If I read it correctly, you have +ve of battery to one end of 0.222Ω resistor, other end of 0.222Ω resistor to -ve of battery (I can't see any other components in this loop).
1. The motor will have no voltage across it, and no current through it.
2. That resistor will get very, very hot. It will be dissipating about 650W.
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2. Omit the link from the 120Ω resistor to the -ve of the battery.
All you have then is the battery, 120Ω resistor and motor in series.
Assuming the motor is about 300mW at 4.5V, that means it's drawing about 70mA. The 120Ω resistor should limit the current nicely[1]. It will dissipate about 500mW. Use at least a 1W rated resistor. It will get quite warm.
[1] (12-4.5)V/120Ω = 0.0625A, or 62.5mA.
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"Love justice, you who rule the world" - Dante Alighieri
One would think all you need is 120Ω in series with the motor.
i.e. GA JD
But maybe if this is a 'list of bits' arranged to 'connect somehow' - the .222 is actually µF and 2 along with the diodes are meant to be across the brushes for RF and back EMF?
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i will get current as u said but what about voltage i want it 5-6volt.so i have connected that 0.222ohm resistance in parallel to battery.if i am wrong plz tell me where.
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