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 resis

Question

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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Answer:

C

Solution:

$V_{d i o d e} = 0.5 volt$

$V_{R} = 1.5 - 0.5 = 1 volt$

$iR = 1$

$R = \frac{1}{i} = \frac{1}{10^{- 2}} = 100 Ω$

TestHub · Accepted Answer

V d i o d e = 0 . 5 volt V R = 1 . 5 - 0 . 5 = 1 volt iR = 1 R = 1 i = 1 10 - 2 = 100 Ω

Physics - Semiconductors Question with Solution | TestHub

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