Q.
Consider the junction diode as ideal. The value of current flowing through $AB$ is
NTA AbhyasNTA Abhyas 2022
Solution:
For an ideal diode, potential drop across diode is zero. Since the current flows from higher potential to lower potential, the direction of a current will be from $A$ to $B$ .
Current in the diode, $I=\frac{\Delta V}{R}$
$\Rightarrow I=\frac{V_{A} - V_{B}}{R}$ $= \frac{4 - \left(\right. - 6 \left.\right)}{1000}$ $= \frac{10}{1000} = 10^{- 2} A$
Current in the diode, $I=\frac{\Delta V}{R}$
$\Rightarrow I=\frac{V_{A} - V_{B}}{R}$ $= \frac{4 - \left(\right. - 6 \left.\right)}{1000}$ $= \frac{10}{1000} = 10^{- 2} A$