The equivalent resistance between $A$ and $C$ of the given circuit, is
The equivalent resistance between $A$ and $C$ of the given circuit, is
$8 \Omega$
$\frac{32}{14} \Omega$
$\frac{4}{3} \Omega$
$\frac{8}{3} \Omega$.
Solution
The given circuit is a balanced Wheatstone bridge circuit. So, the resistance $7 \Omega$ may be regulated. Therefore, we have the equivalent circuit as,
$\therefore$ The quivalent resistance between $A$ and $B$ is $R=\frac{4 \times 8}{4+8}=\frac{32}{12}=\frac{8}{3} \Omega$
$\therefore$ The quivalent resistance between $A$ and $B$ is $R=\frac{4 \times 8}{4+8}=\frac{32}{12}=\frac{8}{3} \Omega$