First slide

Diodes

Question

A light emitting diode has a voltage drop of 5 volt across it and passes a current 2mA when it operates with a 9 volt battery through a limiting resistor R. The value of R is

Easy

A

4kΩ
B

2kΩ
C

400Ω
D

200Ω

Solution

$P o t e n t i a l D r o p a c r o s s r e s i s t o r = V_{B} - V_{L E D} = 9 - 5 = 4 V$
$O h m' s L a w : R = \frac{P . D}{c u r r e n t} = \frac{4}{2 \times 10^{- 3}} = 2000 Ω = 2 k Ω$

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