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(A) The circuit consists of three parallel branches,each containing a diode and a resistor $R$. The total resistance between the junction points of th

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$(i) < (ii)$

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(A) The circuit consists of three parallel branches,each containing a diode and a resistor $R$. The total resistance between the junction points of the diodes is $R_{eq} = R/3$. The total resistance between $A$ and $B$ is $R_{total} = R/4 + R_{eq} + R/4 = R/2 + R/3 = 5R/6$ if all diodes are conducting.Case $(i): V_A = -10\,V, V_B = -5\,V$. The potential difference $V_A - V_B = -5\,V$. Since $V_A Case $(ii): V_A = -4\,V, V_B = -12\,V$. The potential difference $V_A - V_B = 8\,V$. Since $V_A > V_B$,the diode $D_2$ is forward-biased,while $D_1$ and $D_3$ are reverse-biased. The equivalent resistance is $R_2 = R/4 + R + R/4 = 1.5R$.Comparing the two,$R_1 = R$ and $R_2 = 1.5R$. Thus,$R_1 < R_2$,which means $(i) < (ii)$.

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:
$(i) V_A = -10\,V, V_B = -5\,V$
$(ii) V_A = -4\,V, V_B = -12\,V$

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