Explanation

(b) The circuit can be shown as given below
The equivalent resistance between D and C.
$R_{DC} = \frac{15 \times (15 + 15)}{15 + (15 + 15)}$
$\ \ \ \ \ \ \ \ \ = \frac{15 \times 30}{15 + 30}$
$\ \ \ \ \ \ \ \ \ \ = \frac{15 \times 30}{45} = 10\Omega$
Now, between A and B, the resistance of upper part ADCB,
$R_{1} = 15 + 10 + 15 = 40\Omega$
Between A and B, the resistance of middle part AOB
$R_{2} = 15 + 15 = 30\Omega$
Therefore, equivalent resistance between A and B
$\frac{1}{R^{'}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}$
$\ \ \ \ \ \ = \frac{1}{40} + \frac{1}{30} + \frac{1}{15}$
$\ \ \ \ \ \ = \frac{3 + 4 + 8}{120}$
$\ \ \ \ \ \ = \frac{15}{120}$
$\Longrightarrow \ \ R^{'} = \frac{120}{15} = 8\Omega$