Two wires A & B 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 a 10 V battery. Calculate the ratio of their resistances.
Answers
Answer:
Answer
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(
2
2d
)
2
=πd
2
Resistance, R=ρ
A
L
=ρ
πd
2
L
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(
2
3d
)
2
=
4
9
×πd
2
=
4
9
×A
Resistance, R
′
=ρ
A
2L
′
=ρ
4
9
A
2L
=ρ
9A
8L
=
9
8
ρ
A
L
=
9
8
R
If current I flows through the combination then the potential difference across R
′
is, V
′
=IR
′
=
9
8
IR
So, the ratio of potential difference across the wires in series is,
V:V
′
=IR:
9
8
IR=9:8