Saved Bookmarks
| 1. |
A light emitting diode (LED) has a voltage drop of 2 volt across it and passes a current of 10 mA when it operates with a 6 volt battery through a limiting register R. The value of R is …… |
|
Answer» `40kOmega` As LED is connected to a battery through a resistance in series. The CURRENT flowing, 10 mA is THESAME. The voltage drop across `LED=2V` As the battery has 6 V, the potential DIFFERENCE across `R=4V` `therefore iR=4V rArr R=(4V)/(10xx10^(-3)A)` `=400Omega` |
|