Question

state the change in the ammeter reading observed when a resistance wir in a circuit is replaced by a wire of same material and same length but comparatively double cross sectional area keeping the source of potential difference constant

Answer 1

The ammeter will show reading double the previous one....

We have a wire of resistance R

thus, its resistance is given by R = ρL/A ---------------------(1)

Now we replace the wire with another wire which is the same material and same length but double the cross sectional area,

so its resistance is given by R'= ρL/2A  

                                                  = 1/2(ρL/A)

                                              R'  = 1/2R { from (1)}

Thus the resistance of the new wire is half less than the previous wire.

Now according to ohms law, we know:

V/I = R

Thus, Resistance is inversely proportional to the current flowing through it.

If we take the ammeter reading of the wire with resistance R to be say, I

then the resistance of wire with R' will show ammeter reading as 2I

as R' ∝ 1/I

∴  1/2R ∝ 1/I

∴    R/2 = V/I

∴ V/R = 2I

Thus the ammeter reading will show double the increase in current when the new wire is placed.

[* also note, since material is same, therefore value of ρ is same]