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

Option 1 -

100Ω

Option 2 -

200Ω

Option 3 -

50Ω

Option 4 -

300Ω

0 3 Views | Posted a month ago
Asked by Shiksha User

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    a month ago
    Correct Option - 1


    Detailed Solution:
    Kindly go through the solution

     

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alok kumar singh

ibattery= 1 0 ? 3 1 0 0 0  = 7mA

i2kW = 3 2 0 0 0  =1.5 mA

iz = (7 – 1.5) mA

= 5.5mA

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