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(A) The circuit is a Wheatstone bridge. We check the ratio of the resistors in the arms: $\frac{R_{AC}}{R_{AD}} = \frac{5}{10} = 0.5$ and $\frac{R_{CB

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$6$

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(A) The circuit is a Wheatstone bridge. We check the ratio of the resistors in the arms: $\frac{R_{AC}}{R_{AD}} = \frac{5}{10} = 0.5$ and $\frac{R_{CB}}{R_{DB}} = \frac{4}{8} = 0.5$.Since $\frac{R_{AC}}{R_{AD}} = \frac{R_{CB}}{R_{DB}}$,the bridge is balanced.Therefore,no current flows through the central $9\,\Omega$ resistor,and it can be removed from the circuit.Now,the $5\,\Omega$ and $4\,\Omega$ resistors are in series: $R_1 = 5 + 4 = 9\,\Omega$.The $10\,\Omega$ and $8\,\Omega$ resistors are in series: $R_2 = 10 + 8 = 18\,\Omega$.These two branches are in parallel between $A$ and $B$. The equivalent resistance $R_{AB}$ is given by:$R_{AB} = \frac{R_1 	imes R_2}{R_1 + R_2} = \frac{9 	imes 18}{9 + 18} = \frac{162}{27} = 6\,\Omega$.

Five resistors are connected as shown in the diagram. The equivalent resistance between $A$ and $B$ is .............. $\Omega$.

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