WBJEE · Physics · Semiconductors
A junction diode has a resistance of \(25 \Omega\) when forward biased and \(2500 \Omega\) when reverse biased. The current in the diode, for the arrangement shown will be
- A \(\frac{1}{15} \mathrm{~A}\)
- B \(\frac{1}{7} \mathrm{~A}\)
- C \(\frac{1}{25} \mathrm{~A}\)
- D \(\frac{1}{180} \mathrm{~A}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{7} \mathrm{~A}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{R}_{\mathrm{eq}}=25+10=35 \Omega\) Because diode is forward biased. So \(I=\frac{V}{R_{e q}}=\frac{5}{35}=\frac{1}{7} \mathrm{~A}\)
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