In the following circuit,PN- junction diodes | vedclass

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(C) The equivalent resistance $R_{AB}$ is determined by the state of the diodes (forward or reverse biased).$(i)$ $V_A = -10 \ V$ and $V_B = -5 \ V$.

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

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(C) The equivalent resistance $R_{AB}$ is determined by the state of the diodes (forward or reverse biased).$(i)$ $V_A = -10 \ V$ and $V_B = -5 \ V$. The potential difference $V_A - V_B = -5 \ V$. Current flows from $B$ to $A$. Diodes $D_1$ and $D_3$ are reverse biased (open circuit),and $D_2$ is forward biased (short circuit). The circuit consists of $R/4$ in series with $R$ and $R/4$. Thus,$R_{AB} = R/4 + R + R/4 = 1.5R$.$(ii)$ $V_A = -5 \ V$ and $V_B = -10 \ V$. The potential difference $V_A - V_B = +5 \ V$. Current flows from $A$ to $B$. Diode $D_2$ is reverse biased,while $D_1$ and $D_3$ are forward biased. The circuit consists of $R/4$ in series with the parallel combination of $R$ and $R$ (which is $R/2$) and $R/4$. Thus,$R_{AB} = R/4 + R/2 + R/4 = R$.$(iii)$ $V_A = -4 \ V$ and $V_B = -12 \ V$. The potential difference $V_A - V_B = +8 \ V$. Current flows from $A$ to $B$. Similar to case $(ii)$,$D_2$ is reverse biased and $D_1, D_3$ are forward biased. Thus,$R_{AB} = R/4 + R/2 + R/4 = R$.Comparing the resistances: $R_{AB}(i) = 1.5R$ and $R_{AB}(ii) = R_{AB}(iii) = R$. Therefore,the increasing order is $(ii) = (iii) < (i)$.

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

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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 $B,$ the correct increasing order of resistance between $A$ and $B$ will be:$(i) -10 \ V, -5 \ V$$(ii) -5 \ V, -10 \ V$$(iii) -4 \ V, -12 \ V$