Two metallic wires A and B are connected. A has length l and radius r while B has length 2l and radius 2r. Find the ratio of effective resistance of series combination of both and resistance of wire A?
Answers
WIRE A WIRE B
length = l length = 2l
radius = r radius= 2r
Let the length of resistance of Wire A (R1) = p×l÷πr² ................(1)
Let the length of resistance of Wire B (R2) = p×2l÷π(2r)²
=p×2l÷π4r²
=2/4{p×l÷πr²}
=1/2[p×l÷πr²] ..................(2)
By ep (1) & (2)
we get R1 = 1/2 R1
As the wire are connected in series
= resistance of wire A+ resistance of wire B
=R1 + R2
= R1 + 1/2 R1
=3/2 R1
RATIO= 3/2 R1 ÷ R1
=3/2