Four wires AB, BC, CD, DA of resistance 4, Om | vedclass

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(D) The circuit consists of a rectangle $ABCD$ with a diagonal $BD$. When calculating the equivalent resistance between points $A$ and $B$,we can anal

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$\frac{8}{3}\, \Omega$

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(D) The circuit consists of a rectangle $ABCD$ with a diagonal $BD$. When calculating the equivalent resistance between points $A$ and $B$,we can analyze the circuit structure.$1$. The wires $BC$ and $CD$ are in series,so their combined resistance is $R_{BC} + R_{CD} = 4\, \Omega + 4\, \Omega = 8\, \Omega$.$2$. This combination of $8\, \Omega$ is in parallel with the diagonal wire $BD$ which also has a resistance of $8\, \Omega$. The equivalent resistance of this parallel part is $R_p = \frac{8 	imes 8}{8 + 8} = \frac{64}{16} = 4\, \Omega$.$3$. Now,this $4\, \Omega$ equivalent resistance is in series with the wire $AD$ $(4\, \Omega)$,giving $4\, \Omega + 4\, \Omega = 8\, \Omega$.$4$. Finally,this $8\, \Omega$ is in parallel with the wire $AB$ $(4\, \Omega)$. The total equivalent resistance $R_{eq}$ is $\frac{8 	imes 4}{8 + 4} = \frac{32}{12} = \frac{8}{3}\, \Omega$.

Four wires $AB, BC, CD, DA$ of resistance $4\, \Omega$ each and a fifth wire $BD$ of resistance $8\, \Omega$ are joined to form a rectangle $ABCD$,where $BD$ is a diagonal. The effective resistance between the points $A$ and $B$ is:

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