A letter 'A' is constructed of a uniform wire | vedclass

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(D) The letter '$A$' consists of two sides of length $20\,cm$ and a cross-piece of length $10\,cm$. Let the cross-piece be connected at a distance $x$

Vedclass · Accepted Answer

$26.7$

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(D) The letter '$A$' consists of two sides of length $20\,cm$ and a cross-piece of length $10\,cm$. Let the cross-piece be connected at a distance $x$ from the apex $A$. Since the apex angle is $60^{\circ}$ and the triangle formed by the apex and the cross-piece is equilateral,the length of the cross-piece is equal to the distance from the apex to the cross-piece. Thus,$x = 10\,cm$.The resistance of each segment is calculated as follows:Resistance of top segments $AD$ and $AE$ is $10\,\Omega$ each.Resistance of the cross-piece $DE$ is $10\,\Omega$.Resistance of the bottom segments $DB$ and $EC$ is $(20 - 10) = 10\,\Omega$ each.The circuit can be simplified: the segment $AD$ and $AE$ are in series with each other,and this combination is in parallel with the cross-piece $DE$. However,looking at the circuit from terminals $B$ and $C$,the path is $B-D$,then the parallel combination of $(D-A-E)$ and $(D-E)$,and finally $E-C$.Resistance of branch $DAE = 10 + 10 = 20\,\Omega$.Resistance of branch $DE = 10\,\Omega$.Equivalent resistance of $DAE$ and $DE$ in parallel: $R_p = \frac{20 	imes 10}{20 + 10} = \frac{200}{30} = 6.67\,\Omega$.Total resistance between $B$ and $C$: $R_{BC} = R_{BD} + R_p + R_{EC} = 10 + 6.67 + 10 = 26.67\,\Omega \approx 26.7\,\Omega$.

$A$ letter $'A'$ is constructed of a uniform wire with resistance $1.0\,\Omega/cm$. The sides of the letter are $20\,cm$ and the cross-piece in the middle is $10\,cm$ long. The apex angle is $60^{\circ}$. The resistance between the ends of the legs is close to ................ $\Omega$.

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