MCQ
Consider the junction diode as ideal. The value of current flowing through $AB$ is


- ✓$10^{-2}\;A$
- B$10^{-1}\;A$
- C$10^{-3}\;A$
- D$0\;A$

$\therefore {{I}_{AB}}=\frac{{{V}_{A}}-{{V}_{B}}}{{{R}_{AB}}}=$ $\frac{4\text{V}-(-6\text{V})}{1\text{k}\Omega }=$ $\frac{10}{1000}\text{A}={{10}^{-2}}\text{A}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(A)$ the mean free path of the molecules decreases.
$(B)$ the mean collision time between the molecules decreases.
$(C)$ the mean free path remains unchanged.
$(D)$ the mean collision time remains unchanged.
