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What resister do you need for a 6.3v 250mA bulb on an 9v battery or 6 AA's?
Wiki User
∙ 11y ago
-- The bulb draws 250 mA when energized by 6.3 volts,
so the resistance of the bulb is (6.3/0.25) = 25.2 ohms.
-- You need a resistor that will drop 2.7 volts when in series with 25.2 ohms
and energized by 9 volts.
R/(R + 25.2) = 2.7/9
R + 25.2/R = 9/2.7
1 + 25.2/R = 9/2.7
25.2/R = 6.3/2.7
6.3 R = (25.2) x (2.7)
R = (25.2) x (2.7)/6.3 = 10.8 ohms
Check:
-- 10.8 ohms and 25.2 ohms in series total 36 ohms.
-- When energized by 9 volts, the circuit draws 9/36 = 0.25 A = 250 mA.
-- The drop across the 10.8-ohm resistor is 10.8 x 0.25 = 2.7 volts.
-- The bulb has (9 - 2.7) = 6.3 volts across it, and 250 mA of current through it.
-- Yay !
Wiki User
∙ 11y ago
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