The cause of the potential barrier in a $p-n$ junction diode is

  • A
    depletion of positive charges near the junction
  • B
    concentration of positive charges near the junction
  • C
    depletion of negative charges near the junction
  • D
    concentration of positive and negative charges near the junction

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The circuit contains two diodes,each with a forward resistance of $50\, \Omega$ and an infinite reverse resistance. If the battery voltage is $6\, V$,the current through the $120\, \Omega$ resistance is $mA$.

For a $p-n$ junction,the intensity of the electric field is $1 \times 10^{6} \text{ V/m}$ and the width of the depletion region is $5000 \text{ Å}$. The value of the potential barrier is $\dots \text{ V}$.

$A$ junction diode has a resistance of $50 \, \Omega$ when forward biased and $5000 \, \Omega$ when reverse biased. The current in the arrangement shown in the figure will be

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