Question 135: Current Electricity

Question

A wire of resistance R is bent into an equilateral triangle and an identical wire is bent into $a$ square. The ratio of resistance between the two end points of an edge of the triangle to that of the square is

Accepted Answer

We know, $$R = ho {l \over A} \Rightarrow R \propto l$$
Side length of triangle $$ = {l \over 3}$$
Side length of square $$ = {l \over 4}$$
So,
$$ \Rightarrow {\left( {{R_{eq}}} ight)_1} = {{{{2R} \over 3} 	imes {R \over 3}} \over {{{2R} \over 3} + {R \over 3}}}$$ and $${\left( {{R_{eq}}} ight)_2} = {{{{3R} \over 4} 	imes {R \over 4}} \over {{{3R} \over 4} + {R \over 4}}}$$
$$ = {{2{R^2}} \over 9} 	imes {3 \over {3R}} = {{2R} \over 9}$$ and $${\left( {{R_{eq}}} ight)_2} = {{3{R^2}} \over {16}} - R = {{3R} \over {16}}$$
Hence, $${{{{\left( {{R_{eq}}} ight)}_1}} \over {{{\left( {{R_{eq}}} ight)}_2}}} = {{{{2R} \over 9}} \over {{{3R} \over {16}}}} = {{32} \over {27}}$$

Answer

$8 / 9$

Answer

$9 / 8$

Answer

$32 / 27$

Answer

$27 / 32$

Current Electricity

A wire of resistance R is bent into an equilateral triangle and an identical wire is bent into a square. The ratio of resistance between the two end points of an edge of the triangle to that of the square is

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