A letter $A$ is constructed of a uniform wire with resistance $1.0\,\Omega \,per\,cm$. The sides of the letter are $20\, cm$ and the cross piece in the middle is $10\, cm$ long. The apex angle is $60^o$. The resistance between the ends is .............. $\Omega$
Diffcult
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The circuits are shown above.
Using trignometry, $\frac{5}{x}=\sin 30=0.5$
$\Longrightarrow x=10 \mathrm{cm}$
Thus we get $A B=C D=20-10=10 \mathrm{cm}$
since all parts are of same length, thus resistance of each part $R=10 \times 1=10 \Omega$ Equivalent circuit is also shown above.
Equivalence resistance between $\mathrm{B}$ and $\mathrm{E}, R_{B E}=(10+10)\|10=20\| 10$
$\therefore R_{e q}=\frac{20 \times 10}{20+10}=6.7 \Omega$
Equivalent resistance between $\mathrm{A}$ and $\mathrm{D}, R_{A D}=10+6.7+10=26.7 \Omega$
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