Resistors of 10, Omega, 5, Omega, 7, Omega, a | vedclass

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(B) $1$. The resistors $7\, \Omega$ (on $CD$) and $3\, \Omega$ (on $DA$) are in series, so their equivalent resistance is $R_{CDA} = 7 + 3 = 10\, \Ome

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$5$

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(B) $1$. The resistors $7\, \Omega$ (on $CD$) and $3\, \Omega$ (on $DA$) are in series, so their equivalent resistance is $R_{CDA} = 7 + 3 = 10\, \Omega$.$2$. This $R_{CDA}$ is in parallel with the diagonal resistor of $10\, \Omega$ connected across $AC$. The equivalent resistance of this parallel combination is $R_{AC} = \frac{10 	imes 10}{10 + 10} = 5\, \Omega$.$3$. Now, this $5\, \Omega$ resistance is in series with the $5\, \Omega$ resistor on side $BC$. So, the equivalent resistance of the branch $ABC$ is $R_{ABC} = 5 + 5 = 10\, \Omega$.$4$. Finally, this $10\, \Omega$ branch is in parallel with the $10\, \Omega$ resistor on side $AB$. The total equivalent resistance between $A$ and $B$ is $R_{eq} = \frac{10 	imes 10}{10 + 10} = 5\, \Omega$.

Resistors of $10\, \Omega, 5\, \Omega, 7\, \Omega$, and $3\, \Omega$ are connected along the sides $AB, BC, CD$, and $DA$ of a square, respectively. A resistor of $10\, \Omega$ is connected across the diagonal $AC$. What is the equivalent resistance between points $A$ and $B$ in $\Omega$?

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