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