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Applied Electricity Two
Cross sectional area is calculated by using the formula:
                          ∏R  2



               ∏ Pi is a constant = 3.14

           R is the radius and it is squared
           which means multiplied by itself

           The Radius is half of the diameter
Applied Electricity Two




If you increase the diameter you increase the radius.

      If the diameter is 300 mm the radius is 150 mm
           If the diameter is increased to 500 mm
                     the radius is 250 mm
Applied Electricity Two




If you doubled the diameter of the conductor
   you would change the cross sectional area
                by how much?
Applied Electricity Two


      Example:
 A = ∏R2
   =3.14 (150x 150)
   =70685 mm   2

         or
  70685 x 10-6m2
         or
  70.685 x 10-3m 2
Applied Electricity Two

       Example:

A=    ∏R 2

 =3.14 (300x 300)
 =282,743 mm   2

       or
282743 x 10-6m   2

       or
   .282743 m 2
Applied Electricity Two



After doubling the diameter we have an area of
                    282743
             before the area was
                     70685
            divide 282743 by 70685
                      =4
          the area has increased by 4
Applied Electricity Two


By doubling the diameter (or radius) we have increased
                        the area
                          by 4.
   If we had increased the diameter (or radius) by 3
           we would increase the area by 9.

The increase is multiplied by itself because of the square
                    in the equation.
Applied Electricity Two



         A is measured in m2 ( meters squared)
    So if we are given a cross sectional area in mm2

we need to take into account that in the equation it will
     need to be converted to m2 (meters squared)
                        2.5 mm2
               in the equation looks like
                        2.5 x 10-6

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Cross Sectional Area

  • 1. Applied Electricity Two Cross sectional area is calculated by using the formula: ∏R 2 ∏ Pi is a constant = 3.14 R is the radius and it is squared which means multiplied by itself The Radius is half of the diameter
  • 2. Applied Electricity Two If you increase the diameter you increase the radius. If the diameter is 300 mm the radius is 150 mm If the diameter is increased to 500 mm the radius is 250 mm
  • 3. Applied Electricity Two If you doubled the diameter of the conductor you would change the cross sectional area by how much?
  • 4. Applied Electricity Two Example: A = ∏R2 =3.14 (150x 150) =70685 mm 2 or 70685 x 10-6m2 or 70.685 x 10-3m 2
  • 5. Applied Electricity Two Example: A= ∏R 2 =3.14 (300x 300) =282,743 mm 2 or 282743 x 10-6m 2 or .282743 m 2
  • 6. Applied Electricity Two After doubling the diameter we have an area of 282743 before the area was 70685 divide 282743 by 70685 =4 the area has increased by 4
  • 7. Applied Electricity Two By doubling the diameter (or radius) we have increased the area by 4. If we had increased the diameter (or radius) by 3 we would increase the area by 9. The increase is multiplied by itself because of the square in the equation.
  • 8. Applied Electricity Two A is measured in m2 ( meters squared) So if we are given a cross sectional area in mm2 we need to take into account that in the equation it will need to be converted to m2 (meters squared) 2.5 mm2 in the equation looks like 2.5 x 10-6