Question

a potential difference V is applied to a conductor of length L,diameter D.How are the following i) electric field E,drift velocity v,resistance R affected when a) V is doubled b) L is doubled c) D is doubled

Answer 1

i). In the first case E= V/l so E is directly proportional to V so E will be doulbled. Since V= IR So , E= IR/l so Resistance is doubled. & Vd is directly proportional to E so Vd is dubled. So in 1st case everything is doubled.

2). If I is doubled then E= IR/l. so electric field is directly proportinal to I So E is doubled. From above relation u can get El/I = R so R is inversely proportinal to I so if I is doubled then resistance wil be half. & Since I= AneVd. so Vd will be doubled.

3. In third case There wil be no effect on E since E doesn't depend upon the diameter of the conductor. But R will be 1/4 . & Vd will be half Was this answer helpful

Answer 2

Answer:

Electric Field, E=  potential difference, V / length of the conductor, L

​Hence, the electric field is directly proportional to potential difference and inversely proportional to the length of the conductor.

And, Resistance, R= (Resistivity × length of the conductor, L)/ Area of the cross section, A

Hence, the resistance is directly proportional to the length of the conductor and inversely proportional to the area f cross section.

And, area of cross section can be written as  $A = \frac{\pi D^{4} }{4}$ .

Hence area of cross section is directly proportional to the square of the diameter of the conductor.

Case 1: when V is doubled:

As V and E are directly proportional to each other,  so E is doubled.

On halving the voltage, the current will also get halved, and by Ohm's law V=IR, such that R will remain unchanged.

Case 2: L is doubled:

As L and E are inversely proportional to each other, E will get halved.

And as L and R are directly proportional to each other, R will get doubled.

Case 3: D is doubled:

As E is independent of the area of cross-section, so E will remain unchanged.

Resistance is inversely proportional to area of cross-section and area of cross-section is directly proportional to the square of the diameter of the conductor, hence resistance is inversely proportional to square of the diameter of the conductor.

hence when D is doubled, area of cross section will become 4 times more, and resistance will become $(\frac{1}{4} )th$  times the original.