Question

Six resistances are connected together as shown in the figure. Calculate the equivalent resistance between points A and B.

Answer 1

The resistors R₂, R₃ and R₄ in series.

∴ R' = R₂ + R₃ + R₄

= 2 + 3 + 5 = 10 Ω

Now R' and R₅ are in parallel.

∴ ${\displaystyle \frac{1}{\text{R}\prime \prime }=\frac{1}{\text{R}\prime }+\frac{1}{{\text{R}}_5}}$

${\displaystyle =\frac{1}{10}+\frac{1}{10}=\frac{2}{10}=\frac{1}{5}}$

∴ R'' = 5 Ω

Now R₁, R” and R₆ in series between the points A and B. The equivalent resistance between A and B is
R = R₁ + R” + R₆ = 2 + 5 + 5 = 12 Ω