The equivalent resistance of the arrangement | vedclass

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(B) $1$. First,simplify the upper branch: The resistors $8\,\Omega$,$16\,\Omega$,and $16\,\Omega$ are in parallel. Their equivalent resistance $R_1$ i

Vedclass · Accepted Answer

$8$

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(B) $1$. First,simplify the upper branch: The resistors $8\,\Omega$,$16\,\Omega$,and $16\,\Omega$ are in parallel. Their equivalent resistance $R_1$ is given by $\frac{1}{R_1} = \frac{1}{8} + \frac{1}{16} + \frac{1}{16} = \frac{2+1+1}{16} = \frac{4}{16} = \frac{1}{4}$,so $R_1 = 4\,\Omega$.$2$. This $R_1$ is in series with the $20\,\Omega$ resistor,so the total resistance of the upper branch is $R_{upper} = 4\,\Omega + 20\,\Omega = 24\,\Omega$.$3$. Next,simplify the lower branch: The resistors $9\,\Omega$ and $18\,\Omega$ are in parallel. Their equivalent resistance $R_2$ is given by $\frac{1}{R_2} = \frac{1}{9} + \frac{1}{18} = \frac{2+1}{18} = \frac{3}{18} = \frac{1}{6}$,so $R_2 = 6\,\Omega$.$4$. This $R_2$ is in series with the $6\,\Omega$ resistor,so the total resistance of the lower branch is $R_{lower} = 6\,\Omega + 6\,\Omega = 12\,\Omega$.$5$. Finally,the upper branch $(24\,\Omega)$ and lower branch $(12\,\Omega)$ are in parallel between points $A$ and $B$. The equivalent resistance $R_{AB}$ is $\frac{1}{R_{AB}} = \frac{1}{24} + \frac{1}{12} = \frac{1+2}{24} = \frac{3}{24} = \frac{1}{8}$,which gives $R_{AB} = 8\,\Omega$.

The equivalent resistance of the arrangement of resistances shown in the figure between the points $A$ and $B$ is ............... $\Omega$.

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