Given: Two wires of the same material having lengths in the ratio of
1:2 and diameters in the ratio of
2:3 are connected in series with an accumulator.
To find the ratio of the potential difference across the two wires.
Solution:
Let ρ be the resistivity of the material of the wire.
First wire:
Length = L
Diameter = 2d
Area of Cross-section, A = π(2d2)2=πd2
Resistance, R=ρLA=ρLπd2
If current I flows through the combination then the potential difference across R is, V=IR
Second wire:
Length = 2L
Diameter = 3d
Area of Cross-section, A′=π(3d2)2=94×πd2=94×A
Resistance, R′=ρ2LA′=ρ2L94A=ρ8L9A=89ρLA=89R
If current I flows through the combination then the potential difference across R′ is, V′=IR′=89IR
So, the ratio of potential difference across the wires in series is,
V:V′=IR:89IR=9:8