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$0\;A$

$\therefore {{I}_{AB}}=\frac{{{V}_{A}}-{{V}_{B}}}{{{R}_{AB}}}=$ $\frac{4\text{V}-(-6\text{V})}{1\text{k}\Omega }=$ $\frac{10}{1000}\text{A}={{10}^{-2}}\text{A}$
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$_0{n^1}\, \to {\,_1}{H^1}\, + {\,_{ - 1}}{e^0}\, + \,[\,\,]$
Then the particle in the bracket is