Cross Sectional Area

27485[/snapback]

what is the geometrical term for this CSA ?

27491[/snapback]

From the web

"Cross Sectional Area

(of a conductor) --- the sum of the cross- sectional areas of its component wires with that of each wire being measured perpendicular to its individual axis. This measure is expressed in square inches or more commonly in circular mils."

Regards

Bellman

Cross Sectional Area

27485[/snapback]

what is the geometrical term for this CSA ?

27491[/snapback]

From the web

"Cross Sectional Area

(of a conductor) --- the sum of the cross- sectional areas of its component wires with that of each wire being measured perpendicular to its individual axis. This measure is expressed in square inches or more commonly in circular mils."

Regards

Bellman

27493[/snapback]

Circular mil... So if CSA is 1mm it is 0,7854 square mm --> CSA0,2 mm --> 0,157 square mm.. Right?

Suppose a wire has a cross-sectional area of one millimeter squared (1 mm2). This is 0.000001 meter squared (10-6 m2). If the current density in this wire is 1 A/m2, then the wire carries 10-6 A, or one microampere (1 µA), a tiny current. Suppose this same wire carries a current of one ampere (1 A), which is an entirely plausible scenario. Then the current density in the wire is 1,000,000 amperes per meter squared (106 A/m2).

Sometimes, larger units of current density are specified. For example, one ampere per millimeter squared (A/mm2) represents a current of 1 A flowing through a conductor with a cross-sectional area of 1 mm. This unit is equal to 1,000,000 (106) A/m2. One milliampere per millimeter squared (mA/mm2) represents a current of 1 mA flowing through a conductor with a cross-sectional area of 1 mm. This unit is equal to 1,000 (103) A/m2.

Determination of current density is straightforward in direct-current (DC) and low-frequency alternating-current (AC) circuits, because the current is distributed uniformly throughout the cross section of a solid conductor. But at radio frequencies (RF), more current flows near the outer surface of a solid conductor than near its center. This is known as skin effect, and it dramatically reduces the conductivity of wires in RF applications as compared with DC and low-frequency AC circuits. At RF, current density is sometimes near zero near the center of a solid conductor, and quite high near the outer periphery. The average current density can nevertheless be calculated according to the following formula:

D = I/X

where D is the current density in amperes per meter squared, I is the current in amperes, and X is the cross-sectional area of the conductor in meters squared.

Regards

Bellman

Ill come back and read this later on when I cant sleep

No wonder Finland has the highest rate of suicide in the world

Suppose a wire has a cross-sectional area of one millimeter squared (1 mm2). This is 0.000001 meter squared (10-6 m2). If the current density in this wire is 1 A/m2, then the wire carries 10-6 A, or one microampere (1 µA), a tiny current. Suppose this same wire carries a current of one ampere (1 A), which is an entirely plausible scenario. Then the current density in the wire is 1,000,000 amperes per meter squared (106 A/m2).

Sometimes, larger units of current density are specified. For example, one ampere per millimeter squared (A/mm2) represents a current of 1 A flowing through a conductor with a cross-sectional area of 1 mm. This unit is equal to 1,000,000 (106) A/m2. One milliampere per millimeter squared (mA/mm2) represents a current of 1 mA flowing through a conductor with a cross-sectional area of 1 mm. This unit is equal to 1,000 (103) A/m2.

Determination of current density is straightforward in direct-current (DC) and low-frequency alternating-current (AC) circuits, because the current is distributed uniformly throughout the cross section of a solid conductor. But at radio frequencies (RF), more current flows near the outer surface of a solid conductor than near its center. This is known as skin effect, and it dramatically reduces the conductivity of wires in RF applications as compared with DC and low-frequency AC circuits. At RF, current density is sometimes near zero near the center of a solid conductor, and quite high near the outer periphery. The average current density can nevertheless be calculated according to the following formula:

D = I/X

where D is the current density in amperes per meter squared, I is the current in amperes, and X is the cross-sectional area of the conductor in meters squared.

Regards

Bellman

27501[/snapback]

I am familiar with this current density and that higher the freq goes, less current goes in the middle and more on the skin. But CSA 1mm is not 1 square mm. CSA 1mm is simply d² , but 1 square mm is ∏r² ...

Ill come back and read this later on when I  cant sleep

No wonder Finland has the highest rate of suicide in the world 

27507[/snapback]

You look like you havent slept for a month!

Ill come back and read this later on when I  cant sleep

No wonder Finland has the highest rate of suicide in the world 

27507[/snapback]

You look like you havent slept for a month!

27519[/snapback]

Only a month?

I am familiar with this current density and that higher the freq goes, less current goes in the middle and more on the skin. But CSA 1mm is not 1 square mm. CSA 1mm is simply  d² , but 1 square mm is  ∏r² ...

27517[/snapback]

Off out for a beer now (I do have a life)

I'll answer your question when i get back and my mind is suitably refreshed...

Regards

Bellman

Good for you : . I am waiting for call out's.

Georg,

CSA = ∏r 2

So as you correctly posted earlier

Cat5 seems to be 0,14 square milli metres.. --> diameter is approx 0,42 mm..

27239[/snapback]

Diameter = 0.42 Radius = 0.21

3.142 x (0.21 x 0.21) =

3.142 x 0.0441 =

0.14mm2 CSA