three register 4 ohm each are connected in the form of equilateral triangle find the distance between the corners

Dear student,

Suppose we need to find the equivalent resistance between A and C.

Resistance in AB and BC are in series. So, they make net = 4 + 4 = 8 Ω

This 8 Ω and AC are in parallel.

So, the required equivalent resistance is = (8)(4)/(8+4) = 2.67 Ω


  • 0
the answer is 8/3.
two resistance are in series 4+4=8.
now this resistance and 3rd resistance is in parallel.....8*4/(8+4)....
solve this u will get 8/3
  • 1
Hiiii heyyy
  • 0
Good m.orning cute Pye vinayak..
  • 0
hiii janab
  • 0
  • 0


The resistances are connected in the form of an equilateral triangle. So, it is understood that the two resistors are in series and the third resistors with both its end connected to the other two resistors is in parallel to the two.
So for series combination of resistors,
Rs = 4 + 4 = 8 Ω 
So this series combination is in parallel with the third resistor.
Thus, the effective resistance in the two corners of the triangle can be calculated using
1 over R subscript p space equals space 1 over R subscript 1 plus 1 over R subscript 2
fraction numerator begin display style 1 end style over denominator begin display style R subscript p end style end fraction space equals space 1 over 8 plus 1 fourth space equals space 3 over 8

T h u s comma space R subscript p space equals space 8 over 3 space equals space 2.66 space capital omega
Thus, the effective resistance between the two corners is 2.66 Ω
  • 1
  • 0
What are you looking for?