Question

The equivalent resistance between A and B in the network in

• A. $\frac{4}{3}\Omega$
• B. $\frac{3}{2}\Omega$
• C. $3\Omega$
• D. $2\Omega$

Answer 1

The correct option is A $\frac{4}{3}\Omega$

The following circuit is equivalent to
The equivalent of resistor $8\Omega$ and $2\Omega$ is
$\frac{1}{R_1}=\frac{1}{8}+\frac{1}{2}$
$R_{!}=\frac{8}{5}\Omega$
The equivalent of $4\Omega$ of $6\Omega$ is
$\frac{1}{R_2}=\frac{1}{4}+\frac{1}{6}$
$R_2=\frac{12}{5}\Omega$
$R_1$ & $R_2$ are in series, therefore their equivalent
$R_3=\frac{12}{5}+\frac{8}{5}$
=$4\Omega$
hence net resistance between A & B is
$\frac{1}{R}=\frac{1}{3}+\frac{1}{6}+\frac{1}{4}$
=$\frac{4+2+4}{12}$
$R=\frac{4}{3}\Omega$