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

Answer: (B)

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, $I=\frac{\Delta V}{R}$
$\Rightarrow I=\frac{V_A-V_B}{R}$ $=\frac{4-(-6)}{1000}$ $=\frac{10}{1000}=10^{-2}A$