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(B) The circuit consists of five resistors,each of $2\,\Omega$. Let the resistance of each resistor be $R = 2\,\Omega$. By redrawing the circuit betwe

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(B) The circuit consists of five resistors,each of $2\,\Omega$. Let the resistance of each resistor be $R = 2\,\Omega$. By redrawing the circuit between points $A$ and $B$,we can identify the series and parallel combinations. Resistors $OC$ and $CE$ are in series,so their equivalent resistance is $R_{OC} + R_{CE} = 2 + 2 = 4\,\Omega$. This combination is in parallel with resistor $DE$ $(2\,\Omega)$,so the equivalent resistance of this branch is $R_p = (4 	imes 2) / (4 + 2) = 8 / 6 = 4/3\,\Omega$. Finally,this is in series with resistors $AO$ and $OD$,which are also in series with each other $(2 + 2 = 4\,\Omega)$. However,looking at the simplified diagram provided in the solution image,we have: $R_{OC} = 2\,\Omega$,$R_{OD} = 2\,\Omega$,$R_{CD} = 2\,\Omega$,$R_{CE} = 2\,\Omega$,$R_{DE} = 2\,\Omega$. Total resistance $R_{AB} = 2 + 2 + 2 = 6\,\Omega$ is incorrect based on the diagram. Correct analysis: The path $A-O-C-E-B$ has $R_{AO} + R_{OC} + R_{CE} = 2+2+2 = 6\,\Omega$. The path $A-O-D-E-B$ has $R_{AO} + R_{OD} + R_{DE} = 2+2+2 = 6\,\Omega$. These two paths are in parallel with the branch $CD$ $(2\,\Omega)$. Actually,the simplest interpretation of the provided diagram is that the equivalent resistance is $R_{AB} = 2\,\Omega$.

In the given circuit,each resistor has a resistance of $2\,\Omega$. The equivalent resistance between $A$ and $B$ is ............ $\Omega$.

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