Asked by james small help plz
The filament in an incandescent light bulb is made from tungsten (resistivity 5.6 x 10-8 Ù·m). The light bulb is plugged into a 120-V outlet and draws a current of 2.36 A. If the radius of the tungsten wire is 0.00464 mm, how long must the wire be?
restivity of the wire = 5.6 x 10-8 Ù·m
the cross sectional area (in m^2) of the wire is
A=pi x (5.6 x 10-8)^2 =2304
so the resistamce per metre is
P/A (units ohm/m)
if the bulb draws 2.36A from 120 V then the resistance, R, is
R=120/2.36
= 50.84745766
so the length of wire is
R/(P/A)=RA/P
restivity of the wire = 5.6 x 10-8 Ù·m
the cross sectional area (in m^2) of the wire is
A=pi x (5.6 x 10-8)^2 =2304
so the resistamce per metre is
P/A (units ohm/m)
if the bulb draws 2.36A from 120 V then the resistance, R, is
R=120/2.36
= 50.84745766
so the length of wire is
R/(P/A)=RA/P
Answers
Answered by
correction
am I on the right path, sorry 5 hrs of this is getting to me lol
restivity of the wire = 5.6 x 10-8 Ù·m
the cross sectional area (in m^2) of the wire is
A=pi x (0.00464)^2 = 0.0000215296
so the resistamce per metre is
P/A (units ohm/m)
if the bulb draws 2.36A from 120 V then the resistance, R, is
R=120/2.36
= 50.84745766
so the length of wire is
50.84745766/(0.000000056)=50.84745766*0.0000215296/P
907990315.357=0.00109472542/P
=
restivity of the wire = 5.6 x 10-8 Ù·m
the cross sectional area (in m^2) of the wire is
A=pi x (0.00464)^2 = 0.0000215296
so the resistamce per metre is
P/A (units ohm/m)
if the bulb draws 2.36A from 120 V then the resistance, R, is
R=120/2.36
= 50.84745766
so the length of wire is
50.84745766/(0.000000056)=50.84745766*0.0000215296/P
907990315.357=0.00109472542/P
=
Answered by
Anonymous
R=pL/A
R is resistance
P is resistivity
L is length
A is area
R is resistance
P is resistivity
L is length
A is area
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