A uniform metallic wire has a resistance of 18   Ω and is bent into an equilateral triangle. Then,…

A uniform metallic wire has a resistance of 18 Ω and is bent into an equilateral triangle. Then, the resistance between any two vertices of the triangle is:
  1. 2 Ω
  2. 12 Ω
  3. 8 Ω
  4. 4 Ω

Solution

Resistance,

R=ρlAR α lsay Reff=l=18as l1=l2=l3=l/3⇒R1=R2=R3=R=18/3=6

So from diagram

2R=12 and R=6also we know RAB=R2RRRR2R+RRRAB=6×126+12=4 Ω

Asked in: JEE Main 2019 (10 Jan Shift 1)

Practice more Current Electricity questions on Aicharya