Calculate the equivalent resistance between P and Q from the following diagram:

In the circuit, there are two parts. In the first part, two resistors of 10 Ω each are connected in series. If the equivalent resistance of this part is R'_s then

R'_s = (10 + 10) Ω = 20 Ω

In the second part, resistance R'_s = 20 Ω and 5 Ω are connected in parallel. If the equivalent resistance of this part is R_p then

$\dfrac{1}{R$p} = \dfrac{1}{20} + \dfrac{1}{5} \\[0.5em] \dfrac{1}{Rp} = \dfrac{1 + 4}{20} \\[0.5em] \dfrac{1}{Rp} = \dfrac{5}{20} \\[0.5em] \dfrac{1}{Rp} = \dfrac{1}{4} \\[0.5em] \Rightarrow R_p = 4 Ω \\[0.5em]

∴ Equivalent resistance of second part = 4 Ω

In the last part, 3 Ω, R_p = 4 Ω and 2 Ω are connected in series. If the equivalent resistance of this part is R_s then

R_s = (3 + 4 + 2) Ω = 9 Ω

∴ Equivalent resistance between the points P and Q = 9 Ω