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

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(C) The circuit consists of three parallel branches,each containing a diode and a resistor $R$. The total resistance between $A$ and $B$ is $R_{AB} =

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

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(C) The circuit consists of three parallel branches,each containing a diode and a resistor $R$. The total resistance between $A$ and $B$ is $R_{AB} = \frac{R}{4} + R_{eq} + \frac{R}{4} = \frac{R}{2} + R_{eq}$,where $R_{eq}$ is the equivalent resistance of the three parallel branches.$(i)$ For $V_A = -10 \ V$ and $V_B = -5 \ V$,the potential difference across the branches is $V_A - V_B = -5 \ V$. Diode $D_2$ is forward-biased,while $D_1$ and $D_3$ are reverse-biased. Thus,$R_{eq} = R$. Total resistance $R_{AB} = \frac{R}{2} + R = 1.5R$.$(ii)$ For $V_A = -5 \ V$ and $V_B = -10 \ V$,the potential difference is $V_A - V_B = +5 \ V$. Diodes $D_1$ and $D_3$ are forward-biased,while $D_2$ is reverse-biased. Thus,$R_{eq} = \frac{R}{2}$. Total resistance $R_{AB} = \frac{R}{2} + \frac{R}{2} = R$.$(iii)$ For $V_A = -4 \ V$ and $V_B = -12 \ V$,the potential difference is $V_A - V_B = +8 \ V$. Diodes $D_1$ and $D_3$ are forward-biased,while $D_2$ is reverse-biased. Thus,$R_{eq} = \frac{R}{2}$. Total resistance $R_{AB} = \frac{R}{2} + \frac{R}{2} = R$.Comparing the values: $(ii) = R$,$(iii) = R$,and $(i) = 1.5R$. 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$