Question

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

A) \[0A\]
B) ${{10}^{-2}}A$
C) ${{10}^{-1}}A$
D) ${{10}^{-3}}A$

Answer 1

Hint:When the voltage is applied forward biased, then the ideal diode acts like a perfect conductor and when the voltage is applied reverse biased, then the ideal diode acts like a perfect insulator.

Formula Used:
Current $i=\dfrac{E}{R}$

Complete step by step answer:
Diode:- An electric component that allows the flow of current in only one direction.
The most common type of diode uses a p-n junction. In this type of diode one material (n) in which electrons are charged carriers and second material (p) in which holes are charged carriers. At the interface, a depletion region is formed across which electrons diffuse to fill holes in the p-side.
When the junction is forward biased, the electrons easily cross the junction. The current flows normally as a conductor but when the junction is reverse biased the depletion region is wide and electrons cannot easily move across. The current remains very small until a certain voltage (breakdown voltage) is reached and current suddenly increased symbol of diode is
                     

The most common of a diode is to allow an electric current to pass in one direction.
Given,

Potential difference between A and B is,
$V=4-\left( -6 \right)=4+6=10volt$
$\because $ Positive terminals of the battery are connected to the positive side of the diode, and the negative terminal is connected to the negative side, so it is forward biased, and works as a conductor.
Current $i=\dfrac{V}{R}$
$\begin{align}
  & V=10volt \\
 & R=1K\Omega ={{10}^{3}}\Omega \\
\end{align}$
$\begin{align}
  & i=\dfrac{V}{R} \\
 & \Rightarrow i=\dfrac{10}{{{10}^{3}}} \\
 & \Rightarrow i=10\times {{10}^{-3}} \\
 & \Rightarrow i={{10}^{-2}}Amp \\
\end{align}$

Note: Sometimes we think that $-6V$ is greater than $+4V$. The difference between them is $2V$ but this is wrong, because there are possible terminals in the sign. So the potential difference is $10V$.