The equivalent resistance between the points | vedclass

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(B) The given circuit is symmetric. By observing the symmetry,we can see that the circuit consists of three parallel branches between points $P$ and $

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$1$

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(B) The given circuit is symmetric. By observing the symmetry,we can see that the circuit consists of three parallel branches between points $P$ and $Q$. Each branch contains two resistors of resistance $r$ connected in series.$1$. Resistance of each branch = $r + r = 2r$.$2$. Since there are three such branches in parallel,the equivalent resistance $R_{eq}$ is given by:$\frac{1}{R_{eq}} = \frac{1}{2r} + \frac{1}{2r} + \frac{1}{2r} = \frac{3}{2r}$$R_{eq} = \frac{2r}{3}$$3$. Given $r = \frac{3}{2} \Omega$,substituting this value:$R_{eq} = \frac{2}{3} 	imes \left(\frac{3}{2}ight) = 1 \Omega$.

The equivalent resistance between the points $P$ and $Q$ in the network given here is equal to ................ $\Omega$ (given $r = \frac{3}{2} \Omega$).

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