In the arrangement of resistances shown below | vedclass

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(B) The given circuit consists of two parallel branches connected between points $A$ and $B$.$1$. The upper branch consists of three resistors in seri

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$20$

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(B) The given circuit consists of two parallel branches connected between points $A$ and $B$.$1$. The upper branch consists of three resistors in series: $5\,\Omega$,$10\,\Omega$,and $15\,\Omega$. The equivalent resistance of this branch is $R_1 = 5 + 10 + 15 = 30\,\Omega$.$2$. The lower branch consists of three resistors in series: $10\,\Omega$,$20\,\Omega$,and $30\,\Omega$. The equivalent resistance of this branch is $R_2 = 10 + 20 + 30 = 60\,\Omega$.$3$. Since these two branches are in parallel,the effective resistance $R_{AB}$ is given by:$\frac{1}{R_{AB}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{30} + \frac{1}{60} = \frac{2+1}{60} = \frac{3}{60} = \frac{1}{20}$.Therefore,$R_{AB} = 20\,\Omega$.

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

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