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Thanx, but may i know more about it?Use a DPDT relay and put the bulb on the NC contacts
battery positive terminal to a resistor. the resistor to the bulb. the bulb to the battery negative terminal. place a switch in parallel with the bulb.
Here's how I do this. With the switch closed, the supply current is only 2uA. With the switch open, the current through the load is determined only by the load resistance and supply voltage, with no current wasted.
The FET can be almost anything that has a Vg < (supply voltage/2), a Vd> (supply voltage), and an On-resistance small compared to the load resistance.