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Two manganin wires whose lengths are in the ratio 1:2 and whose resistance are in the ratio 1:2 are connected in series with a battery. What will be the ratio of drift velocities of free electrons in the two wires?

Two manganin wires whose lengths are in the ratio 1:2 and whose resistance are in the ratio 1:2 are connected in series with a battery. What will be the ratio of drift velocities of free electrons in the two wires?

Grade:12

3 Answers

Adarsh
733 Points
6 years ago
The equation used for solving this problem is I=neAvd
The current through the two wires are same as they are connected in series connection.
Now R = ρl/A
=> A α l/R
 => A1/A2 =( l1/l2)*( R2/R2) = (1/2) *(2/1)=1/1
So the area of the two wires are same.
Also e is electronic charge which is constant
For two wires of same material carrying same current the value of n(no. of electrons per unit volume) will also be same.
So the ratio of drift velocities of free electrons in the two wires will be 1:1
 
Please approve if my answer is correct.
Hence
 
 
Adarsh
733 Points
6 years ago
The word hence is not there it came out as a typing mistake.
Adarsh
733 Points
5 years ago
Please approve if my answer finds useful.

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