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