In the arrangement of resistances shown below | vedclass

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(A) The circuit consists of three sections connected in series. First,consider the first section: the upper branch has $5\,\Omega$ and the lower branc

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$20$

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(A) The circuit consists of three sections connected in series. First,consider the first section: the upper branch has $5\,\Omega$ and the lower branch has $10\,\Omega$. The vertical resistor is $10\,\Omega$. However,this is a bridge-like structure. Let's simplify the circuit by identifying series and parallel combinations. Looking at the circuit,the resistors $5\,\Omega$ and $10\,\Omega$ (top) are in series with $15\,\Omega$,giving $5+10+15 = 30\,\Omega$. The bottom resistors $10\,\Omega$,$20\,\Omega$,and $30\,\Omega$ are in series,giving $10+20+30 = 60\,\Omega$. The vertical resistors are connected between these two main branches. By symmetry or nodal analysis,the effective resistance is calculated as the parallel combination of the two main branches: $R_{eq} = \frac{30 	imes 60}{30 + 60} = \frac{1800}{90} = 20\,\Omega$.

In the arrangement of resistances shown below,the effective resistance between points $A$ and $B$ is ............... $\Omega$.

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