Question

For the circuit given below, calculate the equivalent resistance between the points P and Q.

Answer 1

7,8,9 are in parallel

$\frac{1}{R_{{eq}_1}}=\frac{1}{7}+\frac{1}{8}+\frac{1}{9}{R}_{{eq}_1}=\frac{7\times 8+8\times 9+7\times 9}{7\times 8\times 9}\Omega =\frac{191\Omega }{504}$

5,4 are in parallel

$\frac{1}{R_{{eq}_2}}=\frac{1}{5}+\frac{1}{4}{R}_{{eq}_2}=\frac{20\Omega }{9}$

$R_{{eq}_1}$ and $R_{{eq}_2}$ are in series

Net resistance $=R_{{eq}_1}+R_{{eq}_2}=\frac{191}{504}+\frac{20}{9}{R}_{net}=\frac{191+20\times 56}{504}{R}_{net}=\frac{191+1120}{504}{R}_{net}=2.6\Omega$