In the following circuit,PN-junction diodes D | vedclass

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(C) The equivalent resistance $R_{AB}$ is calculated based on the biasing of the diodes.$(i)$ For $V_A = -10 \ V$ and $V_B = -5 \ V$,the potential dif

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$(ii) = (iii) < (i)$

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(C) The equivalent resistance $R_{AB}$ is calculated based on the biasing of the diodes.$(i)$ For $V_A = -10 \ V$ and $V_B = -5 \ V$,the potential difference across the diodes makes $D_1$ and $D_3$ reverse-biased (open circuit) and $D_2$ forward-biased (short circuit).$R_{AB} = \frac{R}{4} + R + \frac{R}{4} = 1.5R$.$(ii)$ For $V_A = -5 \ V$ and $V_B = -10 \ V$,$D_2$ is reverse-biased and $D_1, D_3$ are forward-biased.The two parallel branches each have resistance $R$,so their equivalent is $R/2$.$R_{AB} = \frac{R}{4} + \frac{R}{2} + \frac{R}{4} = R$.$(iii)$ For $V_A = -4 \ V$ and $V_B = -12 \ V$,the potential difference is $8 \ V$ (with $A$ at higher potential than $B$),which is similar to case $(ii)$.Thus,$R_{AB} = R$.Comparing the values: $R < 1.5R$,so $(ii) = (iii) < (i)$.

In the following circuit,$PN$-junction diodes $D_1, D_2,$ and $D_3$ are ideal. For the following potentials of $A$ and $B$:
$(i) -10 \ V, -5 \ V$
$(ii) -5 \ V, -10 \ V$
$(iii) -4 \ V, -12 \ V$
The correct increasing order of resistance between $A$ and $B$ will be:

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In the following circuit,$PN$-junction diodes $D_1, D_2,$ and $D_3$ are ideal. For the following potentials of $A$ and $B$:$(i) -10 \ V, -5 \ V$$(ii) -5 \ V, -10 \ V$$(iii) -4 \ V, -12 \ V$The correct increasing order of resistance between $A$ and $B$ will be: