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(A) To find the equivalent resistance between $A$ and $B$,we simplify the circuit step-by-step.$1$. The $30\,\Omega$ and $6\,\Omega$ resistors are in

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$6$

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(A) To find the equivalent resistance between $A$ and $B$,we simplify the circuit step-by-step.$1$. The $30\,\Omega$ and $6\,\Omega$ resistors are in parallel. Their equivalent resistance is $R_p = \frac{30 	imes 6}{30 + 6} = \frac{180}{36} = 5\,\Omega$.$2$. The circuit now consists of a $5\,\Omega$ resistor in series with a $4\,\Omega$ resistor,giving a total branch resistance of $5 + 4 = 9\,\Omega$.$3$. The other branch consists of a $10\,\Omega$ resistor in series with an $8\,\Omega$ resistor,giving a total branch resistance of $10 + 8 = 18\,\Omega$.$4$. These two branches ($9\,\Omega$ and $18\,\Omega$) are connected in parallel across terminals $A$ and $B$.$5$. The equivalent resistance $R_{eq}$ is $\frac{9 	imes 18}{9 + 18} = \frac{162}{27} = 6\,\Omega$.

The equivalent resistance across $AB$ would be ............ $\Omega$

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