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 \[{{R}_{eq}}=\frac{R'}{2}=\frac{1.946}{2}r=0.973r\]You need to login to perform this action.
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