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  • In the following circuits  - junction diodes <img alt="\mathrm{D_{1},D_{2}}" src="https://entra

In the following circuits \mathrm{PN} - junction diodes \mathrm{D_{1},D_{2}} and \mathrm{D_{3}} are ideal for the following potentials of \mathrm{A} and \mathrm{B}. The correct increasing order of resistance between \mathrm{A} and \mathrm{B} will be

(i) -10 \mathrm{~V},-5 \mathrm{~V}\\ (ii) -5 \mathrm{~V},-10 \mathrm{~V}\\ (ii) -4 \mathrm{~V},-12 \mathrm{~V}

Option: 1

\text { (i) }<\text { (ii) }<\text { (iii) }


Option: 2

\mathrm{ { (iii) < (ii) < (i) }}


Option: 3

\mathrm{\text { (ii) }= { (iii) < (i) }}


Option: 4

\mathrm{\text { (i) }=\text { (iii) }<\text { (ii) }}


Answers (1)

best_answer

\mathrm{1. \ \ V_A=-10 \mathrm{~V} \text { and } V_B=-5 \mathrm{~V}}
Diodes \mathrm{D_{1}} and \mathrm{D_{3}} are reverse biased and \mathrm{D_{2}} is forward biased

\mathrm{\Rightarrow R_{A B}=R+\frac{R}{4}+\frac{R}{4}=\frac{3}{2} R} \\\\ \mathrm{2.\ \ \ When\ V_A=-5 \mathrm{~V}\ and\ V_B=-10 \mathrm{~V}}
Diodes \mathrm{D_{2}} is reverse biased \mathrm{D_{1}} and \mathrm{D_{3}} are forward biased

\mathrm{\Rightarrow R_{A B}=\frac{R}{4}+\frac{R}{2}+\frac{R}{4}=R}
\mathrm{(iii)} In this case equivalent resistance between \mathrm{A} and \mathrm{B} is also \mathrm{R}
Hence, \mathrm{(ii)=(iii)< (i)} 

Posted by

Pankaj

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