The existing in a 100 watt lightbulb is 0.900 A . The filament within the bulb is 0.280 mm in diameter.1-What is the present density in the filament? (with units)2-What is the electron current in the filament? (with units)

The ideas used to solve this difficulty are existing density and also electric current.

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Initially usage the current density acts on the filament within the bulb, by using existing passing with the bulb and also area the cross section of the bulb, then calculate the electron existing in the filament by using present in the irradiate bulb and also charge of the electron.


The expression because that the present density in the filament acts inside the pear is,

*

Here, J is the present density, i is the existing flows in the filament, and A is the area that the filament within the bulb.

The area the the bulb is,

*

Here, d is the diameter the the filament within the bulb.

The expression because that the electron existing passing with the filament is,

*

Here,

*
is the electron current and also e is the charge of one electron.


(1)

The area the the bulb is,

*

Substitute

*
because that d to find A.

*

The present density in the filament acts within the bulb is,

*

Substitute

*
for I and
*
because that A to uncover J.

*

(2)

The electron current passing v the filament is,

*

Substitute

*
for I and
*
because that e to uncover
*
.

*

Ans: component 1

The present density in the filament is

*
.

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Part 2

The electron current flow in the filament is

*
.


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