Question
Download Solution PDFAn LED has a rating of 2 V and 10 mA. If it is connected to a 6V battery, the minimum value of series resistance is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFThe given connection is represented with the help of the following diagram:
RS = Series LED resistance
For a minimum value of series resistance, the maximum current will flow through the LED.
Since, the maximum current rating of the LED = 10 mA, the minimum resistance for this maximum current can be calculated using KVL as:
6 - 2 - ImaxRS = 0
4 = 10m × RS
RS = 400 Ω
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