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Question: n the given star network the equivalent resistance between \(A\) and \(F\) is: ![](https://www.ved...

n the given star network the equivalent resistance between AA and FF is:

(A) 1.944R1.944\,R
(B) 0.973R0.973\,R
(C) 0.486R0.486\,R
(D) 0.243R0.243\,R

Explanation

Solution

In the given network first the resistance of the branch B to J is determined. By using that resistance value, the resistance of the triangle BCDBCD is determined. That resistance is the same for the remaining triangle of DEFDEF, FGHFGH and HIJHIJ. By using this, the resistance of the AF is determined.

Complete step by step solution
Assume that the line from the A meets the line BJ at the centre and the meeting point is L, then the equation is given by,
BJ=2×LJBJ = 2 \times LJ
Assume the triangle AEJAEJ is the right angle triangle, and the angle of JJ is given as 72{72^ \circ } in the diagram, then above equation is written as,
BJ=2×Rcos72BJ = 2 \times R\cos {72^ \circ }
The value of the cos72\cos {72^ \circ } from the trigonometry is 0.3090.309, by substituting this value in the above equation, then the above equation is written as,
BJ=2×0.309RBJ = 2 \times 0.309R
By multiplying the terms, then the above equation is written as,
BJ=0.62RBJ = 0.62R
Now, the resistance of RB{R_B} in the branch of BCDBCD, then the above equation is written as,
RB=2R×BJ2R+BJ{R_B} = \dfrac{{2R \times BJ}}{{2R + BJ}}
By substituting the value of the BJBJ in the above equation, then the above equation is written as,
RB=2R×0.62R2R+0.62R{R_B} = \dfrac{{2R \times 0.62R}}{{2R + 0.62R}}
By multiplying the terms in the numerator, then the above equation is written as,
RB=1.24R22R+0.62R{R_B} = \dfrac{{1.24{R^2}}}{{2R + 0.62R}}
By adding the terms in the denominator, then the above equation is written as,
RB=1.24R22.62R{R_B} = \dfrac{{1.24{R^2}}}{{2.62R}}
By cancelling the same terms, then the above equation is written as,
RB=1.24R2.62{R_B} = \dfrac{{1.24R}}{{2.62}}
On dividing the above equation, then the above equation is written as,
RB=0.473R{R_B} = 0.473R
The net resistance of the AFAF is given by,
RAF=R+2RB2{R_{AF}} = \dfrac{{R + 2{R_B}}}{2}
By substituting the value of the RB{R_B}, then the above equation is written as,
RAF=R+(2×0.473R)2{R_{AF}} = \dfrac{{R + \left( {2 \times 0.473R} \right)}}{2}
By multiplying the terms in the above equation, then the above equation is written as,
RAF=R+0.946R2{R_{AF}} = \dfrac{{R + 0.946R}}{2}
By adding the terms in the above equation, then
RAF=1.946R2{R_{AF}} = \dfrac{{1.946R}}{2}
By dividing the terms, then the above equation is written as,
RAF=0.973R{R_{AF}} = 0.973R

Hence, the option (B) is the correct answer.

Note: Hence the equivalent resistance between the AA and FF is given by the product of the 0.9730.973 and the resistance of RR. The resistance of RR is the same in two triangles. So, the equivalent resistance between the AA and FF depends only on the resistance of RR.