Question

In the network shown, the equivalent resistance between $A$ and $B$ is

• A. $\frac{4}{3}\Omega$
• B. $\frac{3}{4}\Omega$
• C. $\frac{24}{17}\Omega$
• D. $\frac{17}{24}\Omega$

Answer 1

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

From point A,

$8\Omega$ and $2\Omega$ are parallel.

$\frac{1}{R_1}=\frac{1}{8}+\frac{1}{2}=\frac{5}{8}$

$\therefore R_1=\frac{8}{5}$

From point B,

$4\Omega$ and $6\Omega$ are parallel.

$\frac{1}{R_2}=\frac{1}{4}+\frac{1}{6}=\frac{5}{12}$

$\therefore R_2=\frac{12}{5}$

Now,

$R_1=\frac{8}{5}\Omega ,R_2=\frac{12}{5}\Omega$ are in series

$\therefore R=\frac{8}{5}+\frac{12}{5}=4\Omega$

Now,

$4\Omega ,3\Omega ,6\Omega$ are parallel to each other.

$\therefore \frac{1}{R_{AB}}=\frac{1}{4}+\frac{1}{3}+\frac{1}{6}=\frac{9}{12}=\frac{3}{4}$

$\therefore R_{AB}=\frac{4}{3}\Omega$