MHT CET · Physics · Semiconductors
In an ideal junction diode, the current flowing through \(\mathrm{PQ}\) is (resistance is 2 kilo-ohms)
- A \(2 \times 10^{-3} \mathrm{~A}\)
- B \(2 \times 10^{-2} \mathrm{~A}\)
- C \(4 \times 10^{-3} \mathrm{~A}\)
- D \(10^{-3} \mathrm{~A}\)
Answer & Solution
Correct Answer
(C) \(4 \times 10^{-3} \mathrm{~A}\)
Step-by-step Solution
Detailed explanation
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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