First slide

Diodes

Question

A light emitting diode has a voltage drop of 2V across it and passes a current of 10mA . When it operates with a 6V battery through a limiting resistor R , the value of R is :

Easy

A

$40 k Ω$
B

$0.4 k Ω$
C

$200 k Ω$
D

$400 k Ω$

Solution

As LED is connected to a battery through a resistance in series
The current flowing, 10mA is the same
The voltage drop across LED = 2V
As the battery has voltage of 6V, the potential difference across R = 4V
$∴ V = i R$
$\Rightarrow R = \frac{4}{i} = \frac{4}{10 \times 10^{- 3}} A$
$∴ R = 400 Ω$

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