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(B) In the given circuit,the leftmost triangle is connected to the rest of the circuit at a single point (the intersection of the two diagonal resisto

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$(2/3)R$

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(B) In the given circuit,the leftmost triangle is connected to the rest of the circuit at a single point (the intersection of the two diagonal resistors). Since there is no closed path for current to flow through the leftmost triangle,no current will flow through those three resistors.Therefore,the circuit simplifies to two resistors of resistance $R$ connected in parallel between terminals $A$ and $B$.The equivalent resistance $R_{eq}$ is given by:$R_{eq} = \frac{R 	imes R}{R + R} = \frac{R^2}{2R} = \frac{R}{2}$.Wait,re-evaluating the circuit: The two resistors connected to $A$ and $B$ are in parallel with the combination of the rest. Actually,the two resistors connected directly to $A$ and $B$ are in series with each other,and that combination is in parallel with the vertical resistor $R$ connected between $A$ and $B$. The leftmost part is a dead end. Thus,the effective circuit is a resistor $R$ in parallel with a series combination of two resistors $R$ and $R$. $R_{eq} = \frac{R 	imes (R+R)}{R + (R+R)} = \frac{R 	imes 2R}{3R} = \frac{2R}{3}$.

The equivalent resistance between $A$ and $B$ in the figure is

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