c
(c) \((i)\) \(V_A = -10V\) and \(V_B\) \(= -5V\)

Diodes \(D_1\) and \(D_3\) are reveres biased and \(D_2\) is forward biased.
\( \Rightarrow {R_{AB}} = R + \frac{R}{4} + \frac{R}{4} = \frac{3}{2}R\).
\((ii)\) When \(V_A = -5V\) and \(V_B = -10V\)
Diodes \(D_2\) is reverse biased \(D_1\) and \(D_3\) are forward biased
\( \Rightarrow {R_{AB}} = \frac{R}{4} + \frac{R}{2} + \frac{R}{4}\)\(=R.\)
\((iii)\) In this case equivalent resistance between \(A\) and \(B\) is also \(R\).
Hence \((ii) = (iii) < (i).\)