- #1
satishinamdar
- 22
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A conducter with two diameters d1 and d2 (one single piece), carries a current.having lengths L.
What is same
n=free electrons in one cubic mtr
e=constant electronic charge=1.6X10^-19columbs
Vd=drift velocity
I solved as
Let total electrons in rod A be N
therefore I=N*e*Vd*A/A*L
Let total electrons in rod B be N'
Therefore I'=N'*e*Vd*A'/A'*L
THEREFORE I is not equal to I'
BECAUSE DRIFT VELOCITY REMAINS CONSTANT
What is same
n=free electrons in one cubic mtr
e=constant electronic charge=1.6X10^-19columbs
Vd=drift velocity
I solved as
Let total electrons in rod A be N
therefore I=N*e*Vd*A/A*L
Let total electrons in rod B be N'
Therefore I'=N'*e*Vd*A'/A'*L
THEREFORE I is not equal to I'
BECAUSE DRIFT VELOCITY REMAINS CONSTANT