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(C) The circuit consists of three parallel branches,each containing a diode in series with a resistor $R$. The entire combination is in series with tw

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

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(C) The circuit consists of three parallel branches,each containing a diode in series with a resistor $R$. The entire combination is in series with two resistors of $R/4$ at terminals $A$ and $B$.$(i)$ $V_A = -10 V, V_B = -5 V$: Here,$V_A $(ii)$ $V_A = -5 V, V_B = -10 V$: Here,$V_A > V_B$. The potential at the anodes of $D_1$ and $D_3$ is higher than their cathodes,making them forward-biased. $D_2$ is reverse-biased. The equivalent resistance is $R_{eq} = R/4 + (R || R) + R/4 = R/4 + R/2 + R/4 = R$.$(iii)$ $V_A = -4 V, V_B = -12 V$: Here,$V_A > V_B$. Similar to case $(ii)$,$D_1$ and $D_3$ are forward-biased,and $D_2$ is reverse-biased. The equivalent resistance is $R_{eq} = R/4 + (R || R) + R/4 = R$.Comparing the values: $1.5 R > R$. Thus,$(ii) = (iii) < (i)$.

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 equivalent resistance between $A$ and $B$ will be:
$(i) V_A = -10 V, V_B = -5 V$
$(ii) V_A = -5 V, V_B = -10 V$
$(iii) V_A = -4 V, V_B = -12 V$

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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 equivalent resistance between $A$ and $B$ will be:$(i) V_A = -10 V, V_B = -5 V$$(ii) V_A = -5 V, V_B = -10 V$$(iii) V_A = -4 V, V_B = -12 V$