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Q.
The effective resistance across the points $A$ and $I$ is
Current Electricity
Solution:
Since the potential at $B, G$ is same, the potential at $C$ $H$ and $F$ is same and the potential at $D$ and $E$ is same.
$\therefore $ The circuit can be reduced to
The equivalent resistance of (i) is
$\frac{1}{R_{1}}=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{4}{2}=\frac{2}{1}$ or $R_{1}=\frac{1}{2}\, \Omega$
The equivalent resistance of (II) is
$\frac{1}{R_{2}}=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{4}{2}=\frac{2}{1}$ or $R_{2}=\frac{1}{2}\, \Omega$
The effective resistance between $A$ and $I$ is
$R_{ eff }=R_{1}+R_{2}=\frac{1}{2}+\frac{1}{2}=1$ or $R_{ eff }=1\, \Omega$
