First slide

Diodes

Question

A light emitting diode has a 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

Easy

A

40kΩ
B

4kΩ
C

200Ω
D

400Ω

Solution

$P o t e n t i a l D r o p a c r o s s R = E - V_{l e d} = 6 - 2 = 4 V$
$O h m' s L a w : R = \frac{P . D}{c u r r e n t} = \frac{6 - 2}{10 \times 10^{- 3}} = 400 Ω$

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