In an ideal junction diode, the current flowing through $\mathrm{PQ}$ is (resistance is 2 kilo-ohms)
In an ideal junction diode, the current flowing through $\mathrm{PQ}$ is (resistance is 2 kilo-ohms)
$2 \times 10^{-3} \mathrm{~A}$
$2 \times 10^{-2} \mathrm{~A}$
$4 \times 10^{-3} \mathrm{~A}$
$10^{-3} \mathrm{~A}$
Solution
The diode is forward biased. The potential difference between $\mathrm{P}$ and $\mathrm{Q}$ is $(3-(-5))=8 \mathrm{~V}$.
$\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{8}{2 \times 10^3}=4 \times 10^{-3} \mathrm{~A}$
~
