it.

This is a much more extensive test than testing against the earthing conductors of the circuit itself. It is also harder to pass, because there are more possible earth leakage paths.

In saying that, it is useful to "pre-verify" a circuit by testing against its own earth. But it is recommended that the final insulation resistance test be performed against the entire earthing system.

Question 5

A cable has an IR test of 77 MΩ and is 85m long. What is the IR of 22 m of this cable?

Solution

We have a sample length (85 m), and a sample resistance (77 MΩ).

We can calculate \( \rho \).

\( \rho = R_1 L_1 = 77 \cdot 85 = \) 6545 MΩm.

The insulation resistance of the 22 m length of cable can be calculated using \( \rho \).

\( R_2 = \frac{\rho}{L_2} = \frac{6545}{22} = \) 297.5 MΩ #.

Question 6

A cable is 45 m long. You test a 1 m off-cut and and get an IR test result of 12 MΩ. At that point in time your IR tester goes faulty. What could you reasonably expect the IR of the 45 m length to be?

Solution

Apart from the rights and wrings of continuing with a faulty piece of test equipment, here is the solution.

We have a sample length (1 m), and a sample resistance (12 MΩ).

We can calculate \( \rho \).

\( \rho = R_1 L_1 = 12 \cdot 1 = \) 12 MΩm.

The insulation resistance of the 45 m length of cable can be calculated using \( \rho \).

\( R_2 = \frac{\rho}{L_2} = \frac{12}{45} = \) 0.267 MΩ #.

As a side note, this cable is likely useless for anything except scrap, given that the maximum length it can be while still being compliant is 12 m.

Question 7

An old cable has an IR of 0.35 MΩ and is measured at 45 m long. What length of this cable would give a test result of 2 MΩ.

Solution

We have a sample length (45 m), and a sample resistance (0.35 MΩ).

We can calculate \( \rho \).

\( \rho = R_1 L_1 = 0.35 \cdot 45 = \) 15.75 MΩm.

The length of the cable required to give an insulation resistance of 2 MΩ can be calculated using \( \rho \).

\( L_2 = \frac{\rho}{R_2} = \frac{15.75}{2} = \) 7.875 m #.

As a side note, this cable is likely useless for anything except scrap, given that the maximum length it can be while still being compliant is 15.75 m.

Question 8

A cable on a drum is of unknown length. You cut 1 m off it and the IR test result for this off cut is 180 MΩ. You then test the remaining cable on the drum and record 2 MΩ. What is the length of cable on the drum?

Solution

We have a sample length (1 m), and a sample resistance (180 MΩ).

We can calculate \( \rho \).

\( \rho = R_1 L_1 = 180 \cdot 1 = \) 180 MΩm.

The length of the cable required to give an insulation resistance of 2 MΩ can be calculated using \( \rho \).

\( L_2 = \frac{\rho}{R_2} = \frac{180}{2} = \) 90 m #.