Q. In the following circuits $PN$-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) $- 10\ V,\ - 5V$ (ii) $- 5V,\ - 10\ V$ (ii) $- 4V,\ - 12V$
A) (i) < (ii) < (iii)
B) (iii) < (ii) < (i)
C) (ii) = (iii) < (i)
D) (i) = (iii) < (ii)
Explanation
(c) $V_{A} = - 10V$ and $V_{B} = - 5V$
Diodes $D_{1}$ and $D_{3}$ are reverse biased and $D_{2}$ is forward biased
$\Rightarrow R_{AB} = R + \frac{R}{4} + \frac{R}{4} = \frac{3}{2}R$
When $V_{A} = - 5V$ and $V_{B} = - 10V$
Diodes $D_{2}$ is reverse biased $D_{1}$ and $D_{3}$ are forward biased
$\Rightarrow R_{AB} = \frac{R}{4} + \frac{R}{2} + \frac{R}{4} = R$
(iii) In this case equivalent resistance between $A$ and $B$ is also $R$
Hence (ii) = (iii) < (i)
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.