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Q.

A light emitting diode (LED) has voltage drop of 2 volt across it and passes a current of 10mA when it operates with a 6 volt battery through a limiting resistor R. The value of R is 100k $Ω$ find k

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answer is 4.

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Detailed Solution

The voltage drop across the resistor will be 4V.

$R = \frac{Δ V}{I} = \frac{4}{10^{- 2}} = 400 Ω = 100 \times 4 Ω, K = 4$

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