Question

A 12 cm wire given a shape of a right angled triangle ABC having side 3cm 4 cm 5 cm as shown in figure the ratio of resistance of side AB BC CA in

Answer 1

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Answer 2

Answer:

$R_{ab} : R_{bc} : R_{ca}$ = 27 : 32 : 35

Solution:  

Let the r and A be the resistivity and the area of cross section respectively.

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Resistance of the side AB = 3r/A

Resistance of the side BC = 4r/A

Resistance of the side CA = 5r/A

Resistivity between A and B = $R_{ab} = R_1(R_2+R_3)/R_1+R_2+R_3$

                        = [3r/A(4r/A+5r/A)]/[3r/A + 4r/A + 5r/A]

                        = 27r/12A

Resistivity between B and C = $R_{bc} = R_2(R_1+R_3)/(R_1+R_2+R_3)$

                        = [4r/A (3r/A+5r/A)]/[3r/A + 4r/A + 5r/A]

                        = 32r/12A

Resistivity between C and A = $R_{ca} = R_3(R_1+R_2)/R_1+R_2+R_3$

                        = [5r/A (3r/A+4r/A)]/[3r/A + 4r/A + 5r/A]

                        = 35r/12A

$R_{ab} : R_{bc} : R_{ca}$ = 27r/12A : 32r/12A : 35r/12A

$R_{ab} : R_{bc} : R_{ca}$ = 27/12 : 32/12 : 35/12

$R_{ab} : R_{bc} : R_{ca}$= 27 : 32 : 35