A) \[660\,\,\Omega \]
B) \[350\,\,\Omega \]
C) \[475\,\,\Omega \]
D) \[500\,\,\Omega \]
Correct Answer: A
Solution :
Let the value of resistance be R which is to be put in series. \[{{({{I}_{f}})}_{peak}}=5\,\,mA=5\times {{10}^{-3}}A\] \[{{({{I}_{f}})}_{peak}}\] is maximum current in circuit and potential drop across p-n junction diode is 0.7V. It is knee voltage and if emf of applied battery is E then \[E={{V}_{knee}}+{{({{I}_{f}})}_{peak}}R\] ... (i) Putting above values in Eq. (i), we get \[4=0.7+5\times {{10}^{-3}}R\] or \[5\times {{10}^{-3}}R=3.3\] or \[R=600\,\,\Omega \]
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