Some of the important electrical properties of insulators are:

(a) Insulation resistance

(b) Volume and surface resistance

(c) Dielectric strength and dielectric loss

(d) Breakdown voltage i.e., the voltage above which the material ceases to act as insulator.

**(a) Insulation Resistance:**

The primary function of an insulating material is to confine the current to a particular path in the conductor and to provide a barrier from other conducting elements. If it has to provide a harrier for the unwanted path for the current, it should have high insulation resistance. When an insulating material is subjected to electrical pressure due to the Ohms law, there is always a small amount of current that will flow through it even when the resistance is too high. This current is called leakage current. The magnitude of the leakage current that flows through the insulator plays an important role in the electrical engineering. Insulation resistance is of two types:

(a) Volume resistance

(b) Surface resistance

**(b) Volume and surface resistance**

**Volume Resistance: **

It is the resistance offered to the flows of current through a unit cube of a material. It is measured in Ohm-cm or Ohm-m. Volume resistance can be expressed as:

R

_{v}= ρ_{v}*l*/aWhere ρ

_{v}is the volume resistivity in Ohm-ml is the length of the current path through the material in metres

a is the area of cross-section of the current path in metre

^{2}(perpendicular to the current path)R

_{v }is the volume resistance in ohms.For a cube of a unit dimensions, the volume resistance is equal to volume resistivity (R

_{v}= ρ_{v}).In practice the units of volume resistivity is called specific volume conductivity. In the case of parallel plate configuration it is known as insulation resistance but in many cases the configurations are cylindrical. These are considered to be of the shape of two co-axial cylinders with radius R1, R2 and R1 gives the distance between potential points (L) using the same formula as before we get.

R

_{v}= ρ_{v}L/aBut a = ½[2πR

_{2}L + 2πR_{1}L] = πL [R_{2}+ R_{1}]Where R

_{2}and R_{1}are external and internal radiusL = R

_{2}- R_{1}Therefore, R

_{v}= ρ_{v}[(R_{2}- R_{1})/ πL(R_{2}+ R_{1})]

**Surface Resistivity:**

It is the resistance offered to the flow of current of a unit square over the surface of the insulating material and is expressed in ohm/m

^{2}. The resistance offered is called surface resistance and is given byR

_{s}= ρ_{s}[*l*_{s}/a_{s}]Where,

*l*_{1}= AB + BC + CD – length of currenta

_{s}= YThe unit of surface resistivity is ohm.

The calculation of insulation resistance of the material R

_{i}is given byR

_{i}= R_{v}R_{s}/( R_{v }+ R_{s}) ohm**Factors Affecting Insulation Resistance:**

The following factors affect the resistance of an insulating material:

(a) Insulation resistance decreases with the temperature due to atomic dislocation.

(b) Moisture decreases the surface resistance and causes insulation breakdown when insulators are exposed to moisture.

(c) The resistance varies with the change in applied voltage above a certain potential.

(d) When the insulation serves for long period its resistance will decrease with the age of service.

**Surface and Volume Resistivity of Insulating Materials**

Material | Surface Resistivity | Volume Resistivity MΩ cm | |

at 50% Humidity MΩ | at 90% Humidity MΩ | ||

Wood | 4 x 10 ^{6} | 7 x 10 ^{3} | 4 x 10 ^{7} |

Mica | 2 x 10 ^{7} | 8 x 10 ^{3} | 2 x 10 ^{11} |

Glass | 5 x 10 ^{4} | 2 x 10 | 2 x 10 ^{7} |

Porcelain (U galvanized) | 6 x 10 ^{5} | 5 x 10 | 3 x 10 ^{8} |

Sulphur | 7 x 10 ^{9} | 1 x 10 ^{8} | 1 x 10 ^{11} |

Marble | 3 x 10 ^{3} | 2 x 10 | 3 x 10 ^{5} |

Hard Rubber | 3 x 10 ^{9} | 2 x 10 ^{3} | 1 x 10 ^{12} |

Sealing wax | 2 x 10 ^{9} | 9 x 10 ^{7} | 8 x 10 ^{2} |

Amber | 6 x 10 ^{8} | 1 x 10 ^{8} | 5 x 10 ^{10} |

Celluloid | 5 x 10 ^{4} | 2 x 10 ^{2} | 2 x 10 ^{4} |

## 0 on: "Electrical Properties of Insulators"