question_answer

In the following star circuit diagram (figure), the equivalent resistance between the points A and H will be

A)            1.944 r

B)            0.973 r

C)            0.486 r

D)            0.243 r

Correct Answer: B

Solution :

                   Resistance of CD arm = 2r cos 72o = 0.62r Resistance of CBFC branch \[\frac{1}{R}=\frac{1}{2r}+\frac{1}{0.62r}=\frac{1}{r}\left( \frac{2.62}{2\times 0.62} \right)\] \[\frac{1}{R}=\frac{2.62}{1.24r}\]        \\[R=\frac{1.24r}{2.62}\] Equivalent \[R'=2R+r=2\times \frac{1.24r}{2.62}+r\] \[=r\,\left( \frac{2.48}{2.62}+1 \right)=1.946r\] Because the star circuit is symmetrical about the line AH \ Equivalent resistance between A and H \[\frac{1}{{{R}_{eq}}}=\frac{1}{R'}+\frac{1}{R'}\]Þ \[{{R}_{eq}}=\frac{R'}{2}=\frac{1.946}{2}r=0.973r\]