Five equal resistances each of value R are co | vedclass

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(D) $1$. For the equivalent resistance between points $B$ and $D$: The circuit can be viewed as three parallel branches connected between $B$ and $D$.

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Both $(b)$ and $(c)$

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(D) $1$. For the equivalent resistance between points $B$ and $D$: The circuit can be viewed as three parallel branches connected between $B$ and $D$. One branch is the resistor $R$ directly between $B$ and $D$. The other two branches consist of two resistors in series $(R+R = 2R)$ connected in parallel to the first. Thus,$\frac{1}{R_{BD}} = \frac{1}{2R} + \frac{1}{R} + \frac{1}{2R} = \frac{1+2+1}{2R} = \frac{4}{2R} = \frac{2}{R}$. Therefore,$R_{BD} = \frac{R}{2}$.$2$. For the equivalent resistance between points $A$ and $C$: The circuit forms a balanced Wheatstone bridge where the ratio of resistances in the arms is equal $(\frac{R}{R} = \frac{R}{R})$. The central resistor between $B$ and $D$ carries no current. Thus,the equivalent resistance is $R_{AC} = \frac{(R+R)(R+R)}{(R+R)+(R+R)} = \frac{2R \cdot 2R}{4R} = R$.

Five equal resistances each of value $R$ are connected in a form shown alongside. The equivalent resistance of the network

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