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(Solved): The limit of the current that the bulb can carry is 3.2 mA. Find the value of Rs to be connected in ...
The limit of the current that the bulb can carry is 3.2 mA. Find
the value of Rs to be connected in series with the bulb so that the
current through the bulb is exactly 3.2 mA.
2\v R? = 600 R? = 4000 KBULB = 1000 R? ww ? R? M Rs 3