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When a diode is forward biased, it has a voltage drop of 0.5 V. The safe limit of current through the diode is 10 mA. If a battery of emf 1.5 V is used in the circuit, the value of minimum resistance to be connected in series with the diode so that the current does not exceed the safe limit is :

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$200 Ω$

$50 Ω$

$100 Ω$

$300 Ω$

answer is C.

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Detailed Solution

$\begin{array}{l} A p p l y i n g K V L, \\ 1.5 - 0.5 - i R = 0 \\ \Rightarrow R = \frac{1}{10 \times 10^{- 3}} = 100 Ω \end{array}$

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