MCQ
Consider the junction diode as ideal. The value of current flowing through $AB$ is
- ✓$10^{-2} A$
- B$0 A$
- C$10^{-1} A$
- D$10^{-3} A$
✓
Answer
Correct option: A.
$10^{-2} A$
An ideal diode does not offer any resistance during forward biasing.
$\therefore I =\frac{V_A-V_B}{R}$
$=\frac{4-(-6)}{1000} A$
$=10^{-2} A$
$\therefore I =\frac{V_A-V_B}{R}$
$=\frac{4-(-6)}{1000} A$
$=10^{-2} A$
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